Unit 7, lesson 96

Series Theorems

The basic facts every series obeys, including the nth term test for divergence, adding convergent series, ignoring finitely many terms, and the Cauchy criterion.

  • Theory part 191
  • Practice part 192
  • Infinite Series

Before reaching for any special test, a few general facts already settle many series. Some series obviously cannot add up because their terms do not even get small. Others can be split into pieces you already know. These rules are short, but you will use them in almost every problem from here on.

Key terms

The nth term test for divergence

Definition If ∑n=1∞an\sum_{n=1}^{\infty} a_n converges, then lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0. Equivalently, if lim⁡n→∞an\lim_{n \to \infty} a_n does not exist or exists and is not 00, then ∑an\sum a_n diverges.

The proof is one line. If sN→Ss_N \to S, then also sN−1→Ss_{N-1} \to S, so

aN=sN−sN−1→S−S=0.a_N = s_N - s_{N-1} \to S - S = 0.

In plain words You cannot add infinitely many numbers and get a finite total unless the numbers eventually become tiny.

Example For ∑n2n+1\sum \frac{n}{2n + 1} the terms tend to 12≠0\frac{1}{2} \ne 0, so the series diverges. For ∑(−1)n\sum (-1)^n the terms have no limit at all, so it diverges too.

Terms tending to zero is not enough

Definition The condition an→0a_n \to 0 is necessary for convergence of ∑an\sum a_n but not sufficient. There are series with an→0a_n \to 0 that diverge. So the nth term test can only prove divergence. It never proves convergence.

In plain words If the terms do not go to zero, you are done and the series diverges. If they do go to zero, you have learned nothing yet.

Example The terms ln⁡(1+1n)\ln\left(1 + \frac{1}{n}\right) tend to ln⁡1=0\ln 1 = 0. Yet the partial sums telescope to sN=ln⁡(N+1)→∞s_N = \ln(N + 1) \to \infty, so ∑ln⁡(1+1n)\sum \ln\left(1 + \frac{1}{n}\right) diverges. The harmonic series ∑1n\sum \frac{1}{n} is another example.

Linearity of convergent series

Definition If ∑an=A\sum a_n = A and ∑bn=B\sum b_n = B both converge and cc is a real number, then

∑n=1∞(an+bn)=A+B,∑n=1∞(an−bn)=A−B,∑n=1∞c an=cA.\sum_{n=1}^{\infty} (a_n + b_n) = A + B, \qquad \sum_{n=1}^{\infty} (a_n - b_n) = A - B, \qquad \sum_{n=1}^{\infty} c\,a_n = cA.

If ∑an\sum a_n converges and ∑bn\sum b_n diverges, then ∑(an+bn)\sum (a_n + b_n) diverges, and for c≠0c \ne 0 the series ∑c bn\sum c\,b_n diverges.

In plain words Convergent series can be added, subtracted and scaled term by term, just like finite sums. This follows from the limit laws applied to partial sums.

Example Using ∑n=1∞12n=1\sum_{n=1}^{\infty} \frac{1}{2^n} = 1 and ∑n=1∞1n(n+1)=1\sum_{n=1}^{\infty} \frac{1}{n(n + 1)} = 1,

∑n=1∞(32n+4n(n+1))=3⋅1+4⋅1=7.\sum_{n=1}^{\infty} \left(\frac{3}{2^n} + \frac{4}{n(n + 1)}\right) = 3 \cdot 1 + 4 \cdot 1 = 7.

Be careful with two divergent series. With an=1a_n = 1 and bn=−1b_n = -1 both diverge, but ∑(an+bn)=∑0=0\sum (a_n + b_n) = \sum 0 = 0 converges. The rule says nothing in that case.

Finite changes

Definition Adding, deleting or changing finitely many terms of a series does not affect whether it converges, although it can change the sum. In particular, for each integer m≥1m \ge 1, the series ∑n=1∞an\sum_{n=1}^{\infty} a_n converges if and only if the tail ∑n=m∞an\sum_{n=m}^{\infty} a_n converges, and then

∑n=1∞an=∑n=1m−1an+∑n=m∞an.\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{m-1} a_n + \sum_{n=m}^{\infty} a_n.

In plain words Convergence depends only on the long run. The first hundred or the first million terms add a fixed finite number to every partial sum and nothing more.

Example Since ∑n=1∞12n=1\sum_{n=1}^{\infty} \frac{1}{2^n} = 1 and the first three terms add to 78\frac{7}{8}, the tail is

∑n=4∞12n=1−78=18.\sum_{n=4}^{\infty} \frac{1}{2^n} = 1 - \frac{7}{8} = \frac{1}{8}.

If you replace those first three terms by 100100, 200200 and 300300, the new series still converges, now to 600+18600 + \frac{1}{8}.

Cauchy criterion

Definition The series ∑an\sum a_n converges if and only if for every ε>0\varepsilon > 0 there is an integer NN such that

∣an+1+an+2+⋯+am∣<εfor all m>n≥N.\lvert a_{n+1} + a_{n+2} + \cdots + a_m \rvert < \varepsilon \quad \text{for all } m > n \ge N.

This is the Cauchy criterion for the sequence of partial sums, since the sum in the bars equals sm−sns_m - s_n.

In plain words A series converges exactly when every block of terms far enough out adds up to something tiny, no matter how long the block is. You can check this without knowing the sum.

Example For the harmonic series the block from n+1n + 1 to 2n2n has nn terms, each at least 12n\frac{1}{2n}, so

1n+1+1n+2+⋯+12n≥n⋅12n=12\frac{1}{n + 1} + \frac{1}{n + 2} + \cdots + \frac{1}{2n} \ge n \cdot \frac{1}{2n} = \frac{1}{2}

for every nn. The criterion fails with ε=12\varepsilon = \frac{1}{2}, so the harmonic series diverges. For ∑12k\sum \frac{1}{2^k}, on the other hand, every block after nn adds to less than 12n\frac{1}{2^n}, which is below any ε>0\varepsilon > 0 once nn is large.

Check yourself

  1. Does ∑n=1∞n2n2+1\sum_{n=1}^{\infty} \frac{n^2}{n^2 + 1} converge?
Show the answer

No. The terms tend to 1≠01 \ne 0, so the nth term test shows divergence.

  1. Find ∑n=1∞(23n−1n(n+1))\sum_{n=1}^{\infty} \left(\frac{2}{3^n} - \frac{1}{n(n + 1)}\right).
Show the answer

Both pieces converge. The first is geometric with first term 23\frac{2}{3} and ratio 13\frac{1}{3}, and the second telescopes, so

∑n=1∞23n=2/31−1/3=1,∑n=1∞1n(n+1)=1.\sum_{n=1}^{\infty} \frac{2}{3^n} = \frac{2/3}{1 - 1/3} = 1, \qquad \sum_{n=1}^{\infty} \frac{1}{n(n + 1)} = 1.

By linearity the sum is 1−1=01 - 1 = 0.

  1. A classmate says ∑1n2\sum \frac{1}{n^2} converges because 1n2→0\frac{1}{n^2} \to 0. What is wrong with the reasoning?
Show the answer

The conclusion is true but the reason is not. Terms tending to 00 never prove convergence, as the harmonic series shows. A valid reason is the comparison with a telescoping series.

How this lesson connects