Examples where series converges but product diverges and vice versa Our professor gives us the following ungraded exercise for our analytic number theory class:
Let $ E $ be a set with one element. Suppose $ (b_n) $ is a sequence with $ |b_n| \leq \lambda < 1 $, and let $ a_n = 1 + b_n $.
1) Find $ (b_n) $ so that $ \sum b_n $ converges, but $ \prod a_n $ diverges.
2) Find $ (b_n) $ so that $ \prod a_n $ converges, but $ \sum b_n $ diverges.
I am not sure how to do this problem. Any help is appreciated.
 A: Suppose
$$
b_{2n-1}=\frac1{\sqrt{n+1}}\quad\text{and}\quad b_{2n}=-\frac1{\sqrt{n+1}}
$$
Then, Dirichlet's Convergence Test guarantees convergence of the series. In fact,
$$
\sum_{k=1}^\infty b_k=0
$$
However,
$$
\left(1+b_{2n-1}\right)\left(1+b_{2n}\right)=1-\frac1{n+1}
$$
Therefore,
$$
\prod_{k=1}^{2n}\left(1+b_k\right)=\frac1{n+1}
$$

$$
a_{2n-1}=\frac{\sqrt{n}+1}{\sqrt{n}}\quad\text{and}\quad a_{2n}=\frac{\sqrt{n}}{\sqrt{n}+1}
$$
Then, Dirichlet's Convergence Test guarantees convergence of the log of the series. In fact,
$$
\prod_{k=1}^\infty a_k=1
$$
However,
$$
\begin{align}
\left(a_{2k-1}-1\right)+\left(a_{2k}-1\right)
&=\frac1{\sqrt{n}}-\frac1{\sqrt{n}+1}\\
&=\frac1{n+\sqrt{n}}
\end{align}
$$
Therefore,
$$
\begin{align}
\sum_{k=1}^{2n}\left(a_k-1\right)
&=\sum_{k=1}^n\frac1{k+\sqrt{k}}\\
&\ge\int_1^{n+1}\frac{\mathrm{d}x}{x+\sqrt{x}}\\
&=2\log\left(\frac{\sqrt{n+1}+1}2\right)
\end{align}
$$
A: In the real setting, the product $\prod (1 + b_n)$ and the series $\sum b_n$ converge or diverge together for the separate cases where $b_n \geqslant 0$ and $-1 < b_n < 0$ for all $n$. 
Even in the real setting, we can find a sequence $(b_n)$ such that the product and series are not simultaneously convergent if we allow $b_n$ to change sign.
For example, let 
$$b_{2n-1}= -\frac{1}{\sqrt{n+1}}, \\ b_{2n}= \frac{1}{\sqrt{n+1}}+\frac{1}{n+1}+  \frac{1}{\sqrt{n+1}(n+1)}.  $$
We have 
$$(1 + b_{2n-1})(1 + b_{2n})= 1 - \frac{1}{(n+1)^2},$$
and
$$P_{2m}= \prod_{n=1}^{m}\left(1+ \frac{1}{(n+1)^2} \right), \\ P_{2m+1}= P_{2m}\left(1 - \frac{1}{\sqrt{m+2}}\right).$$
The partial product $P_{2m}$ converges since the series $\sum(n+1)^{-2}$ converges.
Furthermore
$$\lim_{m \to \infty} P_{2m+1} = \lim_{m \to \infty} P_{2m}\lim_{m \to \infty}\left(1 - \frac{1}{\sqrt{m+2}}\right) = \lim_{m \to \infty} P_{2m}.$$
This shows that the product $\prod (1 + b_n)$ is convergent as partial products with both odd and even number of factors converge to the same limit.
However, the series $\sum b_n$ diverges analogously to the harmonic series.
