In Apostol's Mathematical Analysis there is the following exercise. Show that the Cauchy product of $$\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}$$ with itself is the series $$2\sum_{n=1}^\infty{(-1)^{n+1}\over n+1}\left(1+\frac12+\cdots+\frac1n\right).$$

This doesn't seem to be true though. Writing out the terms, $$\left(\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}\right)\left(\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}\right)=\frac11-\left(\frac11\frac12+\frac12\frac11\right)+\left(\frac11\frac13+\frac12\frac12+\frac13\frac11\right)+\cdots$$ and $$2\sum_{n=1}^\infty{(-1)^{n+1}\over n+1}\sum_{k=1}^n\frac1k=2\left(-\frac12+\frac13\left(1+\frac12\right)-\frac14\left(1+\frac12+\frac13\right)+\cdots\right).$$

I can see that in the first sum, all terms of the form $\frac1n\frac1m$ with $m\ne n$ are doubled, but the terms $\frac1{n^2}$ don't seem to be accounted for. It would be nice if the sum of the missing terms was $0$, but in fact what we seems to have is $$\begin{align} \left(\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}\right)\left(\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}\right) & =2\sum_{n=1}^\infty{(-1)^{n+1}\over n+1}\sum_{k=1}^n\frac1k+\sum_{n=1}^\infty\frac1{n^2}\\ & =2\sum_{n=1}^\infty{(-1)^{n+1}\over n+1}\sum_{k=1}^n\frac1k+{\pi^2\over 6}. \end{align}$$

Any help is appreciated.


The Cauchy product will be

$$\sum_{n = 0}^\infty \sum_{k = 0}^n \frac{(-1)^{k+1}}{k+1} \frac{(-1)^{n-k+1}}{n-k+1} = \sum_{n = 0}^\infty \sum_{k = 0}^n \frac{(-1)^n}{(k+1)(n-k+1)}.$$


\begin{align}\sum_{k = 0}^n \frac{(-1)^n}{(k+1)(n-k+1)} &= (-1)^n \sum_{k = 0}^n \frac{1}{n+2}\left(\frac{1}{n-k+1} + \frac{1}{k+1}\right)\\ &= \frac{(-1)^n}{n+2}\left\{\sum_{k = 0}^n \frac{1}{n-k+1} + \sum_{k = 0}^n \frac{1}{k+1}\right\}\\ &= \frac{(-1)^n}{n+2}\left\{\sum_{j = 0}^n \frac{1}{j+1} + \sum_{k = 0}^n \frac{1}{k+1}\right\} \quad (j = n - k)\\ &= \frac{2(-1)^n}{n+2}\sum_{k = 0}^n \frac{1}{k+1}\\ \end{align}

So the product is reduced to

$$2\sum_{n = 0}^\infty \frac{(-1)^n}{n+2}\sum_{k = 0}^n \frac{1}{k+1} = 2\sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n+1}\left(1 + \frac{1}{2} + \cdots + \frac{1}{n}\right).$$

  • $\begingroup$ Partial fractions. Jeez why didn't I think of that. :) $\endgroup$ – Tim Raczkowski Feb 17 '15 at 23:42

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