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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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It is best not to simplify too much. Write down an expression for the $n$-th term, observe it is nowhere near $0$. Why is it nowhere near $0$? The terms go down and then up, reach a minimum around the middle. – André Nicolas Mar 23 '12 at 22:17
I agree with your observation that the terms are bouncing up and down and not close to zero it is an alternating series but it would be hard to definitively say it does n't converge unless we can use alternating series test which requires a partial sum. Also i really wish that some one could shed some light on how the product generated the second series as i have more such questions which are to follow in the exercises and i think it would be good for me to learn. – Comic Book Guy Mar 23 '12 at 22:25
The bizzare thing i found about abel's rule is that the formula appears to multiply the first expression of the first series with the nth expression of the second series, then 2nd term of the first series and n-1 th term of the second series forming a whole bunch of expressions under the summand and ofcourse n goes to infinity which under the scenario of this question changes a bunch of starting terms to 0. – Comic Book Guy Mar 23 '12 at 22:29
There is nothing bizarre about it. The rule you are using is just saying to change $\sum a_n$ to $\sum a_n \; x^n,$ then change $\sum b_n$ to $\sum b_n \; x^n.$ You then multiply the power series, which gives the rule you are using. Sometimes everything converges when $x=1,$ in which case you have your desired product. Usually, however, if $\sum a_n$ and $\sum b_n$ are not absolutely convergent but have decreasing (in absolute value) terms, the radius of convergence for all three resulting power series is $1,$ and the product series is divergent when $x=1.$ – Will Jagy Mar 23 '12 at 22:48
@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 $0$. – André Nicolas Mar 23 '12 at 23:52
up vote 5 down vote accepted

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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This is very cleverly done. thank you. – Comic Book Guy Mar 24 '12 at 11:08
Just wanted to add there is small typo the max value of $\dfrac {k^2}{4}$ is actually reached at $x=k/2$ – Comic Book Guy Mar 24 '12 at 19:40
@Hardy: Thank you very much, the typo may be small, but it could severely interfere iwth readability of the argument. Fixed. – André Nicolas Mar 24 '12 at 19:46
@André Nicolas This is rather off-topic, but would you mind having a look to this question of mine Divergent series expansion in Apéry's proof of the irrationality of ζ(2) and ζ(3)? – Américo Tavares Mar 24 '12 at 19:59
@Américo Tavares: I will try to look at it, but it will not be soon, Saturday/Sunday are not my own. – André Nicolas Mar 24 '12 at 20:16

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