# 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.

• Logarithm? ${}$ Mar 2, 2016 at 2:04
• Can you elaborate? These are complex sequences.
– user228960
Mar 2, 2016 at 2:22

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}

• Is there something wrong with this answer? If not, then I assume the downvote is making a statement, but without a comment, that statement is not clear.
– robjohn
May 25, 2019 at 1:44
• can you elaborate more on why the infinite product of your first example diverges? No matter how I look at it it seems to have to converge. Sorry for replying a few years later. Thanks in advance :) @robjohn Apr 9, 2022 at 4:14
• @Explorer1234: see Infinite Product: "... the infinite product $$\prod_{n=1}^\infty a_n=a_1a_2a_3\cdots$$ is defined to be the limit of the partial products $a_1a_2\dots a_n$ as $n$ increases without bound. The product is said to converge when the limit exists and is not zero. Otherwise the product is said to diverge."
– robjohn
Apr 9, 2022 at 7:53
• isn't the partial products $1/(n+1)$ in this case? So wouldn't that converge to 0 as $n$ tends to infinity? Or am I missing something? @robjohn Apr 9, 2022 at 7:55
• The fact that $\prod\limits_{n=1}^\infty a_n=0$ and $\prod\limits_{n=1}^\infty \frac1{a_n}=\infty$ are equivalent to $\sum\limits_{n=1}^\infty b_n=-\infty$ and $\sum\limits_{n=1}^\infty -b_n=\infty$ because $\lim\limits_{x\to0^+}\log(x)=-\infty$.
– robjohn
Apr 9, 2022 at 8:07

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.

• Should $P_{2m}$ be $\prod_{n=1}^{m}\left(1- \frac{1}{(n+1)^2} \right)$ instead? Jun 29, 2020 at 23:33