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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!

EditPiAf
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MAN-MADE
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  • There's something wrong...your solution refers to a function series but the first one is a numerical series – Matheman Jun 15 '17 at 07:06
  • @Matheman see http://planetmath.org/abelsummability (although I used $x^{n-1}$ instead of $x^n$, but I don't think that makes any difference. – MAN-MADE Jun 15 '17 at 07:15
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    That still looks OK, but it won't be easy to find a closed form. I'm not sure it's possible at all. –  Jun 15 '17 at 13:18
  • Sums with terms of the form $1/(a^k -1)$ seem to be related to the q-Polygamma-function http://mathworld.wolfram.com/q-PolygammaFunction.html, Wolfram gives the result $$\sum_{k=1}^\infty \frac{1}{3^k-1} = \frac{\log(3/2) - \psi_{1/3}(1) }{\log(3)}$$ – Hyperplane Jun 15 '17 at 13:33
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    So these sums are so called Lambert series which are heavily related to number theory. As an example the Erdös-Borwein constant is defined as $$E=\sum_{n=1}^\infty \frac{1}{2^n-1} = \sum_{n=1}^\infty \frac{d(n)}{2^n}\approx 1.606$$ which is proven to be irrational. To be honest unless $\sum_k 1/(3^k-1) +\sum_k 1/(3^k-1)^2$ somehow simplifies a lot, I too doubt there will be a closed form. – Hyperplane Jun 15 '17 at 14:07
  • @Hyperplane so the simplification to hyperbolic sum is not enough! I don't think it can be simplified further(may be I am wrong). – MAN-MADE Jun 15 '17 at 14:17
  • @Hyperplane It is indeed related to Lambert series(I didn't know that). Thanks for the link. – MAN-MADE Jun 15 '17 at 14:23

2 Answers2

2

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.

Paramanand Singh
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  • Thanks a lot, and thanks for the links, but I am surprised that there is no straight forward result for that series of hyperbolic function I gave. – MAN-MADE Jun 21 '17 at 03:21
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    @MANMAID: The general theory of sums like these is rather difficult and Ramanujan was a master of this theory. He hand calculated many values of $P(q) $ with great computational effort. – Paramanand Singh Jun 21 '17 at 03:26
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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.