Divergent series whose terms converge to zero 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?  
 A: 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?
A: 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,
A: 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).
A: 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. 
A: 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}
$
