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I am trying to show that the series $\dfrac {1} {\sqrt {1}}-\dfrac {1} {\sqrt {2}}+\dfrac {1} {\sqrt {3}}-\ldots $ is convergent, but that its square (formed by Abel's rule) $\dfrac {1} {1}-\dfrac {2} {\sqrt {2}}+\left( \dfrac {2} {\sqrt {3}}+\dfrac {1} {2}\right) -\left( \dfrac {2} {\sqrt {4}}+\dfrac {2} {\sqrt {6}}\right) -\ldots $ is divergent.

Now in order to verify the author's claims. I observed that the first series $\sum _{n=1}^{n=\infty }\dfrac {\left( -1\right) ^{n+1}} {\sqrt {n}}$ converges by Leibniz test or alternating series test, but i am having a hard time firstly verifying the product would create the second series and then finding a generic pattern in it to be able to test for convergence.

Any help would be much appreciated.

Edit I think the product by abels' rule should yield. $\sum _{n=1}^{n=\infty }\frac {\left( -1\right) ^{n+1}} {\sqrt {n}}.\sum _{n=1}^{n=\infty }\frac {\left( -1\right) ^{n+1}} {\sqrt {n}} = \lim _{p\rightarrow \infty }\sum _{n=1}^{n=\infty } \frac {\left( -1\right) ^{n+1}} {\sqrt {n}}.\frac {\left( -1\right) ^{p-(n+1)}} {\sqrt {p - n}}= \lim _{p\rightarrow \infty }\sum _{n=1}^{n=\infty }\frac {\left( -1\right) ^{p}} {\sqrt {p - n}\sqrt {n}}$

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    @Hardy: I hope that the answer I posted is enough to clear things up. If you have troubles, write down explicitly the expression that gives the seventh term, say, and maybe the eighth. I know and you know that the first and last part are equal, and so on, but do not simplify. You know how to do this, you did it already, but simplified. Do not use summation notation, you have incomplete control of it. (The stuff added in the last edit is not right.) Then the calculation I did may make sense. Remember that alternation of signs is useless if the terms don't go to $0$.2012-03-23

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Extended hint: Let $b_n$ be the absolute value of the $n$-th summand in the Abel "square." Then $b_n=a_1a_n+a_2a_{n-1}+a_3a_{n-2}+\cdots +a_na_1,$ where $a_i=\frac{1}{\sqrt{i}}$. We will produce a lower bound for $b_n$.

Note that the function $f(x)=x(k-x)$, where $0\le x\le k$, reaches a maximum value of $k^2/4$ at $x=k/2$.

So the function $i(n+1-i)$, where $1\le i\le n$, is always $\le (n+1)^2/4$. The square root of this is always $\le (n+1)/2$. Thus $a_ia_{n+1-i}=\frac{1}{\sqrt{i}}\cdot\frac{1}{\sqrt{n+1-i}} \ge \frac{2}{n+1}.$

We conclude that every one of the $n$ terms whose sum is $b_n$ is $\ge \frac{2}{n+1}$, and therefore the sum $b_n$ of these $n$ terms is $\ge \frac{2n}{n+1}$. This is always $\ge 1$, which forces divergence.

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    @André Nicolas: Many thanks!2012-03-24