4
$\begingroup$

Using the Mertens' theorem for Cauchy products we know that the Cauchy product of series $$ \sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n+1}\qquad\text{and}\qquad\sum_{n=0}^{\infty}\frac{1}{3^n} $$ does converge. But how can we try to find the value of this sum?

What I only know is that we can write that this sum equals to $$ S=\sum_{n=0}^{\infty}c_n\qquad\text{where}\qquad c_n=\sum_{k=0}^{n}\frac{1}{3^{n-k}}\frac{(-1)^{k+1}}{k+1} $$ but I'm not able to find any way to calculate the sum. Any hints?

$\endgroup$
3
  • 5
    $\begingroup$ If I understand correctly, by MERTENS' Theorem, $S$ is equal to the product of the of the initial sums, right? The first is $-\ln(2)$ and the second is $\frac{3}{2}$ by the geometric series. $\endgroup$ Commented Dec 23, 2018 at 12:38
  • 1
    $\begingroup$ Yes but how to find it using only the definition of Cauchy product of series? $\endgroup$
    – avan1235
    Commented Dec 23, 2018 at 12:44
  • $\begingroup$ If you were presented with this sum as a $S$ and asked to evaluate it. I would think that decomposing it (via cauchy product) would be the easiest way of coming to a conclusion. $\endgroup$
    – Mason
    Commented Dec 23, 2018 at 16:11

1 Answer 1

3
$\begingroup$

Here is my "lengthy" solution (elaborating @ViktorGlombik comment). Let's start with your sum:

$$S=\sum_{n=0}^\infty\frac{1}{3^n}\sum_{k=0}^n\frac{1}{3^{-k}}\frac{(-1)^{k+1}}{k+1}$$

where

$$\sum_{k=0}^n\frac{1}{3^{-k}}\frac{(-1)^{k+1}}{k+1}=(-3)^{n+1}\Phi(-3,1,n+2)-\frac{2}{3}\log(2)$$

where $\Phi(a,b,c)$ is so-called Lerch transcendent.

Also there is one useful identity: $\Phi(-3, 1, n+2) = \frac{1}{\Gamma(1)}\int_0^\infty\frac{e^{-(n+2)t}}{1+3e^{-t}}dt$.

By substitution $u=1+3e^{-t}$ (and $\Gamma(1) = 1$), this integral leads to

$$\Phi(-3, 1, n+2) = \frac{1}{3^{n+2}}\int_1^4 \frac{(u-1)^{n+1}}{u} du.$$

Plugging it back (observe $u\in(1,4)$ for the convergence of the geometric series), we obtain: $$S=\sum_{n=0}^\infty\frac{1}{3^n}\left[(-3)^{n+1}\frac{1}{3^{n+2}}\int_1^4\frac{(u-1)^{n+1}}{u}du-\frac{2}{3}\log(2)\right]=$$ $$=\sum_{n=0}^\infty\frac{1}{3^{n+1}}\int_1^4\frac{(1-u)^{n+1}}{u}du-\log(2)=$$ $$=\int_1^4\frac{1}{u}\sum_{n=0}^{\infty}\left(\frac{1-u}{3}\right)^{n+1}du - \log(2)=$$ $$=\int_1^4\frac{1}{u}\frac{1-u}{u+2}du-\log(2)=-\frac{3}{2}\log(2)+\log(2)-\log(2)=-\frac{3}{2}\log(2).$$

So yes, indeed the sum converges to $S=-\frac{3}{2}\log(2)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .