I've been stuck on this problem over the weekend so I decided to ask for some direction. The problem reads:

"The multiplication theorem for series requires that the two series be absolutely convergent; if this condition is not met, their product may be divergent. Show that the series $\sum_{0}^{\infty}\frac{(-1)^i}{\sqrt{i + 1}}$ gives an example: it is conditionally convergent, but its product with itself is divergent. Estimate the size of the odd terms $c_{2n+1} $ in the product."

I shall refer to $\frac{(-1)^i}{\sqrt{i + 1}}$ as $a_n$. I would imagine this converges because it's an alternating sum. I was thinking of rearranging the terms of the series so that I have all $\sum a_n$= $\sum (a_{n}^{+} - a_{n}^{-})$. However, this is where I am stuck. I am not sure where to take it from there. I was thinking of breaking up the sum into two different series such that $\sum (-1)^i$ and $\sum \frac{1}{\sqrt{i + 1}}$, the first sum would be -1 + 1 + -1 + 1 +.... Depending on how you rearrange this one, it could be -1, 0, or 1 I think.

I think the $\sum \frac{1}{\sqrt{i + 1}}$ diverges. I know the sum of $\frac{1}{n}$ does diverge. I was relating it to that.

However, I fear I am over thinking the whole problem. I need help.

Thanks you for taking the time to read this post and thanks in advanced to those who comment.

  • $\begingroup$ they told you it is conditionally convergent already. All you have to do is actually write the Cauchy product and show it diverges. $\endgroup$ – oldrinb Oct 8 '13 at 4:24
  • $\begingroup$ So I was over thinking it then. $\endgroup$ – Kevin_H Oct 8 '13 at 19:39

The series $$ \sum_{k=0}^\infty\frac{(-1)^k}{\sqrt {k+1}} $$ is convergent by the Leibnitz criterion, because it is alternating and the general term goes to zero. It is not absolutely convergent because $$ \sum_{k=0}^\infty\frac{1}{\sqrt {k+1}}\geq\sum_{k=1}^\infty\frac1{k+1}=+\infty. $$ Finally, $$ \sum_{k=0}^\infty\frac{(-1)^k}{\sqrt {k+1}}\,\cdot\,\sum_{k=0}^\infty\frac{(-1)^k}{\sqrt {k+1}}=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{(-1)^k}{\sqrt {k+1}}\,\frac{(-1)^{n-k}}{\sqrt{n+1- k}}\right)\\=\sum_{n=1}^\infty\,(-1)^n\sum_{k=0}^n\frac1{\sqrt{(k+1)(n+1-k)}} $$ Now note that the expression $(k+1)(n+1-k)$ is a downwards parabola on $k$, and so it has a maximum. Elementary calculus shows this maximum to occurs at $k=n/2$, so $$ (k+1)(n+1-k)\leq(n/2+1)(n/2+1)=(n+1)^2/4. $$ So $$ \sum_{k=0}^n\frac1{\sqrt{(k+1)(n+1-k)}}\geq\sum_{k=0}^n\frac2{\sqrt{(n+1)^2}}=\sum_{k=0}^n\frac2{n+1}=2. $$ The general term does not go to zero, and so the product series is divergent.


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