Unit 7, lesson 95

Telescoping and Harmonic Series

How cancellation gives exact partial sums for telescoping series, why the harmonic series diverges even though its terms shrink to zero, and how this settles many p-series.

  • Theory part 189
  • Practice part 190
  • Infinite Series

Some sums collapse like a folding telescope. Each term cancels part of the next one, and after the dust settles only the two ends remain. Other sums look just as harmless and yet grow past any bound. The harmonic series 1+12+13+⋯1 + \frac{1}{2} + \frac{1}{3} + \cdots is the most famous example, and it marks the border between series that add up and series that do not.

Key terms

Telescoping series

Definition A series ∑n=1∞an\sum_{n=1}^{\infty} a_n is telescoping if there is a sequence (bn)(b_n) with an=bn−bn+1a_n = b_n - b_{n+1} for every n≥1n \ge 1. Then the partial sums are

sN=(b1−b2)+(b2−b3)+⋯+(bN−bN+1)=b1−bN+1,s_N = (b_1 - b_2) + (b_2 - b_3) + \cdots + (b_N - b_{N+1}) = b_1 - b_{N+1},

so the series converges if and only if (bn)(b_n) converges, and in that case

∑n=1∞an=b1−lim⁡n→∞bn.\sum_{n=1}^{\infty} a_n = b_1 - \lim_{n \to \infty} b_n.

In plain words Write each term as a difference of two neighbors. In a finite partial sum all the middle pieces cancel, and you can take the limit of what is left.

Example Partial fractions give

1n(n+1)=1n−1n+1,\frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1},

so bn=1nb_n = \frac{1}{n} and

sN=1−1N+1→1,∑n=1∞1n(n+1)=1.s_N = 1 - \frac{1}{N + 1} \to 1, \qquad \sum_{n=1}^{\infty} \frac{1}{n(n + 1)} = 1.

Always cancel inside the finite sum sNs_N first. Rearranging the infinitely many terms of a series before you know it converges can give nonsense.

Telescoping can also prove divergence. Since

ln⁡(1+1n)=ln⁡(n+1)−ln⁡n,\ln\left(1 + \frac{1}{n}\right) = \ln(n + 1) - \ln n,

the series ∑ln⁡(1+1n)\sum \ln\left(1 + \frac{1}{n}\right) has sN=ln⁡(N+1)−ln⁡1=ln⁡(N+1)→∞s_N = \ln(N + 1) - \ln 1 = \ln(N + 1) \to \infty and diverges.

Harmonic series

Definition The harmonic series is

∑n=1∞1n=1+12+13+14+⋯ .\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots.

Its partial sums

HN=∑n=1N1nH_N = \sum_{n=1}^{N} \frac{1}{n}

increase without bound, so the series diverges, even though 1n→0\frac{1}{n} \to 0.

In plain words The terms do shrink to zero, but too slowly. Enough of them always add up to something sizeable.

Example The partial sums grow, but very slowly, roughly like ln⁡N\ln N.

NNHNH_Nln⁡N\ln N
10102.92902.92902.30262.3026
1001005.18745.18744.60524.6052
100010007.48557.48556.90786.9078

A table alone could never prove divergence. The next section does.

Divergence by grouping

Definition For every integer k≥0k \ge 0 the harmonic partial sums satisfy

H2k≥1+k2.H_{2^k} \ge 1 + \frac{k}{2}.

The proof groups the terms in blocks. For each j≥1j \ge 1 the block

12j−1+1+12j−1+2+⋯+12j\frac{1}{2^{j-1} + 1} + \frac{1}{2^{j-1} + 2} + \cdots + \frac{1}{2^j}

has 2j−12^{j-1} terms, each at least 12j\frac{1}{2^j}, so the block is at least 12\frac{1}{2}. Adding H1=1H_1 = 1 and the blocks for j=1,…,kj = 1, \ldots, k gives the inequality. Since (HN)(H_N) is increasing and its subsequence H2kH_{2^k} is unbounded, HN→∞H_N \to \infty.

In plain words You can split the harmonic series into infinitely many groups that each add to at least one half. Infinitely many halves are infinite.

Example For k=4k = 4, that is N=16N = 16,

H16=1+12+(13+14)+(15+⋯+18)+(19+⋯+116)≥1+4⋅12=3.H_{16} = 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \cdots + \frac{1}{8}\right) + \left(\frac{1}{9} + \cdots + \frac{1}{16}\right) \ge 1 + 4 \cdot \frac{1}{2} = 3.

The actual value is H16≈3.381H_{16} \approx 3.381. The bound is slow, though. To guarantee HN≥51H_N \ge 51 this way you need N=2100N = 2^{100} terms.

Comparison with p-series

Definition For a real number pp, the p-series is

∑n=1∞1np.\sum_{n=1}^{\infty} \frac{1}{n^p}.

It converges when p>1p > 1 and diverges when p≤1p \le 1. Two cases follow directly from comparing partial sums.

If p≤1p \le 1, then np≤nn^p \le n for n≥1n \ge 1, so 1np≥1n\frac{1}{n^p} \ge \frac{1}{n} and the partial sums are at least HNH_N, which tends to ∞\infty.

If p=2p = 2, then for n≥2n \ge 2 we have

1n2≤1n(n−1)=1n−1−1n,\frac{1}{n^2} \le \frac{1}{n(n - 1)} = \frac{1}{n - 1} - \frac{1}{n},

so by telescoping

∑n=1N1n2≤1+(1−1N)<2.\sum_{n=1}^{N} \frac{1}{n^2} \le 1 + \left(1 - \frac{1}{N}\right) < 2.

The partial sums increase and stay below 22, so they converge by the monotone convergence theorem. The general case p>1p > 1 is proved in the integral test.

In plain words A series of positive terms that is at least as big as the harmonic series diverges. One that sits below a convergent telescoping series converges.

Example The series ∑1n\sum \frac{1}{\sqrt{n}} has p=12≤1p = \frac{1}{2} \le 1. Each term 1n\frac{1}{\sqrt{n}} is at least 1n\frac{1}{n}, so its partial sums are at least HNH_N and it diverges. The series ∑1n2\sum \frac{1}{n^2} converges with a sum below 22. Its exact value turns out to be π26≈1.645\frac{\pi^2}{6} \approx 1.645.

Check yourself

  1. Find ∑n=1∞1n(n+2)\sum_{n=1}^{\infty} \frac{1}{n(n + 2)}.
Show the answer

Partial fractions give

1n(n+2)=12(1n−1n+2).\frac{1}{n(n + 2)} = \frac{1}{2}\left(\frac{1}{n} - \frac{1}{n + 2}\right).

Each term cancels with the one two places later, so for N≥2N \ge 2

sN=12(1+12−1N+1−1N+2)→12⋅32=34.s_N = \frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{N + 1} - \frac{1}{N + 2}\right) \to \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}.
  1. Find ∑n=1∞(1n−1n+1)\sum_{n=1}^{\infty} \left(\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}}\right).
Show the answer

It telescopes with bn=1nb_n = \frac{1}{\sqrt{n}}, so sN=1−1N+1→1s_N = 1 - \frac{1}{\sqrt{N + 1}} \to 1. The sum is 11.

  1. Does ∑n=1∞1n0.9\sum_{n=1}^{\infty} \frac{1}{n^{0.9}} converge?
Show the answer

No. Since 0.9≤10.9 \le 1, each term is at least 1n\frac{1}{n}, and the partial sums are at least the harmonic partial sums, which grow without bound.

How this lesson connects

It also uses partial fractions and harmonic sums.