Divergent series whose terms converge to zero

Question

I've just begun to teach my class series. We usually have workshops where they work in groups on a tougher problem, and I was thinking of asking them to come up with a divergent series whose terms converge to $0$. While $\sum 1/n$ is the prototypical example, we haven't done the integral test yet. Do you know of any simpler examples?

Answer 1

The simplest one I can think of:

$$ \begin{align} S& = 1 + \frac12 + \frac12 + \frac13 + \frac13 + \frac13 + \frac14 + \frac14 + \frac14 + \frac14 + \cdots \\ &= 1 + \,1\quad +\quad \,\, 1 \quad \quad\, +\quad \quad \,\, 1\, + \cdots \end{align} $$

In fact why don't you let your students make the suggestions?

Answer 2

The integral test isn't needed: we can argue by contradiction.

Let $S_n = \sum_{k=1}^n 1/k$. Then:

$$S_{2n} - S_n= \sum_{k=1}^{2n} 1/k - \sum_{k=1}^n1/k = \sum_{k=n+1}^{2n}1/k$$

If $(S_n)$ converges, then $\lim(S_{2n} - S_n) = 0$. However,

$$k \le 2n \implies \frac{1}{2n} \le \frac{1}{k}$$

Applying sums from $n+1$ till $2n$:

$$\frac12 \le S_{2n} - S_n$$

Taking $n \to \infty$, $\lim(S_{2n} - S_n) \ge 1/2$. Contradiction,

Answer 3

Consider the series $$ 1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\cdots $$ (here $\frac{1}{2^i}$ appears $2^i$ times). The terms of this series are going to zero, but each $2^i$ terms adds to $1$, so you're adding an infinite number of $1$'s.

(This is quite similar to the comparison test which proves $\sum\frac{1}{n}$ diverges).

Answer 4

The harmonic series is pretty simple and you do not need the integral test to show it diverges. Here is a short proof by contradiction:

Suppose $\sum^{\infty } _{n=1}\frac{1}{n}$ converges to a finite number $L$. Then,

$L=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\cdots +\cdots$ and this is greater than

$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{6}+\frac{1}{6}+\frac{1}{8}+\frac{1}{8}+\cdots$. Now group the terms:

$1+\frac{1}{2}+\left ( \frac{1}{4}+\frac{1}{4} \right )+\left ( \frac{1}{6}+\frac{1}{6} \right )+\left ( \frac{1}{8}+\frac{1}{8} \right )+\cdots $ and this latter item is

$1+\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots =L+\frac{1}{2}$ so what we've shown is

$L\geq L+\frac{1}{2}$ which is absurd. Hence, the series does not converge.

Answer 5

For the harmonic series, do like Cauchy's condensation test:

$\begin{array}\\ 2 \text{ terms } \ge 1/4 &\implies \sum \ge \frac12\\ 4 \text{ terms } \ge 1/8 &\implies \sum \ge \frac12\\ 8 \text{ terms } \ge 1/16 &\implies \sum \ge \frac12\\ ...\\ 2^n \text{ terms } \ge 1/2^{n+1} &\implies \sum \ge \frac12\\ \end{array} $

Answer 6

This question is from some time ago, but I just came across this thread and I know many who've had the same question. Below is an example that I rarely see despite its simplicity. We know

$$ \frac{n+1}{n}\ \to 1$$

and it's positive and decreasing. Therefore,

$$a_n= \log\bigg(\frac{n+1}{n} \bigg) \to 0$$

and $a_n$ is also positive and decreasing. But of course, \begin{align} \sum_{n=1}^{N}\log\bigg(\frac{n+1}{n} \bigg) & =\sum_{n=1}^{N}(\log(n+1) - \log(n) )\\ & = \log(N+1) \end{align}

So the series is telescoping and it diverges. The divergence of the harmonic series is still an important example for many other reasons, so it should still be taught.