The binomial theorem expands when is a counting number. What does or expand to, and for which is the expansion true?

Calculus 2

The binomial series

The binomial theorem expands (1 + x)^n when n is a counting number. What does (1 + x)^{1/3} or (1 + x)^{-1/2} expand to, and for which x is the expansion true?

You need

CountingGrades 7 to 11

Where you are 5 in Taylor and Maclaurin series

Calculus8

Sequences and series in calculus8

Taylor and Maclaurin series5

Taylor polynomials The Taylor remainder and the Lagrange error bound

The binomial series

The common Maclaurin series Using series to evaluate limits and integrals

Why this exists

The binomial theorem expands (1 + x)^m for a counting number m into a polynomial whose coefficients are the counts \binom{m}{n} from counting and Pascal’s triangle. It says nothing about (1 + x)^{1/3} or (1 + x)^{-1/2}: you cannot multiply 1 + x by itself a third of a time, and “choose n from \displaystyle -\frac12” counts nothing.

Yet these functions have Taylor polynomials like any other, and their coefficients turn out to follow the same pattern as the counting numbers, with m replaced by any real k. The result, the binomial series, contains the geometric series as the case k = -1, turns roots and reciprocal powers into sums of powers, and simplifies any formula with a root or a \displaystyle \frac32 power in it when one quantity is small next to another.

Definition

For any real number k and any integer n \ge 0, the generalized binomial coefficient is

\binom{k}{n} = \frac{k(k-1)(k-2)\cdots(k-n+1)}{n!},

with \binom{k}{0} = 1. The top is a product of n factors, starting at k and stepping down by 1. For example,

\begin{aligned} \binom{-2}{3} &= \frac{(-2)(-3)(-4)}{3!} = -4, \\ \binom{1/3}{2} &= \frac{\frac13 \cdot \left(-\frac23\right)}{2} = -\frac19. \end{aligned}

When k is a counting number these agree with the counts: \displaystyle \binom{5}{2} = \frac{5 \cdot 4}{2} = 10 (Exercise 18).

The binomial series for exponent k is the power series

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

Intuition

The series is the Maclaurin series of f(x) = (1 + x)^k. Each derivative brings the current exponent down in front and lowers it by one:

\begin{aligned} f'(x) &= k(1 + x)^{k-1} \\ f''(x) &= k(k-1)(1 + x)^{k-2} \\ f^{(n)}(x) &= k(k-1)\cdots(k-n+1)(1 + x)^{k-n}. \end{aligned}

At x = 0 the power of 1 + x is 1, so f^{(n)}(0) is the falling product, and dividing by n! gives exactly \binom{k}{n}.

Now watch what the falling product does. If k = 3, the factors are 3, 2, 1, 0, -1, \ldots, and from n = 4 on every product contains the factor 0: \displaystyle \binom{3}{4} = \frac{3 \cdot 2 \cdot 1 \cdot 0}{4!} = 0. The series stops, and what is left is 1 + 3x + 3x^2 + x^3, the binomial theorem. If k is not a counting number, the factors k, k - 1, k - 2, \ldots never hit 0, every coefficient is nonzero, and the series goes on forever. Its radius of convergence is then 1 (Figure 1): (1 + x)^{-1/2} is defined for every x > -1, but a power series reaches equally far on both sides of its center, and the blow-up at x = -1 caps the radius at 1.

Figure 1y = (1 + x)^{-1/2} with the partial sums P_3 and P_4 of its binomial series (Example 1). Inside (-1, 1) the partial sums close in on the curve, slowly near -1; past x = 1 consecutive partial sums split apart, because the series diverges there.

Results

The binomial series. For every real k and every x with -1 < x < 1,

(1 + x)^k = \sum_{n=0}^{\infty} \binom{k}{n} x^n.

When it stops. If k = m is a nonnegative integer, \binom{m}{n} = 0 for n > m, the series is the polynomial \sum_{n=0}^{m} \binom{m}{n} x^n, and it equals (1 + x)^m for every x. This is the binomial theorem.

Radius 1. If k is not a nonnegative integer, the binomial series has radius of convergence exactly 1.

Endpoints (stated without proof). For k not a nonnegative integer:

k	at x = -1	at x = 1	the series equals (1 + x)^k on
k > 0	converges	converges	[-1, 1]
-1 < k < 0	diverges	converges	(-1, 1]
k \le -1	diverges	diverges	(-1, 1)

At an endpoint where the series converges, its sum is (1 + x)^k there (a theorem of Abel, also stated without proof).

Rescaling and substitution. For a > 0, factoring out the value at the center gives

\begin{aligned} (a + x)^k &= a^k\left(1 + \frac{x}{a}\right)^k \\ &= \sum_{n=0}^{\infty} \binom{k}{n} a^{k-n} x^n, \end{aligned}

valid for \lvert x \rvert < a. Replacing x by cx^p gives a series valid where \lvert cx^p \rvert < 1.

Why it is true

The coefficients are forced, as the Intuition section showed: they are f^{(n)}(0)/n!.

The radius. When k is not a nonnegative integer, no coefficient is 0, so the ratio test applies to a_n = \binom{k}{n}x^n with x \ne 0. From the definition, \binom{k}{n+1} = \binom{k}{n}\dfrac{k - n}{n + 1}, so

\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = \frac{\lvert k - n \rvert}{n + 1}\lvert x \rvert \to \lvert x \rvert,

because \displaystyle \frac{n - k}{n + 1} \to 1 as n \to \infty. The series converges for \lvert x \rvert < 1 and diverges for \lvert x \rvert > 1 (The ratio and root tests, Power series and the interval of convergence).

The sum is (1 + x)^k. A function can have a convergent Maclaurin series that sums to something else, so this needs an argument. Let g(x) = \sum_{n=0}^{\infty} \binom{k}{n}x^n on (-1, 1). Inside the radius, a power series may be differentiated term by term:

\begin{aligned} g'(x) &= \sum_{n=1}^{\infty} n\binom{k}{n}x^{n-1} \\ &= \sum_{n=0}^{\infty} (n + 1)\binom{k}{n+1}x^n. \end{aligned}

By the identity above, (n + 1)\binom{k}{n+1} = (k - n)\binom{k}{n}. Multiply g' by 1 + x; the factor x shifts the first sum into \sum n\binom{k}{n}x^n, and the coefficient of x^n in (1 + x)g'(x) is

(k - n)\binom{k}{n} + n\binom{k}{n} = k\binom{k}{n}.

So (1 + x)g'(x) = k\,g(x), the same equation (1 + x)^k satisfies. Now let h(x) = g(x)(1 + x)^{-k}. By the product rule,

\begin{aligned} h'(x) &= (1 + x)^{-k-1}\big[(1 + x)g'(x) - k\,g(x)\big] \\ &= 0 \end{aligned}

on (-1, 1), so h is constant there (a function with zero derivative on an interval is constant, by the mean value theorem). Since h(0) = g(0) = 1, g(x) = (1 + x)^k on all of (-1, 1).

Worked examples

Example 1

The coefficients for exponent minus one half

Find a closed form for \binom{-1/2}{n}, write the first five terms of the binomial series for (1 + x)^{-1/2}, and test them at x = 0.2.

Solution

Each of the n factors \displaystyle -\frac12, -\frac32, \ldots, -\frac{2n-1}{2} is negative with denominator 2:

\binom{-1/2}{n} = \frac{(-1)^n \cdot 1 \cdot 3 \cdot 5 \cdots (2n-1)}{2^n\, n!}.

The odd product is (2n)! with the even factors removed, and the even factors 2 \cdot 4 \cdots 2n multiply to 2^n n!. So 1 \cdot 3 \cdots (2n-1) = \dfrac{(2n)!}{2^n n!} and

\begin{aligned} \binom{-1/2}{n} &= \frac{(-1)^n (2n)!}{4^n (n!)^2} \\ &= \frac{(-1)^n}{4^n}\binom{2n}{n}. \end{aligned}

With \binom{2n}{n} = 1, 2, 6, 20, 70 for n = 0, \ldots, 4:

\begin{aligned} \frac{1}{\sqrt{1 + x}} = 1 - \frac{x}{2} &+ \frac{3x^2}{8} - \frac{5x^3}{16} \\ &+ \frac{35x^4}{128} - \cdots \end{aligned}

for -1 < x < 1. By the table in Results (\displaystyle k = -\frac12) the series also converges at x = 1, to \displaystyle \frac{1}{\sqrt2}, and diverges at x = -1 (Exercise 20). At x = 0.2 the five terms give 1 - 0.1 + 0.015 - 0.0025 + 0.0004375 = 0.9129375, against \displaystyle \frac{1}{\sqrt{1.2}} \approx 0.9128709. Figure 1 shows the partial sums P_3 and P_4.

Example 2

Central binomial coefficients

Find the Maclaurin series of (1 - 4x)^{-1/2} and its interval of validity, and use it to evaluate \displaystyle \sum_{n=1}^{\infty} \binom{2n}{n}\left(\frac{4}{25}\right)^n, the sum quoted in Example 1 of The ratio and root tests.

Solution

Substitute -4x for x in Example 1. The n-th term becomes

\frac{(-1)^n}{4^n}\binom{2n}{n}(-4x)^n = \binom{2n}{n}x^n,

since (-1)^n(-4)^n = 4^n. The substitution is valid where \lvert -4x \rvert < 1, so

\frac{1}{\sqrt{1 - 4x}} = \sum_{n=0}^{\infty} \binom{2n}{n}x^n

for \lvert x \rvert < \tfrac14.

The point \displaystyle x = \frac{4}{25} = 0.16 is inside. There \displaystyle 1 - 4x = \frac{9}{25}, and the whole series sums to \displaystyle \frac{1}{\sqrt{9/25}} = \frac53. The requested sum starts at n = 1, so remove the n = 0 term, 1:

\sum_{n=1}^{\infty} \binom{2n}{n}\left(\frac{4}{25}\right)^n = \frac53 - 1 = \frac23.

The ratio test lesson trapped this sum between 0.6278 and 0.6709; the series gives it exactly.

Example 3

A cube root by rescaling

Blerta needs the edge of a cube that holds 8.24 liters, which is \displaystyle \sqrt[3]{8.24} decimeters. Find the Maclaurin series of \displaystyle \sqrt[3]{8 + x} through x^3, with its interval of validity, and use it to estimate \displaystyle \sqrt[3]{8.24}.

Solution

The series for (1 + x)^{1/3} needs small x, so factor out the 8, whose cube root is exact:

\sqrt[3]{8 + x} = 2\left(1 + \frac{x}{8}\right)^{1/3}.

The coefficients for \displaystyle k = \frac13 are 1, \displaystyle \frac13, \displaystyle \binom{1/3}{2} = -\frac19 and \displaystyle \binom{1/3}{3} = \frac{\frac13 \cdot (-\frac23)(-\frac53)}{6} = \frac{5}{81}. With \displaystyle u = \frac{x}{8},

\begin{aligned} \sqrt[3]{8 + x} &= 2\left(1 + \frac{u}{3} - \frac{u^2}{9} + \frac{5u^3}{81} - \cdots\right) \\ &= 2 + \frac{x}{12} - \frac{x^2}{288} + \frac{5x^3}{20{,}736} - \cdots \end{aligned}

The series in u is valid for \lvert u \rvert < 1, so this one for \lvert x \rvert < 8 (and, since k > 0, at x = \pm 8 as well).

For 8.24, x = 0.24 and u = 0.03. The terms inside the bracket are 1, 0.01, -0.0001 and \displaystyle \frac{5}{81}(0.000027) \approx 0.0000017, so

\sqrt[3]{8.24} \approx 2(1.0099017) = 2.0198033.

A calculator gives 2.01980327. For u > 0 the terms after the first alternate in sign and shrink, so by the alternating series error bound the error is less than the next term, \displaystyle 2 \cdot \frac{10}{243}(0.03)^4 \approx 6.7 \times 10^{-8}. The edge is 2.0198 decimeters to the nearest ten-thousandth.

Example 4

The inverse sine from a binomial series

Find the Maclaurin series of \arcsin x and its radius.

Solution

The derivative of \arcsin t is (1 - t^2)^{-1/2}. Substitute -t^2 for x in Example 1; the sign (-1)^n cancels against (-t^2)^n = (-1)^n t^{2n}:

\frac{1}{\sqrt{1 - t^2}} = \sum_{n=0}^{\infty} \frac{1}{4^n}\binom{2n}{n}t^{2n}

for \lvert t \rvert < 1.

Integrate from 0 to x term by term, which keeps the radius, and use \arcsin 0 = 0:

\begin{aligned} \arcsin x &= \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{4^n(2n + 1)}x^{2n+1} \\ &= x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{5x^7}{112} + \cdots \end{aligned}

for \lvert x \rvert < 1. Only odd powers appear because \arcsin is an odd function.

Example 5

Ilir’s workbench lamp

Ilir hangs a small lamp h = 1.2 meters above his workbench (Figure 2). For a small lamp of fixed brightness I, the light falling on the bench at distance x from the foot O is

E(x) = \frac{I\,h}{(h^2 + x^2)^{3/2}},

the inverse-square law for the distance \displaystyle \sqrt{h^2 + x^2} times \displaystyle \cos\theta = \frac{h}{\sqrt{h^2 + x^2}} for the slant. Find a two-term approximation to \displaystyle \frac{E(x)}{E(0)}, and use it to find how far from O the light stays at least 96 percent of its value at O. Compare with the exact distance.

Figure 2A lamp at height h over a bench. Light reaching a point x from the foot O travels \displaystyle \sqrt{h^2 + x^2} and arrives at a slant (Example 5).

Solution

Factor h^2 out of the bracket: \displaystyle (h^2 + x^2)^{3/2} = h^3\left(1 + \frac{x^2}{h^2}\right)^{3/2}. Since \displaystyle E(0) = \frac{I}{h^2},

\frac{E(x)}{E(0)} = (1 + u)^{-3/2}, \quad u = \frac{x^2}{h^2}.

The coefficients for \displaystyle k = -\frac32 are 1, \displaystyle -\frac32, \displaystyle \binom{-3/2}{2} = \frac{(-\frac32)(-\frac52)}{2} = \frac{15}{8}, so

\frac{E(x)}{E(0)} = 1 - \frac32 u + \frac{15}{8}u^2 - \cdots

for 0 \le x < h.

The two-term rule says the light drops by about \displaystyle \frac32 \cdot \frac{x^2}{h^2}: a quadratic fall-off, flat at O. Asking \displaystyle 1 - \frac32 u \ge 0.96 gives \displaystyle u \le \frac{2}{75}, so

x \le 1.2\sqrt{\tfrac{2}{75}} \approx 0.196 \text{ meter}.

Exactly, (1 + u)^{-3/2} = 0.96 means u = 0.96^{-2/3} - 1 \approx 0.027588, so \displaystyle x \approx 1.2\sqrt{0.027588} \approx 0.199 meter. The two-term answer is cautious because for small u > 0 the terms alternate in sign and shrink, so the true ratio lies above \displaystyle 1 - \frac32 u: at x = 0.196 meter the light is about 96.13 percent of its value at O.

Boundaries

Inside the radius only. Unless k is a nonnegative integer, the series equals (1 + x)^k only for \lvert x \rvert < 1, plus whichever endpoints the table in Results allows. The function may be perfectly defined farther out, as (1 + x)^{-1/2} is at x = 3, and the series still diverges there (Exercise 21).

Rescaling needs a > 0. The factoring \displaystyle (a + x)^k = a^k(1 + \frac{x}{a})^k uses a^k, which for a fractional k is a real number only when a > 0; the result is valid for \lvert x \rvert < a.

Slow near the ends. Convergence inside (-1, 1) can be very slow close to \pm 1: in Figure 1, at x = -0.9 the partial sum P_4 still misses (1 + x)^{-1/2} by about 1.

Where this goes next

With the binomial series the list of standard Maclaurin series in the common Maclaurin series is complete, and every combination built there by substituting, multiplying and integrating works with (1 + x)^k too. The Lagrange error bound bounds the error of a partial sum where the alternating series bound does not apply. And using series to estimate integrals integrates expansions such as (1 + x^4)^{-1/2}, whose antiderivatives have no formula, to any accuracy wanted.

Exercises

Coefficients and the series

Compute \binom{3/2}{n} for n = 0, 1, 2, 3, 4.
Show that \binom{-3}{n} = (-1)^n\dfrac{(n+1)(n+2)}{2}, and write the binomial series for (1 + x)^{-3} with its interval of validity.
Write the first four terms of the binomial series for (1 + x)^{2/3}.
Write the first four terms of the binomial series for (1 + x)^{-1/3}.
Use the binomial series to expand (1 + x)^5, and explain why it has exactly six terms.
Find the coefficient of x^4 in the binomial series for (1 + x)^{5/2}, and explain why this series never stops.

Rescaling and substitution

Find the first four terms of the Maclaurin series of (9 + x)^{-1/2} and the interval where \lvert x/9 \rvert < 1.
Find the first four terms of the Maclaurin series of (27 + x)^{2/3} and its radius.
Find the first three nonzero terms and the general term of the Maclaurin series of \displaystyle \dfrac{x^2}{\sqrt{1 + x^3}}, and its radius.
Use Exercise 2 to find the Maclaurin series of \dfrac{1}{(2 + x)^3}, its first four terms, and its radius.
Use Example 2 to evaluate \sum_{n=0}^{\infty} \binom{2n}{n}\dfrac{1}{8^n}, and compare with the sum of its first four terms.

Estimates

Estimate (1.03)^{-2/5} with three terms of a binomial series, and compare with a calculator.
Estimate \displaystyle \sqrt[5]{31} by writing \displaystyle 31 = 32\left(1 - \frac{1}{32}\right) and using three terms. Compare with a calculator and explain why the estimate is too high.
Use the series of Example 4 to estimate \arcsin 0.2 with two and with three nonzero terms, and compare with a calculator.
Fatos lights a vivarium with a small lamp 0.5 meter above its floor. Using the model of Example 5, find \displaystyle \frac{E(x)}{E(0)} at x = 0.125 meter from two terms, from three terms, and exactly.
Use four terms of the series of Example 3 to estimate \displaystyle \sqrt[3]{7.6}, and compare with a calculator.

Reasoning

Explain why the ratio test gives radius 1 for the binomial series when k is not a nonnegative integer, and why the argument fails when k is a nonnegative integer. What is the radius then?
Let m be a nonnegative integer. Show that \binom{m}{n} = \dfrac{m!}{n!\,(m-n)!} for 0 \le n \le m and \binom{m}{n} = 0 for n > m, and conclude that the binomial series for (1 + x)^m is the binomial theorem.
Show that \binom{-1}{n} = (-1)^n and \binom{-2}{n} = (-1)^n(n + 1). Which series, found earlier by other means, are the binomial series for k = -1 and k = -2?
Show that the binomial series for (1 + x)^{-1/2} diverges at x = -1. (Show that its terms there are \dfrac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n)} and that each is at least \dfrac{1}{2n} for n \ge 1.)
A claim: “(1 + x)^{2/3} is defined at x = 3, so its binomial series converges there to 4^{2/3}.” Show that the claim is false, and find a series that does give 4^{2/3}.

Answers

1, \displaystyle \frac32, \displaystyle \frac38, \displaystyle -\frac{1}{16}, \displaystyle \frac{3}{128}. For example \displaystyle \binom{3/2}{3} = \frac{\frac32 \cdot \frac12 \cdot (-\frac12)}{6} = -\frac{1}{16} and \displaystyle \binom{3/2}{4} = \frac{\frac32 \cdot \frac12 \cdot (-\frac12)(-\frac32)}{24} = \frac{3}{128}.
The factors -3, -4, \ldots, -(n+2) give \displaystyle \binom{-3}{n} = \frac{(-1)^n (n+2)!}{2 \cdot n!}, which is \displaystyle (-1)^n\frac{(n+1)(n+2)}{2}. So \displaystyle (1 + x)^{-3} = \sum_{n=0}^{\infty} (-1)^n\frac{(n+1)(n+2)}{2}x^n, which is 1 - 3x + 6x^2 - 10x^3 + \cdots, for -1 < x < 1 (diverges at both ends, since k \le -1).
\displaystyle 1 + \frac{2x}{3} - \frac{x^2}{9} + \frac{4x^3}{81}. \displaystyle \binom{2/3}{2} = \frac{\frac23(-\frac13)}{2} = -\frac19 and \displaystyle \binom{2/3}{3} = \frac{\frac23(-\frac13)(-\frac43)}{6} = \frac{4}{81}.
\displaystyle 1 - \frac{x}{3} + \frac{2x^2}{9} - \frac{14x^3}{81}. \displaystyle \binom{-1/3}{2} = \frac{(-\frac13)(-\frac43)}{2} = \frac29 and \displaystyle \binom{-1/3}{3} = \frac{(-\frac13)(-\frac43)(-\frac73)}{6} = -\frac{14}{81}.
(1 + x)^5 = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5. For n \ge 6 the factors 5, 4, \ldots, 5 - n + 1 include 0, so every later coefficient is 0.
\displaystyle -\frac{5}{128}: \displaystyle \binom{5/2}{4} = \frac{\frac52 \cdot \frac32 \cdot \frac12 \cdot (-\frac12)}{24} = -\frac{15/16}{24}. The factors \displaystyle \frac52 - j are halves of odd integers, never 0, so no coefficient vanishes.
\displaystyle \frac13 - \frac{x}{54} + \frac{x^2}{648} - \frac{5x^3}{34{,}992}, for -9 < x < 9. Write \displaystyle (9 + x)^{-1/2} = \frac13(1 + \frac{x}{9})^{-1/2} and use Example 1 with \displaystyle u = \frac{x}{9}: \displaystyle \frac13 \cdot \frac38 \cdot \frac{1}{81} = \frac{1}{648} and \displaystyle \frac13 \cdot \frac{5}{16} \cdot \frac{1}{729} = \frac{5}{34{,}992}.
\displaystyle 9 + \frac{2x}{9} - \frac{x^2}{729} + \frac{4x^3}{177{,}147}, radius 27. \displaystyle (27 + x)^{2/3} = 9(1 + \frac{x}{27})^{2/3}, and Exercise 3 with \displaystyle u = \frac{x}{27} gives \displaystyle 9 \cdot \frac23 \cdot \frac{1}{27} = \frac29, \displaystyle 9 \cdot \frac19 \cdot \frac{1}{729} and \displaystyle 9 \cdot \frac{4}{81} \cdot \frac{1}{19{,}683}.
\displaystyle x^2 - \frac{x^5}{2} + \frac{3x^8}{8} - \cdots, general term \displaystyle \frac{(-1)^n}{4^n}\binom{2n}{n}x^{3n+2}, radius 1. Substitute x^3 into Example 1 and multiply by x^2; \lvert x^3 \rvert < 1 exactly when \lvert x \rvert < 1.
\displaystyle \frac{1}{(2 + x)^3} = \sum_{n=0}^{\infty} (-1)^n\frac{(n+1)(n+2)}{2^{n+4}}x^n, which is \displaystyle \frac18 - \frac{3x}{16} + \frac{3x^2}{16} - \frac{5x^3}{32} + \cdots, radius 2. \displaystyle (2 + x)^{-3} = \frac18(1 + \frac{x}{2})^{-3}, and \displaystyle \frac18 \cdot \frac{(n+1)(n+2)}{2} \cdot \frac{1}{2^n} = \frac{(n+1)(n+2)}{2^{n+4}}.
\displaystyle \sqrt2 \approx 1.41421. Here \displaystyle x = \frac18 < \frac14 and \displaystyle 1 - 4x = \frac12, so the sum is \displaystyle \frac{1}{\sqrt{1/2}} = \sqrt2. The first four terms give 1 + 0.25 + 0.09375 + 0.0390625 = 1.3828125; the terms shrink only by a factor of about 2 each step, so many more are needed.
0.988252; a calculator gives 0.988246. With \displaystyle k = -\frac25, \displaystyle \binom{-2/5}{2} = \frac{(-\frac25)(-\frac75)}{2} = \frac{7}{25}, so 1 - 0.012 + 0.28(0.0009) = 0.988252.
1.98734375; a calculator gives 1.9873408. \displaystyle \sqrt[5]{31} = 2(1 - \frac{1}{32})^{1/5}, which is about \displaystyle 2(1 - \frac{1}{160} - \frac{2}{25} \cdot \frac{1}{1024}). With \displaystyle k = \frac15 and u < 0, every term after the first is negative (the coefficients alternate and so do the powers of u), so dropping them leaves the estimate too high.
0.2013333 and 0.2013573; a calculator gives \arcsin 0.2 \approx 0.2013579. The terms are 0.2, \displaystyle \frac{0.008}{6} and \displaystyle \frac{3(0.00032)}{40} = 0.000024.
Two terms 0.90625, three terms 0.913574, exactly 0.913075. Here \displaystyle u = (0.125/0.5)^2 = \frac{1}{16}: \displaystyle 1 - \frac32 \cdot \frac{1}{16}, then \displaystyle +\frac{15}{8} \cdot \frac{1}{256}, and \displaystyle (\frac{17}{16})^{-3/2}.
1.9660957; a calculator gives 1.9660951. Here x = -0.4, u = -0.05: 2(1 - 0.0166667 - 0.0002778 - 0.0000077). Every term after the first is negative, so the estimate is slightly high.
With no coefficient zero, \displaystyle \lvert a_{n+1}/a_n \rvert = \frac{\lvert k - n \rvert}{n + 1}\lvert x \rvert \to \lvert x \rvert, so the series converges for \lvert x \rvert < 1 and diverges for \lvert x \rvert > 1. When k = m is a nonnegative integer, a_n = 0 for n > m, the ratio is undefined, and the test cannot be used; the series is a polynomial, converges for every x, and the radius is infinite.
For n \le m, \displaystyle m(m-1)\cdots(m-n+1) = \frac{m!}{(m-n)!}, so dividing by n! gives \displaystyle \frac{m!}{n!(m-n)!}, the count of choices. For n > m the factors run from m down past 0, so one of them is m - m = 0. The series is \sum_{n=0}^{m}\binom{m}{n}x^n, the binomial theorem for (1 + x)^m.
\displaystyle \binom{-1}{n} = \frac{(-1)(-2)\cdots(-n)}{n!} = (-1)^n and \displaystyle \binom{-2}{n} = \frac{(-2)(-3)\cdots(-(n+1))}{n!}, which is (-1)^n(n + 1). So (1 + x)^{-1} = \sum (-x)^n is the geometric series, and (1 + x)^{-2} = \sum (-1)^n(n + 1)x^n is the differentiated geometric series of Representing functions as power series with -x in place of x.
At x = -1 the term is \displaystyle \binom{-1/2}{n}(-1)^n = \frac{1}{4^n}\binom{2n}{n}, and \displaystyle \frac{(2n)!}{4^n(n!)^2} splits as \displaystyle \frac{1 \cdot 3 \cdots (2n-1)}{2^n n!} \cdot \frac{2 \cdot 4 \cdots 2n}{2^n n!}, where the second fraction is 1 and 2^n n! = 2 \cdot 4 \cdots 2n. For j \ge 2, \displaystyle \frac{2j - 1}{2j} \ge \frac{j - 1}{j} (cross-multiplying, 2j^2 - j \ge 2j^2 - 2j). So the term is at least \displaystyle \frac12 \cdot \frac12 \cdot \frac23 \cdots \frac{n-1}{n} = \frac{1}{2n}. Since \displaystyle \sum \frac{1}{2n} diverges (The integral test and p series), so does the series, by comparison.
The claim is false. No \binom{2/3}{n} is zero, and at x = 3 the ratio of consecutive terms tends to 3 > 1, so the terms grow and the series diverges. Rescale instead: \displaystyle 4^{2/3} = (8 - 4)^{2/3} = 4(1 - \frac12)^{2/3}, and the series for (1 + u)^{2/3} at \displaystyle u = -\frac12 converges to it, since \lvert u \rvert < 1.