# Find $S=\sum_{n=1}^{\infty}\frac{1}{n^3 2^n}$ [duplicate]

Mathematica gives: $$S= -\frac{1}{12}\pi^2\log(2)+\frac{\log(2)^3}{6}+ \frac{7}{8}\zeta(3)$$ How can I prove it?

## marked as duplicate by Did, kingW3, Arnaldo, Jack D'Aurizio sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 16 '17 at 15:23

• Complex analysis seems the way! – vidyarthi Jun 16 '17 at 10:10
• You can't prove it, obviously. Look for identities and special values of polylogarithm (trilogarithm, in this case), cf. math.stackexchange.com/questions/555961/… – Professor Vector Jun 16 '17 at 10:17
• @vidyarthi I don't see how. – MereMortal47 Jun 16 '17 at 10:43
• Why is it obvious, @professorvector? – MereMortal47 Jun 16 '17 at 10:43
• I guess you asked how to prove that, because you don't know. If that's wrong, I'm sorry. – Professor Vector Jun 16 '17 at 10:46

Hint:

Let $$f(x):=\sum_{n=1}^\infty\frac{x^n}{n^3}.$$

Then

$$f'(x):=\sum_{n=1}^\infty\frac{x^{n-1}}{n^2},$$

$$(xf'(x))'=\sum_{n=1}^\infty\frac{x^{n-1}}{n},$$

$$(x(xf'(x))')'=\sum_{n=1}^\infty x^{n-1}=\frac1{1-x}.$$

Now backward, integrating from $0$ to $x$,

$$(xf'(x))'=\frac{\log(1-x)}x,$$

$$f'(x)=\frac1x\int_0^x\frac{\log(1-x)}xdx,$$

$$f(x)=\int_0^x\left[\frac1x\int_0^x\frac{\log(1-x)}xdx\right]dx.$$

• but we require complex analysis to evaluate $\int_0^x\frac{\log(1-x)}{x}dx$ isnt it? – vidyarthi Jun 16 '17 at 11:00
• I don't think the integrals here can be easily evaluated because if they could be, then $lim_{x\rightarrow1^{+}}f(x)$ would give $\zeta(3)$, and this solves an open problem. Right? – MereMortal47 Jun 16 '17 at 11:42
• @Asemismaiel: maybe Ramanujan... :) – Yves Daoust Jun 16 '17 at 11:58

$$S=\sum_{n=1}^{\infty}\frac{x^2}{n^3}=\text{Li}_3(x)$$ where appears the polylogarithm function. Have a look here for special values; in your case, $x=\frac 12$.

• Your link gives $\text{Li}_3(x)= -\frac{1}{12}\pi^2\log(2)+\frac{\log(2)^3}{6}+ \frac{7}{8}\zeta(3)$ and it says that it can be evaluated analytically but it doesn't tell me how. I'm asking for a derivation. – MereMortal47 Jun 16 '17 at 11:47