Evaluate $\sum_{n=1}^{\infty}\frac{n}{3^n-1}.$ Evaluate $$\sum_{n=1}^{\infty}\frac{n}{3^n-1}.$$
Solution(Partial):
$|x|<1$
$$\sum_{n=1}^{\infty}\frac{nx^{n-1}}{3^n-1}=\frac{d}{dt}\sum_{n=1}^{\infty}\frac{t^n}{3^n-1}\big{|}_{t=x}$$
$$\sum_{n=1}^{\infty}\frac{t^n}{3^n-1}=\sum_{n=1}^{\infty} t^n\sum_{k=1}^{\infty}\frac{1}{3^{kn}}=\\ \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{t^n}{3^{kn}}=\sum_{k=1}^{\infty}\frac{\frac{t}{3^k}}{1-\frac{t}{3^k}}=\sum_{k=1}^{\infty}\frac{t}{3^k-t}$$
$$\frac{d}{dt}\sum_{k=1}^{\infty}\frac{t}{3^k-t}\big{|}_{t=x}=\sum_{k=1}^{\infty}\frac{1}{3^k-x}+\frac{x}{(3^k-x)^2}$$
as $x\to 1-$
$$\sum_{n=1}^{\infty}\frac{n}{3^n-1}=\sum_{k=1}^{\infty}\frac{1}{3^k-1}+\frac{1}{(3^k-1)^2}=\sum_{k=1}^{\infty}\frac{1}{3^k-1}+\sum_{k=1}^{\infty}\frac{1}{(3^k-1)^2}$$
$$\Rightarrow \sum_{n=1}^{\infty}\frac{n-1}{3^n-1}=\sum_{k=1}^{\infty}\frac{1}{(3^k-1)^2}$$
But $\frac{n-1}{3^n-1}>\frac{1}{(3^n-1)^2}$ for all $n\geq 2$
Next part of the solution after the answer of Professor Vector:
$$\sum_{k=1}^{\infty}\frac{1}{3^k-1}+\frac{1}{(3^k-1)^2}=\sum_{k=1}^{\infty} \frac{3^k}{(3^k-1)^2}=\sum_{k=1}^{\infty} \frac{1}{3^k-2+\frac{1}{3^k}}=\sum_{k=1}^{\infty} \frac{1}{(3^{k/2}-3^{-k/2})^2}\\=\frac{1}{4}\sum_{k=1}^{\infty} \frac{1}{\big{(}\frac{e^{\frac{\log 3}{2}k}-e^{-\frac{\log 3}{2}k}}{2}\big{)}^2}=\sum_{k=1}^{\infty}\frac{1}{4\sinh^2\big{(}\frac{\log 3}{2}k\big{)}}$$
I can not go more further after the last expression. I do not have much experience about Hyperbolic Trigonometric series.
The main post is here. The main poster is inactive for 8 months, so I had to post it here. I could not do it 5 years ago. I could not do it now sadly!
 A: The sum you have encountered is a special case of Ramanujan function $$P(q) =1-24\sum_{n=1}^{\infty}\frac{nq^{n}}{1-q^{n}}, |q|<1\tag{1}$$ and if your sum is $S$ then $P(1/3)=1-24S$. Unfortunately it is difficult to evaluate $P(q)$ in closed form for a general $q$. The following value is well known by the way: $$P(e^{-2\pi}) =\frac{3}{\pi}\tag{2}$$ and Ramanujan proved that $P(q) $ can always be expressed in a closed form (involving value of Gamma function namely $\Gamma(1/4)$) provided $q=\pm e^{-\pi\sqrt{r}} $ for any positive rational number $r$. But again evaluation of the closed form even in this case requires huge computational effort. For more details see this answer.
A: Even though $\frac{n-1}{3^n-1}>\frac{1}{(3^n-1)^2}$ for all $n\geq 2$, your sum
$$\sum_{n=1}^{\infty}\frac{n-1}{3^n-1}=\sum_{k=1}^{\infty}\frac{1}{(3^k-1)^2}$$
 starts with $n=1,$ so that first term may compensate the others. In fact, a quick numerical check shows that both sides essentially coincide (well, 0.26727973753486445 and 0.2672797375348645, let's not be too picky). Your derivation is absolutely correct, as far as I can see.
