I have a question about the convergence of infinite product.

In real mathematics analysis page 198 question 63,

part a) $a_k$=$(-1)^k$/${\sqrt{k}}$, and we need to show series $\sum\limits_{n=1}^\infty a_k$ converges, but $\prod\limits_{n=1}^\infty(1+a_k)$ diverges.

The convergence part is true by alternating series test. While for the product part, if $k$ is $1$, $a_k$ is $-1$, and $1+a_k$ is $0$, which means any partial product or even the total infinite product will always be $0$. So does this zero value implies the divergence here?

part b) uses $b_k$=$e_k$/$k$+$(-1)^k/{\sqrt{k}}$, where $e_k$ is $0$ when $k$ is odd, $1$ when $k$ is even. So I think if this is how the $b_k$ is defined, the infinite product $\prod\limits_{k=1}^\infty(1+b_k)$ will also be $0$, since if $k$ is $1$, $e_1$ is $0$, and $b_1$=$-1$. $1+b_1$=$0$. But part b is to show series diverges, while infinite product converges.

So can someone tell me where is my mistake? Thx!

Here is the question from the textbook.

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  • $\begingroup$ Most likely, there will be typo! The product might be from $n=2$ (and $k=2$). Otherwise, as you mentioned the product converges to 0. $\endgroup$ – chandu1729 Jan 25 '15 at 19:39
  • $\begingroup$ Yes, a previous question says the product starts from 2.... math.stackexchange.com/questions/294148/… thx! $\endgroup$ – Toad Jiang Jan 25 '15 at 20:35

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