The geometric series gives as an infinite polynomial. Can the same be done for , for , for , and for a function whose antiderivative cannot be written down at all?

Calculus 2

Representing functions as power series

The geometric series gives \displaystyle \frac{1}{1-x} as an infinite polynomial. Can the same be done for \displaystyle \frac{x}{9+x^2}, for \ln(5 - x), for \displaystyle \arctan\left(\frac{x}{3}\right), and for a function whose antiderivative cannot be written down at all?

You need

Power series and the interval of convergenceCalculus 2 The geometric series testCalculus 2

Unlocks

The common Maclaurin seriesCalculus 2

Where you are 8 in Sequences and series in calculus

Calculus8

Sequences and series in calculus8

▴ 3 above The nth term test for divergence Convergence tests6 Power series and the interval of convergence

Representing functions as power series

Taylor and Maclaurin series5

When you reach for it

A power series \sum c_n (x-a)^n defines a function on its interval of convergence, found as in Power series and the interval of convergence. This lesson runs the other way: start from a function and find a power series whose sum it is. Two reasons to want one. A series turns a function into arithmetic, so its values can be computed by adding terms. And a series can be differentiated and integrated term by term, which gives series for functions with no simple formula at all, such as an antiderivative that cannot be written with the familiar functions.

Everything here grows from one identity, the geometric series read as a function of its ratio:

\begin{aligned} \frac{1}{1-u} &= \sum_{n=0}^{\infty} u^n \\ &= 1 + u + u^2 + u^3 + \cdots \end{aligned}

for \lvert u \rvert < 1. At u = \pm 1 the terms do not tend to 0, so the identity holds exactly on (-1, 1) and nowhere else. Day 1 builds new series from it by algebra; Day 2 adds calculus. A different route, finding coefficients from derivatives as in Taylor polynomials, belongs to Taylor and Maclaurin series; the series found here are the same ones, because a function has only one power series at a given center (Exercise 24).

The procedure

Day 1: building series by algebra

Substitution. The identity holds for every number u with \lvert u \rvert < 1, so it holds when u is replaced by any expression g(x), for exactly those x with \lvert g(x) \rvert < 1. With u = -x^2, for instance,

\frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n}, \qquad \lvert x \rvert < 1

The interval of convergence of the new series is the solution of \lvert g(x) \rvert < 1, and at the edges \lvert g(x) \rvert = 1 the terms have size 1, so the series diverges there. When g(x) = c\,(x-a)^k, that solution is an interval centered at a.

Getting into the form \displaystyle \frac{1}{1-u}. A denominator b - c\,x with b \ne 0 is factored as \displaystyle b\left(1 - \frac{c}{b}x\right), so that

\frac{1}{b - cx} = \frac{1}{b} \cdot \frac{1}{1 - \frac{c}{b}x}

for \displaystyle \left\lvert \frac{c}{b}x \right\rvert < 1. The constant in front must be the one that makes the 1 appear; it is set by the value of the function at the center.

Multiplying by a constant or a power of x. If \sum a_n converges to S, then \sum k\,a_n converges to kS, because each partial sum is multiplied by k. With k = x^m for a fixed x, multiplying a power series by x^m shifts every exponent up by m and leaves the interval of convergence unchanged (Exercise 11).

Re-centering. For a series centered at a, write the denominator in terms of x - a and factor as before. The function \displaystyle \frac{1}{b - x} becomes \displaystyle \frac{1}{(b-a) - (x-a)}, a geometric series in \displaystyle \frac{x-a}{b-a}, valid for \lvert x - a \rvert < \lvert b - a \rvert. The radius is the distance from the center to b, the point where the function blows up, so the same function has different series, with different intervals, at different centers (Example 2).

Adding. Where \sum a_n and \sum b_n both converge, \sum (a_n + b_n) converges to the sum of the two. Partial fractions split a rational function into pieces of the form \displaystyle \frac{A}{b - cx}, each a geometric series, and the sum converges where both pieces do. In every example of this lesson the radius turns out to be the distance from the center to the nearest point where the denominator is 0, counting complex zeros such as x = 3i for 9 + x^2. That this holds for every rational function written in lowest terms (common factors of the numerator and denominator cancelled first) is a result of complex analysis, stated here without proof.

Day 2: differentiating and integrating term by term

A polynomial is differentiated one term at a time. A power series has infinitely many terms, and it is not automatic that the derivative of an infinite sum is the infinite sum of the derivatives: that is an exchange of two limits. For power series it is true inside the interval.

Term by term differentiation and integration. Suppose f(x) = \sum_{n=0}^{\infty} c_n (x-a)^n has radius of convergence R > 0. Then f is differentiable on (a - R, a + R), and there

f'(x) = \sum_{n=1}^{\infty} n\,c_n (x-a)^{n-1}

\int f(x)\,dx = C + \sum_{n=0}^{\infty} \frac{c_n}{n+1}(x-a)^{n+1}

Both new series have the same radius R.

The full proof needs uniform convergence, which is beyond this lesson, so the theorem is stated here without it. The radius part is the easy half: when \displaystyle \left\lvert \frac{c_{n+1}}{c_n} \right\rvert \to L, the extra factors n and \displaystyle \frac{1}{n+1} change that ratio by \displaystyle \frac{n+1}{n} or \displaystyle \frac{n+1}{n+2}, both of which tend to 1, so the limit and with it R are unchanged. Three consequences are used constantly.

A power series is continuous inside its interval, since it is differentiable there, and it can be differentiated again as often as needed, each time with the same radius.
The constant of integration comes from the center. At x = a every term (x-a)^{n+1} is 0, so the series part is 0 and the antiderivative there equals C. Equivalently, \displaystyle \int_a^x f(t)\,dt = \sum \frac{c_n}{n+1}(x-a)^{n+1}.
Endpoints must be tested again. The radius stays, the endpoints need not. Differentiating multiplies the nth term by about n and can lose an endpoint; integrating divides by about n and can gain one, and it never loses one (a fact stated here without proof). Each endpoint of the new series is tested as in Power series and the interval of convergence.

When a series converges at an endpoint, does it converge to the function’s value there? Abel’s theorem, stated without proof, says that if \sum c_n R^n converges, it equals \lim_{x \to (a+R)^-} f(x), and similarly at a - R. So when the series came from a function that is continuous at that endpoint from inside, the series sums to the function’s value there. Applied to the standard series this gives two results worth knowing, cited here without derivation because the worked examples build scaled versions of them:

\arctan x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} \text{ on } [-1, 1]

\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n} \text{ on } (-1, 1]

Worked examples

Example 1

Day 1: substituting and multiplying by x

Agustina wants a power series centered at 0 for f(x) = \dfrac{x}{9 + x^2}, with its interval of convergence.

Solution

First find a series for \displaystyle \frac{1}{9 + x^2}. Factor out 9, the value of the denominator at the center, to make a 1 appear:

\frac{1}{9 + x^2} = \frac19 \cdot \frac{1}{1 - \left(-\frac{x^2}{9}\right)}

This is the seed with \displaystyle u = -\frac{x^2}{9}, so for \displaystyle \frac{x^2}{9} < 1, that is \lvert x \rvert < 3,

\begin{aligned} \frac{1}{9 + x^2} &= \frac19 \sum_{n=0}^{\infty} \left(-\frac{x^2}{9}\right)^n \\ &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{9^{n+1}} \end{aligned}

Multiplying by x raises each exponent by one and keeps the interval:

\begin{aligned} \frac{x}{9 + x^2} &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{9^{n+1}} \\ &= \frac{x}{9} - \frac{x^3}{81} + \frac{x^5}{729} - \cdots \end{aligned}

At x = \pm 3, x^{2n+1} = \pm 3 \cdot 9^n, so the terms are \displaystyle \pm\frac{(-1)^n}{3}, which do not tend to 0; both endpoints diverge. The interval of convergence is (-3, 3).

Check at x = 1: the series is \displaystyle \sum \frac{(-1)^n}{9^{n+1}}, geometric with first term \displaystyle \frac19 and ratio \displaystyle -\frac19, so its sum is \displaystyle \frac{1/9}{1 + 1/9} = \frac{1}{10}, and \displaystyle f(1) = \frac{1}{10}. Notice that f is perfectly well behaved at x = 3, where \displaystyle f(3) = \frac16; the series stops there because 9 + x^2 is 0 at x = \pm 3i, at distance 3 from the center.

Example 2

Day 1: one function, two centers

Facundo has g(x) = \dfrac{2}{7 - x}. Find its power series centered at 0 and centered at 3, with the interval of each, and check both at x = 5.

Solution

Centered at 0, factor out 7:

\frac{2}{7 - x} = \frac27 \cdot \frac{1}{1 - \frac{x}{7}} = \sum_{n=0}^{\infty} \frac{2x^n}{7^{n+1}}

for \lvert x \rvert < 7. At x = \pm 7 the ratio \displaystyle \frac{x}{7} is \pm 1 and the series diverges, so the interval is (-7, 7).

Centered at 3, write the denominator in terms of x - 3: \displaystyle 7 - x = 4 - (x - 3) = 4\left(1 - \frac{x-3}{4}\right). So

\frac{2}{7 - x} = \frac12 \cdot \frac{1}{1 - \frac{x-3}{4}} = \sum_{n=0}^{\infty} \frac{(x-3)^n}{2 \cdot 4^n}

valid for \lvert x - 3 \rvert < 4. The endpoints again give ratio \pm 1 and diverge, so the interval is (-1, 7).

At x = 5 the first series is geometric with ratio \displaystyle \frac57: its sum is \displaystyle \frac{2/7}{1 - 5/7} = 1. The second has ratio \displaystyle \frac{2}{4} = \frac12 and first term \displaystyle \frac12: its sum is \displaystyle \frac{1/2}{1/2} = 1. Both agree with \displaystyle g(5) = \frac22 = 1. Each radius is the distance from the center to 7, where g blows up: 7 from the center 0, and 4 from the center 3. The series centered at 3 is no use at x = -2, where the one centered at 0 still works.

Example 3

Day 1: adding two series

Benjamin wants a power series centered at 0 for h(x) = \dfrac{3}{(1-x)(1+2x)} and its interval of convergence.

Solution

Partial fractions: \displaystyle \frac{3}{(1-x)(1+2x)} = \frac{A}{1-x} + \frac{B}{1+2x} means 3 = A(1+2x) + B(1-x). At x = 1 this reads 3 = 3A, so A = 1; at \displaystyle x = -\frac12 it reads \displaystyle 3 = \frac32 B, so B = 2. Each piece is geometric:

\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \text{ for } \lvert x \rvert < 1

\frac{2}{1+2x} = \sum_{n=0}^{\infty} 2(-2)^n x^n \text{ for } \lvert x \rvert < \tfrac12

Where both converge, their sum converges to h(x):

\begin{aligned} h(x) &= \sum_{n=0}^{\infty} \left(1 + (-1)^n 2^{n+1}\right) x^n \\ &= 3 - 3x + 9x^2 - 15x^3 + \cdots \end{aligned}

Both pieces converge for \displaystyle \lvert x \rvert < \frac12. For \displaystyle \frac12 \le \lvert x \rvert < 1 the first converges and the second diverges, so the sum diverges: if it converged, subtracting the first series would make the second converge. For \lvert x \rvert \ge 1 the coefficients have size at least 2^{n+1} - 1, so the terms do not tend to 0. The interval of convergence is \displaystyle \left(-\frac12, \frac12\right), and the radius \displaystyle \frac12 is the distance from 0 to the nearer zero of the denominator, \displaystyle x = -\frac12.

As a check, multiply the series by (1-x)(1+2x) = 1 + x - 2x^2: the constant term is 3, the x term is -3 + 3 = 0, the x^2 term is 9 - 3 - 6 = 0, and the x^3 term is -15 + 9 + 6 = 0, leaving 3.

Example 4

Day 2: differentiating the geometric series

Agustina wants power series for \dfrac{1}{(1-x)^2} and \dfrac{x}{(1-x)^2}, and the sum \displaystyle \displaystyle\sum_{n=1}^{\infty} \frac{n\,2^n}{3^n}.

Solution

The derivative of \displaystyle \frac{1}{1-x} is \displaystyle \frac{1}{(1-x)^2}. Differentiate \sum x^n term by term on (-1, 1); the n = 0 term is the constant 1, with derivative 0:

\begin{aligned} \frac{1}{(1-x)^2} &= \sum_{n=1}^{\infty} n\,x^{n-1} = \sum_{n=0}^{\infty} (n+1)\,x^n \\ &= 1 + 2x + 3x^2 + \cdots \end{aligned}

The radius stays 1. At x = \pm 1 the terms have size n + 1, so both endpoints diverge, and the interval is (-1, 1). Multiplying by x:

\frac{x}{(1-x)^2} = \sum_{n=1}^{\infty} n\,x^n, \qquad \lvert x \rvert < 1

This is the general form of the sum that Absolute and conditional convergence found at \displaystyle x = \frac12 from partial sums: here \displaystyle \frac{1/2}{(1/2)^2} = 2 at once. For the sum asked, \displaystyle \frac{n\,2^n}{3^n} = n\left(\frac23\right)^n and \displaystyle \frac23 lies inside (-1, 1), so

\sum_{n=1}^{\infty} n\left(\frac23\right)^n = \frac{2/3}{(1/3)^2} = 6

Example 5

Day 2: a logarithm by integration

Facundo wants a power series centered at 0 for \ln(5 - x), its interval of convergence, and what the series says at x = -5.

Solution

The derivative is \displaystyle \frac{d}{dx}\ln(5-x) = -\frac{1}{5-x}, which is geometric:

-\frac{1}{5-x} = -\frac15 \cdot \frac{1}{1 - \frac{x}{5}} = -\sum_{n=0}^{\infty} \frac{x^n}{5^{n+1}}

for \lvert x \rvert < 5. Integrate term by term from the center 0 to x. The left side gives \ln(5-x) - \ln 5, and each x^n becomes \displaystyle \frac{x^{n+1}}{n+1}:

\begin{aligned} &\ln(5-x) - \ln 5 \\ &\quad = -\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)\,5^{n+1}} \end{aligned}

Shifting the index to m = n + 1 gives

\ln(5-x) = \ln 5 - \sum_{m=1}^{\infty} \frac{x^m}{m\,5^m}

The constant \ln 5 is the value at the center, where every term of the sum is 0. The radius is 5, as for the derivative. Now the endpoints. At x = 5 the sum becomes \displaystyle \sum \frac1m, the harmonic series, which diverges; the function itself is undefined there. At x = -5 it becomes \displaystyle \ln 5 - \sum \frac{(-1)^m}{m}, and \displaystyle \sum \frac{(-1)^m}{m} converges conditionally by the alternating series test. The interval of convergence is [-5, 5): integration gained the endpoint -5, which the derivative’s series (-5, 5) did not have.

Since \ln(5-x) is continuous at x = -5, Abel’s theorem says the series there sums to \ln 10. So \displaystyle \ln 5 + \sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{m} = \ln 10, which gives \displaystyle 1 - \frac12 + \frac13 - \cdots = \ln 2, the value of the alternating harmonic series.

Example 6

Day 2: an inverse tangent by integration

Benjamin wants a power series centered at 0 for \displaystyle \arctan\left(\frac{x}{3}\right) and its interval of convergence.

Solution

By the chain rule the derivative is \displaystyle \frac{1/3}{1 + x^2/9} = \frac{3}{9 + x^2}. Example 1 found \displaystyle \frac{1}{9 + x^2} = \sum \frac{(-1)^n x^{2n}}{9^{n+1}} for \lvert x \rvert < 3, so

\begin{aligned} \frac{3}{9 + x^2} &= \sum_{n=0}^{\infty} \frac{3(-1)^n x^{2n}}{9^{n+1}} \\ &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{3^{2n+1}} \end{aligned}

using 3 \cdot 9^{-(n+1)} = 3^{-(2n+1)}. Integrate term by term from 0; the constant is \arctan 0 = 0:

\begin{aligned} \arctan\left(\frac{x}{3}\right) &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)\,3^{2n+1}} \\ &= \frac{x}{3} - \frac{x^3}{81} + \frac{x^5}{1215} - \cdots \end{aligned}

The radius is 3. At x = 3 the terms are \displaystyle \frac{(-1)^n}{2n+1}, whose sizes decrease to 0, so the series converges by the alternating series test; it converges only conditionally, since \displaystyle \sum \frac{1}{2n+1} diverges by limit comparison with \displaystyle \frac1n. At x = -3 every term changes sign, giving the same verdict. The interval is [-3, 3], closed, though the derivative’s series converged only on (-3, 3). Abel’s theorem at x = 3 gives \displaystyle \arctan 1 = \frac{\pi}{4} = 1 - \frac13 + \frac15 - \cdots.

Figure 1 shows the partial sums \displaystyle P_5(x) = \frac{x}{3} - \frac{x^3}{81} + \frac{x^5}{1215} and P_7(x), which adds \displaystyle -\frac{x^7}{15309}. Inside [-3, 3] they lie along the curve; outside, each is dominated by its last term and leaves.

Figure 1\displaystyle y = \arctan\left(\frac{x}{3}\right) with the partial sums P_5 and P_7 of its series from Example 6. Inside [-3, 3] the partial sums close in on the curve; outside it each is ruled by its last term, so they leave in opposite directions.

Example 7

Day 2: an antiderivative with no formula

Let \displaystyle F(x) = \displaystyle\int_0^x \frac{\arctan(t/3)}{t}\,dt, where the integrand is given the value \displaystyle \frac13 at t = 0. Agustina wants a power series for F and its interval of convergence.

Solution

The function \displaystyle \frac{\arctan(t/3)}{t} has no antiderivative that can be written with the familiar functions (a known result, stated here without proof), so the Fundamental Theorem of Calculus alone cannot evaluate F. A series can. For t \ne 0 in [-3, 3], divide the series of Example 6 by t, term by term, which is multiplying a convergent series by the constant \displaystyle \frac1t:

\frac{\arctan(t/3)}{t} = \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n}}{(2n+1)\,3^{2n+1}}

At t = 0 the right side is its first term \displaystyle \frac13, the value chosen for the integrand, and it is also \displaystyle \lim_{t \to 0} \frac{\arctan(t/3)}{t}, so the integrand is continuous and equals this series on [-3, 3]. Its radius is 3. Integrate term by term from 0 to x; the constant is F(0) = 0:

\begin{aligned} F(x) &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)^2\,3^{2n+1}} \\ &= \frac{x}{3} - \frac{x^3}{243} + \frac{x^5}{6075} - \cdots \end{aligned}

The radius is still 3. At x = \pm 3 the terms have size \displaystyle \frac{1}{(2n+1)^2}, which is at most \displaystyle \frac{1}{n^2} for n \ge 1, so both endpoints converge absolutely by comparison with a p-series. The interval is [-3, 3], and the endpoints that were conditional for \displaystyle \arctan\left(\frac{x}{3}\right) are absolute for F, one more effect of the extra \displaystyle \frac{1}{2n+1}. Adding the first few terms at x = 1 pins F(1) down to many decimal places, and Using series to evaluate limits and integrals measures how many terms a given accuracy needs.

Boundaries

Only inside the interval. Every identity in this lesson holds on the interval of convergence and nowhere else. \displaystyle \frac{2}{7-x} is defined at x = -2, but its series centered at 3 diverges there (Example 2).
The theorem needs R > 0, and it speaks about the open interval (a - R, a + R). At the endpoints the new series must be tested again: integration can gain an endpoint (Examples 5 and 6) and differentiation can lose one (Exercise 22).
The radius does not always show on the graph. \displaystyle \frac{x}{9+x^2} is smooth on the whole line, yet its series centered at 0 stops at \lvert x \rvert = 3, set by the complex zeros \pm 3i of the denominator (Example 1).
Term by term calculus is special to power series. For a general series of functions, the sum of the derivatives need not be the derivative of the sum. The theorem uses the form c_n (x-a)^n and the radius; it is not a rule for every infinite sum.
Abel’s theorem needs convergence at the endpoint. It gives the value of a series at an endpoint only when the series converges there; at x = 5 in Example 5 there is no sum to compute.
Each move needs its own condition. A substitution u = g(x) is valid only where \lvert g(x) \rvert < 1, and a sum of two series is guaranteed only where both converge.

Where this goes next

The series built here are Taylor series, found without computing a single derivative. Taylor and Maclaurin series and Taylor polynomials give the other route, coefficients \displaystyle c_n = \frac{f^{(n)}(a)}{n!}, and the uniqueness of Exercise 24 says the two routes always agree. The common Maclaurin series uses derivatives for the functions that geometric algebra cannot reach, such as e^x, \sin x and \cos x, and The binomial series extends \displaystyle \frac{1}{(1-x)^2} to (1+x)^k for any real k. The Taylor remainder and the Lagrange error bound measures how close a partial sum is to its function, and Using series to evaluate limits and integrals turns integrals like the one in Example 7 into numbers with a guaranteed accuracy.

Exercises

Day 1: building series

For Exercises 1 to 7, find a power series for the function, centered at the given point, and its interval of convergence.

\dfrac{1}{1 + 4x}, centered at 0
\dfrac{5}{2 - x}, centered at 0
\dfrac{x^3}{4 + x^2}, centered at 0
\dfrac{2}{x}, centered at 5
\dfrac{1}{2x + 1}, centered at 2
\dfrac{1}{(1-x)(1-3x)}, centered at 0
\dfrac{1+x}{1-x}, centered at 0
Find the function that \sum_{n=0}^{\infty} \dfrac{(x+1)^{2n}}{3^n} sums to, and the interval where it does.

Day 1: reasoning

Find the series for \displaystyle \frac{1}{1-x^2} centered at 0 twice: by substituting u = x^2, and by partial fractions, writing it as \displaystyle \frac12\left(\frac{1}{1-x} + \frac{1}{1+x}\right). Show the two agree.
Explain why \displaystyle \frac1x has no power series centered at 0, and show that for every a > 0 its series centered at a has radius a.
Explain why multiplying a power series by a nonzero constant, or by x^m for a positive integer m, does not change its interval of convergence.

Day 2: differentiating and integrating

For Exercises 12, 13, 16, 17, 18 and 21, find a power series centered at 0 (centered at 2 in Exercise 21) and its interval of convergence.

\dfrac{1}{(3-x)^2}
\dfrac{1}{(1-x)^3}
Find \displaystyle \displaystyle\sum_{n=1}^{\infty} \frac{n}{5^n}.
Find \displaystyle \displaystyle\sum_{n=1}^{\infty} \frac{n(n+1)}{3^n}. Start from \displaystyle \sum_{n=0}^{\infty} x^{n+1} = \frac{x}{1-x}.
\ln(4 + x)
\arctan(3x)
\ln(9 + x^2)
Find a power series centered at 0 for \displaystyle \displaystyle\int_0^x \frac{dt}{1 + t^5} and its interval of convergence.
Find the function f(x) = \sum_{n=1}^{\infty} \dfrac{x^n}{n\,2^n} in closed form, its interval of convergence, and \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n\,2^n}.
\ln x, centered at 2

Day 2: reasoning

Let f(x) = \sum_{n=1}^{\infty} \dfrac{x^n}{n^2}. Find the intervals of convergence of the series for f, f' and f'', and explain what happens to the endpoints.
Facundo integrates \displaystyle \frac{1}{2-x} = \sum_{n=0}^{\infty} \frac{x^n}{2^{n+1}} term by term and writes -\ln(2-x) = C + S(x) with \displaystyle S(x) = \sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)\,2^{n+1}}. Find C, and explain why x = 0 is the point to use.
Prove that if \sum c_n x^n = \sum d_n x^n for every x in some interval (-r, r) with r > 0, then c_n = d_n for every n.

Answers

\sum_{n=0}^{\infty} (-4)^n x^n on \displaystyle \left(-\frac14, \frac14\right). Substitute u = -4x; \lvert 4x \rvert < 1. At \displaystyle x = \pm\frac14 the terms are 1 or (-1)^n, which do not tend to 0.
\sum_{n=0}^{\infty} \dfrac{5x^n}{2^{n+1}} on (-2, 2). \displaystyle \frac{5}{2-x} = \frac52 \cdot \frac{1}{1 - x/2}; at x = \pm 2 the ratio is \pm 1.
\sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n+3}}{4^{n+1}} on (-2, 2). \displaystyle \frac{1}{4+x^2} = \frac14 \cdot \frac{1}{1 + x^2/4}, then multiply by x^3. At x = \pm 2 the terms have size \displaystyle \frac{2^{2n+3}}{4^{n+1}} = 2.
\sum_{n=0}^{\infty} \dfrac{2(-1)^n (x-5)^n}{5^{n+1}} on (0, 10). \displaystyle \frac2x = \frac{2}{5 + (x-5)}, which is \displaystyle \frac25 \cdot \frac{1}{1 + (x-5)/5}; the radius 5 is the distance from 5 to 0. At x = 0 and x = 10 the terms have size \displaystyle \frac25, so both endpoints diverge.
\sum_{n=0}^{\infty} \dfrac{(-1)^n 2^n (x-2)^n}{5^{n+1}} on \displaystyle \left(-\frac12, \frac92\right). 2x + 1 = 5 + 2(x-2), so \displaystyle u = -\frac{2(x-2)}{5}, and \lvert u \rvert < 1 means \displaystyle \lvert x - 2 \rvert < \frac52.
\sum_{n=0}^{\infty} \dfrac{3^{n+1} - 1}{2}\,x^n on \displaystyle \left(-\frac13, \frac13\right). Partial fractions give \displaystyle -\frac{1/2}{1-x} + \frac{3/2}{1-3x}; the second piece needs \displaystyle \lvert x \rvert < \frac13, and for \displaystyle \frac13 \le \lvert x \rvert < 1 only the first converges. At \displaystyle x = \pm\frac13 the terms have size \displaystyle \frac{3^{n+1}-1}{2 \cdot 3^n} \to \frac32.
1 + \sum_{n=1}^{\infty} 2x^n on (-1, 1). (1+x)\sum x^n = \sum x^n + \sum x^{n+1}, and each x^n with n \ge 1 appears twice. At x = \pm 1 the terms have size 2.
\dfrac{3}{3 - (x+1)^2} on \displaystyle \left(-1-\sqrt3,\, -1+\sqrt3\right). The series is geometric with ratio \displaystyle u = \frac{(x+1)^2}{3}, and \displaystyle \frac{1}{1-u} = \frac{3}{3-(x+1)^2}. It converges for \displaystyle \lvert x + 1 \rvert < \sqrt3; at the endpoints u = 1 and every term is 1.
Substitution gives \sum x^{2n} = 1 + x^2 + x^4 + \cdots on (-1, 1). Partial fractions give \displaystyle \frac12 \sum \left(1 + (-1)^n\right) x^n, whose coefficient is 1 for even n and 0 for odd n, on (-1, 1). They are the same series.
If \sum c_n x^n had radius R > 0, pick 0 < r < R; for \lvert x \rvert \le r its sum has size at most \sum \lvert c_n \rvert r^n, a finite number, because the series converges absolutely at r. But \displaystyle \frac1x is unbounded as x \to 0^+. Centered at a > 0: \displaystyle \frac1x = \frac{1}{a + (x-a)}, which is \displaystyle \frac1a \sum \frac{(-1)^n (x-a)^n}{a^n}, valid for \lvert x - a \rvert < a, with terms of size \displaystyle \frac1a at both endpoints, so the radius is a.
For a fixed x \ne 0, the new series is the old one with each term multiplied by the same nonzero number k or x^m. Its partial sums are the old ones times that number, so one sequence of partial sums converges exactly when the other does. At x = 0 both series converge.
\sum_{n=0}^{\infty} \dfrac{(n+1)\,x^n}{3^{n+2}} on (-3, 3). \displaystyle \frac{1}{(3-x)^2} is the derivative of \displaystyle \frac{1}{3-x} = \sum \frac{x^n}{3^{n+1}}, giving \displaystyle \sum_{n \ge 1} \frac{n\,x^{n-1}}{3^{n+1}}; shift the index. At x = \pm 3 the terms have size \displaystyle \frac{n+1}{9}.
\sum_{n=0}^{\infty} \dfrac{(n+1)(n+2)}{2}\,x^n on (-1, 1). The second derivative of \displaystyle \frac{1}{1-x} is \displaystyle \frac{2}{(1-x)^3}, which is \sum_{n \ge 2} n(n-1)x^{n-2}, or \sum_{n \ge 0} (n+2)(n+1)x^n after a shift; divide by 2. The terms do not tend to 0 at x = \pm 1.
\displaystyle \frac{5}{16}. \displaystyle \sum n\,x^n = \frac{x}{(1-x)^2} at \displaystyle x = \frac15: \displaystyle \frac{1/5}{(4/5)^2} = \frac{5}{16}.
\displaystyle \frac94. Differentiating \displaystyle \frac{x}{1-x} = -1 + \frac{1}{1-x} twice gives \displaystyle \frac{2}{(1-x)^3} = \sum_{n \ge 1} (n+1)n\,x^{n-1}, so \displaystyle \sum n(n+1)x^n = \frac{2x}{(1-x)^3}, and at \displaystyle x = \frac13 this is \displaystyle \frac{2/3}{8/27} = \frac94.
\ln 4 + \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1} x^n}{n\,4^n} on (-4, 4]. Integrate \displaystyle \frac{1}{4+x} = \sum \frac{(-1)^n x^n}{4^{n+1}} from 0; the constant is \ln 4. At x = 4 the sum is \displaystyle \sum \frac{(-1)^{n+1}}{n}, conditionally convergent; at x = -4 it is \displaystyle -\sum \frac1n, divergent.
\sum_{n=0}^{\infty} \dfrac{(-1)^n 3^{2n+1} x^{2n+1}}{2n+1} on \displaystyle \left[-\frac13, \frac13\right]. The derivative is \displaystyle \frac{3}{1+9x^2} = \sum 3(-9)^n x^{2n} for \displaystyle \lvert x \rvert < \frac13; integrate from 0, constant \arctan 0 = 0. At \displaystyle x = \pm\frac13 the terms are \displaystyle \pm\frac{(-1)^n}{2n+1}, conditionally convergent.
\ln 9 + \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1} x^{2n}}{n\,9^n} on [-3, 3]. The derivative \displaystyle \frac{2x}{9+x^2} is twice the series of Example 1, \displaystyle \sum \frac{2(-1)^n x^{2n+1}}{9^{n+1}}; integrating gives \displaystyle \sum \frac{(-1)^n x^{2n+2}}{(n+1)\,9^{n+1}}, and the constant is \ln 9. At x = \pm 3 the terms are \displaystyle \frac{(-1)^{n+1}}{n}, conditionally convergent.
\sum_{n=0}^{\infty} \dfrac{(-1)^n x^{5n+1}}{5n+1} on (-1, 1]. \displaystyle \frac{1}{1+t^5} = \sum (-1)^n t^{5n} for \lvert t \rvert < 1; integrate from 0. At x = 1 the terms are \displaystyle \frac{(-1)^n}{5n+1}, convergent by the alternating series test (conditionally). At x = -1, (-1)^n(-1)^{5n+1} = -1, so the terms are \displaystyle -\frac{1}{5n+1}, divergent by limit comparison with \displaystyle \frac1n.
f(x) = \ln\dfrac{2}{2-x} on [-2, 2), and the sum is \displaystyle \ln\frac23. \displaystyle f'(x) = \sum \frac{x^{n-1}}{2^n} = \frac{1}{2-x} for \lvert x \rvert < 2 and f(0) = 0, so f(x) = \ln 2 - \ln(2-x). At x = 2 the series is harmonic; at x = -2 it is alternating harmonic. The sum asked is \displaystyle f(-1) = \ln\frac23.
\ln 2 + \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}(x-2)^n}{n\,2^n} on (0, 4]. \displaystyle \frac1x = \frac{1}{2 + (x-2)}, which is \displaystyle \sum \frac{(-1)^n (x-2)^n}{2^{n+1}}; integrate from 2, constant \ln 2. At x = 4 the terms are \displaystyle \frac{(-1)^{n+1}}{n}, convergent; at x = 0 they are \displaystyle -\frac1n, divergent.
f: [-1, 1], since at x = \pm 1 the terms have size \displaystyle \frac{1}{n^2}. \displaystyle f'(x) = \sum \frac{x^{n-1}}{n}: [-1, 1), harmonic at 1 and alternating harmonic at -1. \displaystyle f''(x) = \sum \frac{(n-1)\,x^{n-2}}{n}: (-1, 1), terms of size \displaystyle \frac{n-1}{n} \to 1 at both ends. The radius stays 1; each differentiation multiplies the nth term by about n and here costs an endpoint.
C = -\ln 2. At x = 0 every term \displaystyle \frac{x^{n+1}}{(n+1)2^{n+1}} is 0, so the equation reads -\ln 2 = C. At any other point, finding C would need the value of an infinite series.
Let g(x) = \sum (c_n - d_n) x^n, which is 0 on (-r, r), so every derivative of g is 0 there. By term by term differentiation, g^{(k)}(0) = k!\,(c_k - d_k), since every other term of the kth derivative has a positive power of x or was differentiated away. So c_k = d_k for every k.