Divergence of $\prod_{n=2}^\infty(1+(-1)^n/\sqrt n)$.

Question

Looking looking for a verification of my proof that the above product diverges.

$$\begin{align} \prod_{n=2}^\infty\left(1+\frac{(-1)^n}{\sqrt n}\right) & =\prod_{n=1}^\infty\left(1+\frac1{\sqrt {2n}}\right)\left(1-\frac1{\sqrt{2n+1}}\right)\\ & =\prod_{n=1}^\infty\left(1-\frac1{\sqrt{n_1}}+\frac1{\sqrt {2n}}-\frac1{\sqrt{2n(2n+1)}}\right)\\ & =\prod_{n=1}{\sqrt{2n(2n+1)}-\sqrt{2n}+\sqrt{2n+1}-1\over\sqrt{2n(2n+1)}}\\ & \ge\prod_{n=1}^\infty{\sqrt{2n(2n+1)+2n+1}-\sqrt{2n}-1\over\sqrt{2n(2n+1)}},\quad\sqrt x+\sqrt y\ge\sqrt{x+y}\\ & = \prod_{n=1}^\infty{2n-\sqrt{2n}\over\sqrt{2n(2n+1)}}\\ & = \prod_{n=1}^\infty{2n-1\over\sqrt{2n+1}} \end{align}$$

This last product diverges since

$$\lim_{n\to\infty}{2n-1\over\sqrt{2n+1}}=\infty.$$

I'm suspicious because $$\lim_{n\to\infty}\left(1+{(-1)^n\over\sqrt n}\right)=1.$$

Answer 1

Your calculation is wrong, but the conclusion is correct.

$$ \left( 1 + \dfrac{1}{\sqrt{2n}}\right) \left( 1 - \dfrac{1}{\sqrt{2n+1}}\right) = 1 - \dfrac{1}{2n} + O(n^{-3/2})$$

The infinite product diverges, but to $0$, not $+\infty$.

Answer 2

When you render

$\prod_{n=1}^\infty{2n-\sqrt{2n}\over\sqrt{2n(2n+1)}}\\ = \prod_{n=1}^\infty{2n-1\over\sqrt{2n+1}}$

that is wrong. Properly,

$\prod_{n=1}^\infty{2n-\sqrt{2n}\over\sqrt{2n(2n+1)}}\\ = \prod_{n=1}^\infty{\color{blue}{\sqrt{2n}}-1\over\sqrt{2n+1}}.$

From there:

$\prod_{n=1}^\infty\frac{\sqrt{2n}-1}{\sqrt{2n+1}}=\prod_{n=1}^\infty\frac{2n-1}{(\sqrt{2n+1})(\sqrt{2n}+1)}\le\prod_{n=1}^\infty\frac{2n-1}{2n+1},$

where in the last step we use $\sqrt{a}+\sqrt{b}>\sqrt{a+b}$ for all positive $a,b$ (prove this by squaring both sides). Then the last product telescopes to $1/(2m+1)$ as $m\to\infty$, hence zero.

Answer 3

I would like for some approval to my answer, please.

Show that $$\prod_2^\infty(1+\frac{(-1)^k}{\sqrt k})$$ diverges even though$$\sum_2^\infty \frac{(-1)^k}{\sqrt k}$$ converges.

$\prod_2^\infty(1+\frac{(-1)^k}{\sqrt k})$ diverges if and only if $\sum _2^\infty\log(1+\frac{(-1)^k}{\sqrt k})$ diverges.

Indeed, by Leibnitz' converges test $$\sum_2^\infty \frac{(-1)^k}{\sqrt k}$$ converges. Thus, $\sum _2^\infty\log(1+\frac{(-1)^k}{\sqrt k})$ diverges if and only if $\sum _2^\infty(\log(1+\frac{(-1)^k}{\sqrt k})+\frac{(-1)^k}{\sqrt k})$ diverges.

We observe that for all $k\geq 2$, $$ |\log(1+\frac{(-1)^k)}{\sqrt k})-\frac{(-1)^k)}{\sqrt k}|\\ =|\frac{(-1)^{2k}}{2\sqrt k^2}- \frac{(-1)^{3k}}{3\sqrt k^3} + \frac{(-1)^{4k}}{4\sqrt k^4} -\frac{(-1)^{5k}}{5\sqrt k^5}+-...| $$ But since we know that the last series is convergent (to $|\log(1+\frac{(-1)^k)}{\sqrt k})-\frac{(-1)^k}{\sqrt k}|$), then it is equal to $$ =|(\frac{(-1)^{2k}}{2\sqrt k^2}- \frac{(-1)^{3k}}{3\sqrt k^3}) + (\frac{(-1)^{4k}}{4\sqrt k^4} -\frac{(-1)^{5k}}{5\sqrt k^5})+-...|=:A $$ but every addend in this series is positive because for all $l\in\mathbb{N}$: $$ (-1)^{2l}(2l+1)\sqrt k^{2l+1}>(-1)^{2l+1}\cdot 2l\cdot\sqrt k^{2l} $$ ($\sqrt k>1$). Thus, the previous series $A$ obtains: $$ A=(\frac{(-1)^{2k}}{2\sqrt k^2}- \frac{(-1)^{3k}}{3\sqrt k^3}) + (\frac{(-1)^{4k}}{4\sqrt k^4} -\frac{(-1)^{5k}}{5\sqrt k^5})+-... \\ >\frac{(-1)^{2k}}{2\sqrt k^2}- \frac{(-1)^{3k}}{3\sqrt k^3}\geq \frac{1}{2k}-\frac{1}{3k^{1.5}} $$ But because $\sum \frac{1}{2k}$ diverges and $\sum\frac{1}{3k^{1.5}}$ converges then $\sum(\frac{1}{2k}-\frac{1}{3k^{1.5}})$ diverges then $A$ diverges.