Let $a_{2n-1}=-1/\sqrt{n}$ for $n=1,2,\dots$ Show that $\prod (1+a_n)$ converges but that $\sum a_n$ diverges.

Question

Let $a_{2n-1}=-1/\sqrt{n}$, $a_{2n}=1/\sqrt{n}+1/n$ for $n=1,2,\dots$ Show that $\prod (1+a_n)$ converges but that $\sum a_n$ diverges.

What I have found so far is that $\prod_{k=2}^{2n} a_n$=$3(1-\frac{1}{2\sqrt{2}})\cdots (1-\frac{1}{n\sqrt{n}})$

and

$\prod_{k=2}^{2n+1} a_n$=$3(1-\frac{1}{2\sqrt{2}})\cdots (1-\frac{1}{n\sqrt{n}})(1-\frac{1}{\sqrt{n+1}})$

I'm considering using the theorem that if each $a_n \ge 0$, then the product $\prod(1-a_n)$ converges if and only if, the series $\sum a_n$ converges. So since $\sum \frac{1}{n\sqrt{n}}$ converges, I think the above product converges as well, but I can't use this theorem right now because of the odd partial products. How can I resolve this problem? Also, I can clearly see that $\sum a_n$ diverges, but how can I prove this rigorously?

I would greatly appreciate some help.

Answer 1

It seems the following.

Also, I can clearly see that $\sum a_n$ diverges, but how can I prove this rigorously?

Assume that $\sum a_n$ converges to a number $A$. Then $\sum_{n=1}^{2m} a_n$ converges to $A$ too, but $\sum_{n=1}^{2m} a_n$ tends to $+\infty$.

I'm considering using the theorem that if each $a_n \ge 0$, then the product $\prod(1-a_n)$ converges if and only if, the series $\sum a_n$ converges. So since $\sum \frac{1}{n\sqrt{n}}$ converges, I think the above product converges as well, but I can't use this theorem right now because of the odd partial products. How can I resolve this problem?

So we have that the sequence $b_n=\prod_{k=2}^{2n} (1+a_n)$ converges to a number $A$. Then a sequence $\prod_{k=2}^{2n+1} (1+a_n)=b_n (1-\frac{1}{\sqrt{n+1}})$ converges to $A$ too, so the product $\prod_{k=2}^\infty (1+a_n)$ converges to $A$ too.

Answer 2

Taking your definition at face value, the first terms in your infinite sum are $$\sum_{n=1}^{\infty}a_{n} = \frac{-1}{\sqrt{1}} + \frac{1}{\sqrt{1}} + \frac{1}{1} + \frac{-1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + \frac{1}{2} + \cdots,$$ which leads to the conclusion that your infinite sum is just the harmonic series, which diverges by the integral test.

As $a_{1}=-1$ it is clear that $$\prod_{n=1}^{\infty}(1+a_{n})=0.$$ If we consider instead $$L=\prod_{n=3}^{\infty}(1+a_{n}) = \prod_{k=2}^{\infty}\left(1-\frac{1}{\sqrt{k}}\right)\left(1+\frac{1}{\sqrt{k}}+\frac{1}{k}\right) = \prod_{k=2}^{\infty}\left(1-\frac{1}{k^{3/2}}\right)$$ we can look at $$\ln(L) = \sum_{k=2}^{\infty}\ln\left(1-\frac{1}{k^{3/2}}\right),$$ which is a sum consisting entirely of negative terms. As the terms do not alternate in sign we can use a limit comparison test against $$\sum_{k=2}^{\infty}\frac{1}{k^{3/2}}$$ to show that the infinite sum, and hence the infinite product it came from, converges.