Infinite series and partial sums
Adding infinitely many numbers can never be finished, so what could the sum of an infinite list possibly mean, and why do terms that shrink to zero sometimes add up to a number and sometimes grow past every bound?
Where you are 8 in Sequences and series in calculus
Why this exists
Zviad stacks identical books at the edge of a table, one on top of another, and wants the top book to stick out as far as possible past the edge without the stack tipping. With one book the best he can do is half a book length. With more books he can do better, and the best stack (Figure 1) puts the overhangs, from the top down, at \displaystyle \frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \ldots of a book length. With n books the top book reaches
book lengths past the edge. Can the overhang be made as large as he likes, given enough books? The question is about the sum \displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \cdots continued forever, and an infinite sum cannot be computed by adding. It needs a definition.
The definition turns an infinite sum into something already understood: the limit of a sequence (Convergence of sequences). Add the first n terms, as for the finite sums in Arithmetic series and partial sums, and ask what those running totals do as n grows. Day 1 sets up this definition, the laws it inherits from limits of sequences, and the telescoping series, whose running totals collapse to a short formula. Day 2 answers Zviad: the harmonic series grows past every bound, so the overhang can be as large as he likes, but it grows so slowly that a meter of overhang takes more than twelve thousand books.
Definition
In plain words: a series is an infinite sum of terms. Its partial sums are the running totals, and the series converges if those running totals settle on one number, which is then called its sum.
In symbols: given a sequence \{a_n\}, the series \sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots has nth partial sum
(sigma notation as in Sigma notation). If the sequence of partial sums converges, \lim_{n \to \infty} S_n = S, the series converges and its sum is S, written \sum_{n=1}^{\infty} a_n = S. Otherwise the series diverges. If S_n \to \infty, the series diverges to infinity, written \sum_{n=1}^{\infty} a_n = \infty; the sum does not exist as a number.
A series carries two sequences: the terms a_n and the partial sums S_n. They are linked both ways: S_n = S_{n-1} + a_n for n \ge 2, so a_n = S_n - S_{n-1}, and a_1 = S_1. A series may start at any index, \sum_{n=m}^{\infty} a_n, with partial sums a_m + \cdots + a_n.
A telescoping series is one whose terms can be written as differences a_n = b_n - b_{n+k} of a sequence with itself, shifted by a fixed k, so that almost everything in S_n cancels.
The harmonic series is \displaystyle \sum_{n=1}^{\infty} \frac{1}{n}, and its partial sums \displaystyle H_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n} are the harmonic numbers.
Intuition
Why the stack in Figure 1 balances. A stack of books does not tip as long as, for every book, the books resting on it have their combined center of mass over that book (or, for the bottom book, over the table). Build the stack from the top. Place book k + 1 so that its right edge sits exactly under the center of mass of the k books above it. Take that edge as position 0, so the upper k books have their center at 0, and book k + 1, of length 1, has its center at \displaystyle -\frac{1}{2}. All k + 1 books together have their center at
That is where the right edge of the next book down goes, so book k + 1 overhangs the book below it by \displaystyle \frac{1}{2(k+1)}. The top book overhangs the second by \displaystyle \frac{1}{2}, and the table edge sits under the center of all n books. Adding the steps, the top book reaches \displaystyle \frac{1}{2}H_n book lengths past the edge. Already \displaystyle \frac{1}{2}H_4 = \frac{25}{24} > 1: with four books, the top one lies entirely beyond the table.
So the overhang is a partial sum, and it grows without bound exactly when the harmonic series diverges. Each new book adds less, but a total can creep up forever by smaller and smaller steps (Figure 2), and Day 2 shows this one does.
The walls in Figure 2 come from areas. Stand a rectangle of height \displaystyle \frac{1}{k} on each interval [k, k+1] (Figure 3). The rectangles together have area H_n, and they cover the region under \displaystyle y = \frac{1}{x} from 1 to n + 1, whose area is \ln(n + 1). That area grows without bound, so the rectangles’ total must too.
For a series of positive terms, the partial sums only climb. A climbing sequence converges exactly when it has a ceiling (the monotone convergence theorem in Convergence of sequences), so for positive terms the whole question is whether the running totals stay below some fixed number.
Results
Day 1: series as sequences of partial sums
Sum and constant-multiple laws. If \sum_{n=1}^{\infty} a_n = A and \sum_{n=1}^{\infty} b_n = B, then \sum_{n=1}^{\infty} (a_n + b_n) = A + B, \sum_{n=1}^{\infty} (a_n - b_n) = A - B, and \sum_{n=1}^{\infty} c\,a_n = cA for every constant c.
Convergent plus divergent diverges. If \sum a_n converges and \sum b_n diverges, then \sum (a_n + b_n) diverges. If c \ne 0 and \sum a_n diverges, so does \sum c\,a_n.
Finitely many terms do not decide convergence. For each m \ge 2, \sum_{n=1}^{\infty} a_n converges exactly when \sum_{n=m}^{\infty} a_n does, and then
Adding, dropping or changing finitely many terms can change the sum, but never whether there is one.
Telescoping series. If a_n = b_n - b_{n+1}, then S_n = b_1 - b_{n+1}. The series converges exactly when \{b_n\} converges, and if b_n \to L, its sum is b_1 - L. More generally, if a_n = b_n - b_{n+k} and b_n \to L, the sum is b_1 + \cdots + b_k - kL.
Positive terms. If every a_n \ge 0, the partial sums are nondecreasing, so the series converges exactly when its partial sums are bounded above, and otherwise it diverges to infinity.
The terms of a convergent series tend to 0. If S_n \to S then a_n = S_n - S_{n-1} \to S - S = 0. This is the basis of the The nth term test for divergence. The converse is false: Examples 2 and 5 have terms tending to 0 and diverge.
Day 2: the harmonic series
The harmonic series diverges to infinity. In fact \displaystyle H_{2^k} \ge 1 + \frac{k}{2} for every integer k \ge 0.
Integral bounds. For every n \ge 1,
with equality on the right only at n = 1.
The Euler constant. The sequence H_n - \ln n is decreasing and positive, so it converges. Its limit is the Euler constant \gamma \approx 0.5772, and H_n \approx \ln n + \gamma for large n. (Whether \gamma is rational is not known.)
Why it is true
The laws. The partial sums of \sum (a_n + b_n) are \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k, the sum of the two sequences of partial sums, by the commutative and associative properties of a finite sum. The sum law for sequences gives the limit A + B. Likewise \sum_{k=1}^{n} c\,a_k = c\sum_{k=1}^{n} a_k \to cA by the distributive property and the constant-multiple law. For convergent plus divergent: if \sum (a_n + b_n) converged, then \sum b_n = \sum \big((a_n + b_n) - a_n\big) would converge by the difference law. For c \ne 0, apply the constant-multiple law with \displaystyle \frac{1}{c}.
Finitely many terms. For n \ge m, the partial sums satisfy \sum_{k=1}^{n} a_k = C + \sum_{k=m}^{n} a_k with the constant C = a_1 + \cdots + a_{m-1}. Two sequences that differ by a constant converge or diverge together, and their limits differ by C.
Telescoping. In S_n = (b_1 - b_2) + (b_2 - b_3) + \cdots + (b_n - b_{n+1}), each b_j with 2 \le j \le n appears once with each sign, leaving b_1 - b_{n+1}. Since b_{n+1} is the sequence \{b_n\} with one term dropped, S_n converges exactly when \{b_n\} does, with limit b_1 - L. With a shift of k, the survivors are b_1 + \cdots + b_k at the front and b_{n+1} + \cdots + b_{n+k} at the back, and the k back terms each tend to L.
Positive terms. S_{n+1} = S_n + a_{n+1} \ge S_n, and the monotone convergence theorem applies. A nondecreasing sequence with no upper bound passes every number and stays past it, which is divergence to infinity.
The harmonic series diverges (grouping). After the first two terms, group the rest in blocks that end at a power of 2:
The block ending at 2^j has 2^{j-1} terms, each at least its last term \displaystyle \frac{1}{2^j}, so the block is at least \displaystyle 2^{j-1} \cdot \frac{1}{2^j} = \frac{1}{2}. With \displaystyle \frac{1}{2} and k - 1 blocks after the leading 1, \displaystyle H_{2^k} \ge 1 + \frac{k}{2}. The partial sums are nondecreasing and H_{2^k} exceeds every bound, so H_n \to \infty.
Integral bounds (Figure 3). On [k, k+1] the decreasing function \displaystyle \frac{1}{x} is at most \displaystyle \frac{1}{k}, and less except at x = k, so \displaystyle \int_k^{k+1} \frac{dx}{x} < \frac{1}{k}. Adding over k = 1, \ldots, n,
On [k - 1, k] the function is at least \displaystyle \frac{1}{k}, so \displaystyle \frac{1}{k} < \int_{k-1}^{k} \frac{dx}{x} for k \ge 2, and adding over k = 2, \ldots, n gives H_n - 1 < \ln n for n \ge 2. At n = 1 both sides of H_1 \le 1 + \ln 1 equal 1. The lower bound is a second proof of divergence. It is the same fact as the divergence of the improper integral \displaystyle \int_1^{\infty} \frac{1}{x}\,dx in Improper integrals: the rectangles cover the region under the curve, so their total area cannot stay finite while the area under the curve grows past every bound.
The Euler constant. Let D_n = H_n - \ln n. Then
since the integrand is positive on [n, n+1). So \{D_n\} decreases, and D_n > \ln(n+1) - \ln n > 0 by the lower bound. Decreasing and bounded below, it converges. The value 0.5772\ldots is computed, not derived here.
Worked examples
Day 1: a telescoping sum with a gap of two
Find the sum of \sum_{n=1}^{\infty} \dfrac{1}{n^2 + 4n + 3}, and the first n for which S_n is within 0.01 of it.
SolutionThe bottom factors as (n+1)(n+3). Partial fractions (partial fractions) split the term: from 1 = A(n + 3) + B(n + 1), setting n = -1 gives \displaystyle A = \frac{1}{2}, and n = -3 gives \displaystyle B = -\frac{1}{2}. So
Write out the partial sum. Each negative piece \displaystyle \frac{1}{n+3} cancels the positive piece two terms later:
The survivors are the first two positive pieces and the last two negative ones. Check: \displaystyle S_1 = \frac{1}{2}\left(\frac{5}{6} - \frac{1}{3} - \frac{1}{4}\right) = \frac{1}{8}, which is \displaystyle a_1 = \frac{1}{1 + 4 + 3}. As n \to \infty the last two pieces tend to 0, so
The gap is \displaystyle \frac{5}{12} - S_n = \frac{1}{2}\left(\frac{1}{n+2} + \frac{1}{n+3}\right), which decreases in n. At n = 97 it is \displaystyle \frac{1}{2}\left(\frac{1}{99} + \frac{1}{100}\right) \approx 0.01005; at n = 98 it is \displaystyle \frac{1}{2}\left(\frac{1}{100} + \frac{1}{101}\right) \approx 0.00995. So S_n is within 0.01 of the sum from n = 98 on.
Day 1: Gulshan’s logarithms, shrinking terms that add up to infinity
Gulshan considers \sum_{n=1}^{\infty} \ln\dfrac{n+1}{n}. Do its terms tend to 0? Does it converge? Find the first n with S_n > 10.
SolutionThe term is \displaystyle \ln\left(1 + \frac{1}{n}\right), and \displaystyle 1 + \frac{1}{n} \to 1. The logarithm is continuous at 1, so the terms tend to \ln 1 = 0.
The log of a quotient is a difference, a_n = \ln(n + 1) - \ln n, so the series telescopes with b_n = -\ln n, or directly:
Since \ln(n + 1) \to \infty, the series diverges to infinity, although its terms tend to 0. Shrinking terms are necessary for convergence, never enough.
S_n > 10 means n + 1 > e^{10} \approx 22{,}026.47. The first such n is 22{,}026.
Day 1: Ravshan’s partial sums, and the terms behind them
Ravshan knows only that the partial sums of a series are S_n = 2 - \dfrac{2n + 3}{3^n}. Find the terms, and the sum.
SolutionThe first term is \displaystyle a_1 = S_1 = 2 - \frac{5}{3} = \frac{1}{3}. For n \ge 2, a_n = S_n - S_{n-1}; over the common denominator 3^n,
The formula does not cover n = 1: it gives \displaystyle \frac{4}{3}, not \displaystyle \frac{1}{3}. The difference formula needs an S_0, and the given expression would set S_0 = 2 - 3 = -1, not the empty sum 0. So the series is \displaystyle \frac{1}{3} + \frac{8}{9} + \frac{12}{27} + \frac{16}{81} + \cdots. Check: \displaystyle S_2 = 2 - \frac{7}{9} = \frac{11}{9} = \frac{1}{3} + \frac{8}{9}.
For the sum, \displaystyle \frac{2n + 3}{3^n} \to 0, because an exponential outgrows a polynomial (Comparing growth rates at infinity). So the series converges and its sum is \lim S_n = 2. Dropping the first term then gives \displaystyle \sum_{n=2}^{\infty} \frac{4n}{3^n} = 2 - \frac{1}{3} = \frac{5}{3}.
Day 1: the laws, a dropped head, and a divergent part
(a) Find \sum_{n=1}^{\infty} \dfrac{6}{n(n+3)}. (b) Find the sum of the series, from n = 1, with terms \dfrac{6}{n(n+3)} - 3\left(\dfrac{2}{5}\right)^n. (c) Find \sum_{n=4}^{\infty} \dfrac{6}{n(n+3)} in two ways. (d) Decide whether the series with terms \dfrac{6}{n(n+3)} + \ln\dfrac{n+1}{n} converges.
Solution(a) Partial fractions give \displaystyle \frac{6}{n(n+3)} = 2\left(\frac{1}{n} - \frac{1}{n+3}\right), a telescoping series with \displaystyle b_n = \frac{2}{n} and a shift of 3. The survivors at the front are b_1 + b_2 + b_3, and b_n \to 0, so the sum is \displaystyle 2\left(1 + \frac{1}{2} + \frac{1}{3}\right) = \frac{11}{3}.
(b) The second series is geometric with ratio \displaystyle \frac{2}{5}, whose sum is \displaystyle \frac{2/5}{1 - 2/5} = \frac{2}{3} (Infinite geometric series). Both series converge, so the difference and constant-multiple laws apply: the sum is \displaystyle \frac{11}{3} - 3 \cdot \frac{2}{3} = \frac{5}{3}.
(c) Dropping the first three terms subtracts them from the sum:
Directly, from n = 4 the telescoping survivors are \displaystyle b_4 + b_5 + b_6 = 2\left(\frac{1}{4} + \frac{1}{5} + \frac{1}{6}\right) = \frac{37}{30}. The two ways agree.
(d) The first series converges (part a) and the second diverges (Example 2), so the sum of the two diverges, by the rule that convergent plus divergent diverges.
Day 2: the harmonic series by grouping
Show from the grouping argument that H_{1024} \ge 6, and compare with the true value and with the first n for which H_n \ge 6.
Solution1024 = 2^{10}. Write H_{1024} as \displaystyle 1 + \frac{1}{2} followed by blocks ending at 4, 8, 16, \ldots, 1024: nine blocks. The block from \displaystyle \frac{1}{2^{j-1}+1} to \displaystyle \frac{1}{2^j} has 2^{j-1} terms, none smaller than \displaystyle \frac{1}{2^j}, so it is at least \displaystyle \frac{1}{2}. Then
Since the partial sums increase, every H_n with n \ge 1024 is at least 6.
The bound is crude: H_{1024} \approx 7.509, and a direct computation shows that H_n first reaches 6 at n = 227. Replacing every term of a block by the smallest one loses a lot, but never the divergence.
Day 2: Zviad’s stack, a meter past the edge
Zviad’s books are 20 centimeters long. How far past the edge does the top book reach with 4 books? Use the integral bounds to say how many books a 1 meter overhang needs at least, and how many are certainly enough.
SolutionThe overhang is \displaystyle \frac{1}{2}H_n book lengths, which is 10H_n centimeters. With 4 books, \displaystyle H_4 = \frac{25}{12}, so the overhang is \displaystyle \frac{125}{6} \approx 20.83 centimeters, more than one book length.
One meter needs 10H_n \ge 100, that is H_n \ge 10.
At least. H_n \le 1 + \ln n, so H_n \ge 10 forces \ln n \ge 9, that is n \ge e^9 \approx 8103.08. At least 8104 books are needed.
Enough. H_n > \ln(n + 1), and Example 2 found that \ln(n+1) > 10 first holds at n = 22{,}026. That many books certainly suffice.
The answer lies between 8104 and 22{,}026. Adding the terms by computer pins it down: H_n first reaches 10 at n = 12{,}367. So the overhang has no ceiling, because the harmonic series diverges, but every extra half meter multiplies the number of books by about e^5 \approx 148.
Day 2: how slowly, the Euler constant
Use H_n - \ln n \to \gamma \approx 0.5772 to estimate H_{1{,}000{,}000}, and the overhang of a million 20 centimeter books.
SolutionThe sequence D_n = H_n - \ln n decreases toward \gamma (Results, Day 2). The first values show the approach:
For large n, then, H_n \approx \ln n + \gamma. With n = 10^6, \ln 10^6 = 6\ln 10 \approx 13.8155, so
which agrees with the value found by adding all million terms, 14.392727\ldots, to four decimal places. A million of Zviad’s books reach 10H_n \approx 143.9 centimeters past the edge: under a meter and a half. The partial sums grow like \ln n, without bound and very slowly.
Boundaries
- A series is not its terms. Terms tending to 0 say nothing about convergence by themselves (Examples 2 and 5). Terms not tending to 0 do decide it, which is the The nth term test for divergence.
- Telescoping needs a limit for b_n. \sum (b_n - b_{n+1}) converges exactly when \{b_n\} does. Cancelling in S_n is always allowed, because S_n is a finite sum; cancelling inside the infinite sum before forming S_n is not. With b_n = n, the terms are all -1, and the series diverges.
- The laws need convergent series. \sum (a_n + b_n) = \sum a_n + \sum b_n holds when both sums on the right exist. Two divergent series can have a convergent sum: \sum 1 and \sum (-1) diverge, and their term-by-term sum is \sum 0 = 0.
- Dropping terms changes the sum. Only convergence is unaffected; the sum changes by the dropped total (Example 4).
- The difference formula a_n = S_n - S_{n-1} holds for n \ge 2. The first term is S_1, and a formula for S_n need not give S_0 = 0 (Example 3).
Where this goes next
Most series have no formula for S_n. For them the question is whether the series converges, and the Convergence tests answer it. The first is the The nth term test for divergence: terms that do not tend to 0 force divergence. The rectangles of Figure 3 become the integral test of The integral test and p series, which decides \displaystyle \sum \frac{1}{n^p} for every power p and puts the harmonic series at the exact boundary p = 1. The The geometric series test packages the geometric case. Later, a Taylor and Maclaurin series is a series whose terms depend on x, and its partial sums are the polynomials that approximate a function.
Exercises
Day 1: partial sums and terms
- A series has partial sums S_n = \dfrac{n^2}{n^2 + 1}. Find a_1, a formula for a_n, and the sum.
- A series has partial sums S_n = 2 + \dfrac{(-1)^n}{n}. Find a_1, a formula for a_n when n \ge 2, and the sum.
- Write the first four partial sums of \sum_{n=1}^{\infty} (-1)^{n+1} and decide whether the series converges.
Day 1: telescoping series
Find a formula for S_n and the sum, or show that the series diverges.
- \sum_{n=1}^{\infty} \dfrac{4}{n^2 + 4n}
- \sum_{n=2}^{\infty} \dfrac{2}{n^2 - 1}
- \displaystyle \sum_{n=1}^{\infty} \left(\dfrac{1}{\sqrt{n}} - \dfrac{1}{\sqrt{n+1}}\right)
- \sum_{n=1}^{\infty} \big(\arctan(n+1) - \arctan n\big)
- \sum_{n=1}^{\infty} \ln\dfrac{n(n+2)}{(n+1)^2}
Day 1: the laws and finitely many terms
- Suppose \sum_{n=1}^{\infty} a_n = 7 and \sum_{n=1}^{\infty} b_n = -2, with a_1 = 4 and a_2 = 1. Find \sum_{n=1}^{\infty} (3a_n - 2b_n) and \sum_{n=3}^{\infty} a_n.
- Find \sum_{n=4}^{\infty} \dfrac{1}{n^2 - 9}.
- Decide whether the series with terms \displaystyle \dfrac{1}{\sqrt{n}} - \dfrac{1}{\sqrt{n+1}} + \ln\dfrac{n+2}{n+1}, n \ge 1, converges.
- Show that \displaystyle \sum \ln\frac{n+1}{n} and \displaystyle \sum \ln\frac{n+2}{n+1} both diverge, yet the series with terms \ln\dfrac{n+1}{n} - \ln\dfrac{n+2}{n+1}, n \ge 1, converges, and find its sum.
Day 2: the harmonic series
- Find H_7 exactly and check that it lies between \ln 8 and 1 + \ln 7.
- Find the smallest k for which the grouping bound \displaystyle H_{2^k} \ge 1 + \frac{k}{2} guarantees H_{2^k} \ge 8, and the corresponding n = 2^k.
- Ravshan stacks 8 books, each 24 centimeters long, in the harmonic stack. How far past the table edge does the top book reach?
- With Ravshan’s 24 centimeter books, use the integral bounds to find how many books a 60 centimeter overhang needs at least, and a number of books that is certainly enough.
- Estimate H_{10{,}000} with H_n \approx \ln n + \gamma.
- Use H_n - \ln n \to \gamma to find \lim_{n \to \infty} \left(\dfrac{1}{n+1} + \dfrac{1}{n+2} + \cdots + \dfrac{1}{2n}\right).
- Show that \sum_{n=1}^{\infty} \dfrac{1}{2n - 1} diverges by comparing its partial sums with H_n.
- Show that \sum_{n=1}^{\infty} \dfrac{1}{n + 100} diverges by writing its partial sums with harmonic numbers.
Reasoning
- Show that the partial sums of \sum_{n=1}^{\infty} \dfrac{1}{n^2} satisfy S_n < 2, and explain why the series converges.
- Prove that if \sum a_n converges and \sum b_n diverges, then \sum (a_n + b_n) diverges.
- Gulshan claims that if \sum a_n and \sum b_n both diverge, then \sum (a_n + b_n) diverges. Give a counterexample.
- Explain why a series of positive terms either converges or diverges to infinity, and never diverges in any other way.
- Show that \displaystyle H_{2n} - H_n \ge \frac{1}{2} for every n, and use it to give another proof that \{H_n\} cannot converge.
Answers
- \displaystyle a_1 = \frac{1}{2}, \displaystyle a_n = \frac{2n - 1}{(n^2 + 1)(n^2 - 2n + 2)}, sum 1. Over the common denominator, the top of S_n - S_{n-1} is n^2\big((n-1)^2 + 1\big) - (n-1)^2(n^2 + 1), which is n^2 - (n-1)^2 = 2n - 1; at n = 1 the formula also gives \displaystyle \frac{1}{2}. S_n \to 1.
- a_1 = 1, \displaystyle a_n = \frac{(-1)^n(2n - 1)}{n(n-1)} for n \ge 2, sum 2. \displaystyle S_n - S_{n-1} = (-1)^n\left(\frac{1}{n} + \frac{1}{n-1}\right), and \displaystyle \frac{(-1)^n}{n} \to 0.
- Diverges. S_1 = 1, S_2 = 0, S_3 = 1, S_4 = 0; the partial sums alternate between 1 and 0, and no number is within \displaystyle \frac{1}{2} of both.
- \displaystyle \frac{25}{12}. \displaystyle \frac{4}{n(n+4)} = \frac{1}{n} - \frac{1}{n+4}, so S_n is \displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} minus the four fractions \displaystyle \frac{1}{n+1}, \ldots, \frac{1}{n+4}, which tend to 0.
- \displaystyle \frac{3}{2}. \displaystyle \frac{2}{n^2 - 1} = \frac{1}{n-1} - \frac{1}{n+1}, so \displaystyle S_n = 1 + \frac{1}{2} - \frac{1}{n} - \frac{1}{n+1} (summing from 2 to n).
- 1. \displaystyle S_n = 1 - \frac{1}{\sqrt{n+1}} \to 1.
- \displaystyle \frac{\pi}{4}. \displaystyle S_n = \arctan(n + 1) - \arctan 1 \to \frac{\pi}{2} - \frac{\pi}{4}.
- -\ln 2. The term is b_n - b_{n+1} with \displaystyle b_n = \ln\frac{n}{n+1} \to 0, so \displaystyle S_n = \ln\frac{1}{2} - \ln\frac{n+1}{n+2} \to -\ln 2.
- 25 and 2. By the laws, 3 \cdot 7 - 2(-2) = 25. Dropping two terms, 7 - 4 - 1 = 2.
- \displaystyle \frac{49}{120}. \displaystyle \frac{1}{n^2 - 9} = \frac{1}{6}\left(\frac{1}{n-3} - \frac{1}{n+3}\right), a shift of 6; from n = 4 the survivors are \displaystyle \frac{1}{6}\left(1 + \frac{1}{2} + \cdots + \frac{1}{6}\right) = \frac{1}{6} \cdot \frac{49}{20}.
- Diverges. The first part telescopes to 1 (Exercise 6); the partial sums of the second are \displaystyle \ln\frac{n+2}{2} \to \infty. Convergent plus divergent diverges.
- \ln 2. Partial sums \ln(n+1) and \displaystyle \ln\frac{n+2}{2} both tend to \infty. The difference is b_n - b_{n+1} with \displaystyle b_n = \ln\frac{n+1}{n} \to 0, so the sum is b_1 = \ln 2.
- \displaystyle H_7 = \frac{363}{140} \approx 2.5929, between \ln 8 \approx 2.0794 and 1 + \ln 7 \approx 2.9459.
- k = 14, n = 16{,}384. \displaystyle 1 + \frac{k}{2} \ge 8 needs k \ge 14. (In fact H_n first reaches 8 at n = 1674.)
- \displaystyle \frac{2283}{70} \approx 32.61 centimeters. The overhang is 12H_8 centimeters, and \displaystyle H_8 = \frac{761}{280}.
- At least 55; 148 is enough. 12H_n \ge 60 means H_n \ge 5. From H_n \le 1 + \ln n, need n \ge e^4 \approx 54.60. From H_n > \ln(n+1), n + 1 \ge e^5 \approx 148.41 suffices, so n = 148. (The exact answer is 83.)
- About 9.7876. \ln 10^4 + \gamma \approx 9.21034 + 0.57722; the true value is 9.78761 to five decimals.
- \ln 2. The sum is H_{2n} - H_n, which equals (H_{2n} - \ln 2n) - (H_n - \ln n) + \ln 2. This tends to \gamma - \gamma + \ln 2.
- Each term \displaystyle \frac{1}{2k-1} \ge \frac{1}{2k}, so the nth partial sum is at least \displaystyle \frac{1}{2}H_n, which grows without bound.
- The nth partial sum is H_{n+100} - H_{100}; H_{100} is a constant and H_{n+100} \to \infty.
- For k \ge 2, \displaystyle \frac{1}{k^2} < \int_{k-1}^{k} \frac{dx}{x^2}, so \displaystyle S_n \le 1 + \int_1^n \frac{dx}{x^2} = 2 - \frac{1}{n} < 2. The partial sums are increasing and bounded above, so they converge by the monotone convergence theorem. (The sum is \displaystyle \frac{\pi^2}{6}, stated without proof.)
- If \sum (a_n + b_n) converged, the difference law would make \sum \big((a_n + b_n) - a_n\big) = \sum b_n converge, a contradiction.
- a_n = 1 and b_n = -1: both diverge (partial sums n and -n), while \sum (a_n + b_n) = \sum 0 = 0.
- Positive terms make S_n increasing. If the partial sums are bounded above they converge (monotone convergence). If not, for every M some S_N > M, and every later S_n \ge S_N > M, which is divergence to infinity. Oscillation is impossible because S_n never goes down.
- \displaystyle H_{2n} - H_n = \frac{1}{n+1} + \cdots + \frac{1}{2n} has n terms, each at least \displaystyle \frac{1}{2n}, so it is at least \displaystyle \frac{1}{2}. If H_n \to L, then H_{2n} \to L too (a subsequence), so H_{2n} - H_n \to 0, which contradicts \displaystyle H_{2n} - H_n \ge \frac{1}{2}.