I am trying to calculate the infinite sum of a geometric sequence times a harmonic sequence, so a series formatted in this way:
$$ \sum_{n=1}^\infty \frac{1}{n}\cdot q^n $$
I simply know that it is possible because Wolfram gave me a precise value when I was checking if my answer about series convergence was correct.
My specific case was $ \sum_{n=1}^\infty \frac{1}{n}\cdot(\frac{3}{5})^n $, and Wolfram solved it as $ \ln\frac{5}{2} $, but clearly I'd prefer a more generic explanation, thanks in advance!
Let $S$ denote the infinite series.
$S =\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n} q^n =\displaystyle\sum_{n=0}^{\infty} \dfrac{1}{n+1} q^{n+1}$.
Noting that$ \dfrac{1}{n+1} q^{n+1}$ is the integral w.r.t $q$ of $q^n$, we have: $S = \displaystyle \sum_{n=0}^{\infty} \displaystyle \int q^n dq $.
Interchanging the integral and sum signs, we get : $S = \displaystyle \int \Big (\displaystyle \sum_{n=0}^{\infty} q^n\Big) dq$.
Note that the series is geometric, it remains to compute its integral, which is not hard, and we are done.