How to calculate the infinite sum of $\frac{1}{(n+1)2^n}$? [duplicate]

This question already has an answer here:

I've got a test tomorrow, and in one of the practice problems I'm asked to solve

$$\sum_{n=0}^{\infty}\frac{1}{(n+1)2^n} = ?$$

Now, I couldn't figure out a way to transform this in to some sort of mixture of a geometric and arithmetic partial sum, which would allow me to take the limit. Hence I asked WolframAlpha, but they returned the partial sum with something called the "Lerch transcendent", which is definitely not something we needed to learn.

This leaves me to asking here about how to tackle this problem.

marked as duplicate by José Carlos Santos calculus 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(); } ); }); }); Apr 9 at 9:45

Hint:

$$|x|<1 \implies\frac1{1-x}=\sum_{n=0}^\infty x^n\implies\int \frac{dx}{1-x}=\sum_{n=0}^\infty \frac{x^{n+1}}{n+1}=x\sum_{n=1}^\infty\frac{x^n}{n+1}\implies\ldots$$ ...and watch closely the integration constant...

For $$-1\le x<1$$

$$\ln(1-x)=-\sum_{r=1}^\infty\dfrac{x^r}r$$

For $$x\ne0,$$

$$\dfrac{\ln(1-x)}x=-\sum_{r=1}^\infty\dfrac{x^{r-1}}r$$

Hint: Solve the more general problem: calculate $$\;\sum_{n=0}^\infty \frac{x^n}{n+1}$$, and rewrite first the latter as $$\frac1x\sum_{n=0}^\infty \frac{x^{n+1}}{n+1}=\frac1x\sum_{n=1}^\infty \frac{x^n}{n}.$$