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this series has been driving me crazy, the thing is I need to determine if the series converges or not, and if it does I need to find the value at which converges.

$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{2^n n (3n-1)}}}$

I know that this series converges by the ratio test, but I can't find yet at which value converges, I tried to find the partial sums but I couldn't get anywhere.

I would appreciate any advice that you guys could give me. Thank so much.

David
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    Perhaps try approaching via partial fraction decomposition. $\frac{1}{2^nn(3n-1)}=\frac{1}{2^n}(\frac{1}{n(3n-1)})=\frac{3}{2^n(3n-1)}- \frac{1}{2^nn}$. You could then split this into two sums that seem more manageable. Do you know how to find the sum of $\sum\limits_{n=1}^\infty \frac{1}{k^nn}$? – JMoravitz Nov 24 '15 at 01:58
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    Have a look here:http://math.stackexchange.com/questions/1543454/calculating-sum-n-1-infty-frac12nn3n-1/1543494#1543494 – Olivier Oloa Nov 24 '15 at 01:59
  • I am not so sure if it has a closed form. For example, replacing $2$ with $x$ we have that the series -after partial decomposition - is taking the form:

    $$\sum_{n=1}^{\infty}\frac{3}{3(n-1)}x^n - \sum_{n=1}^{\infty}\frac{x^n}{n}= \sum_{n=1}^{\infty}\frac{3}{(3n-1)}x^n + \ln (1-x)$$

    However, the first sum does not seem to have a closed form.

     – Tolaso Nov 24 '15 at 01:59
  • Well the first one does have a closed form as pointed in the link. – Tolaso Nov 24 '15 at 02:00
  • Well I was thinking as you say using partial fraction, but then I don't know if you guys can help me, the only way I know to find the sum of a series is by using partial sums or if it is a geometric series, there are any other ways of finding the sum? – David Nov 24 '15 at 02:10

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$$ P=\sum_{n=1}^{\infty}\frac{1}{2^n n(3n-1)} =\sum_{n=1}^{\infty}\frac{3}{2^n 3n(3n-1)} =\sum_{n=1}^{\infty}\frac{3}{2^n}\left(\frac{1}{3n-1}-\frac{1}{3n}\right)\\ S_m=\sum_{n=1}^{\infty}\frac{1}{(3n-m)\cdot2^n}\\ f_m(x)=\sum_{n=1}^{\infty}\frac{x^{3n-m}}{3n-m}\\ {f_m}'(x)=\sum_{n=1}^{\infty}x^{3n-m-1}=\frac{1}{x^{m+1}}\sum_{n=0}^{\infty}x^{3n}=\frac{1}{x^{m+1}(1-x^3)}\\ f_1(x)=\ln x - \frac{\ln (1-x^3)}{3}\\ f_0(x)=\frac{1}{6}\left(\ln (x^2+x+1)-2\ln (1-x)-2\sqrt 3 \tan ^{-1}\left(\frac{2x+1}{\sqrt 3}\right)\right)\\ P=3\left(S_1-S_0\right)=3\left(2^{-\frac{1}{3}}\cdot f_1(2^{-\frac{1}{3}})-f_0(2^{-\frac{1}{3}})\right) $$

Kay K.
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