$$\sum_{n=1}^\infty \frac{1}{n^2 2^n}$$

I am new in series thus I tried a pair of methods to compute but I couldn't

  • $\begingroup$ If you are only required to check its convergence, compare this with a suitable geometric series. If your aim is to find its closed form, then you can utilize the reflection formula for the dilogarithm function. $\endgroup$ – Sangchul Lee Mar 21 '16 at 2:41
  • $\begingroup$ Specifically you want $Li_2(1/2)$, and the third identity in the link implies $2Li_2(1/2)=\frac{\pi^2}{6}-\ln(1/2)^2$. $\endgroup$ – Ian Mar 21 '16 at 2:51

The idea is to consider the power serie

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

The value of your serie is then $f(\frac{1}{2} )$

Now, how to find $f$? Derivate !

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

Now you multiply by $x$ to find an usual power serie :

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


$$f(x) = \int \frac{\ln(1-x)}{x} dx$$

$$f(1/2) - f(1) = \int_{\frac{1}{2}}^1 \frac{\ln(1-x)}{x} dx = [\ln(1-x)\ln(x) ]_{1/2}^1 + \int_{\frac{1}{2}}^1 \frac{\ln(x)}{1-x} dx $$

$$= -\ln(1/2)^2 - \int_0^{\frac{1}{2}} \frac{\ln(1-x)}{x} dx $$

$$= -\ln(1/2)^2 - ( f(1/2)-f(0) ) $$

But $f(1) = \frac{\pi^2}{6}$, so

$$2 f(1/2) = \frac{\pi^2}{6} -\ln(1/2)^2 $$

And you have the result

  • $\begingroup$ I understood your solution and that "understanding and implementing your idea" thing works but your solution is not true. Because after "taking derivative of second equation, you missed shifting indexes and thus your solution starts true but entering a wrong way, see more at p5 math.wisc.edu/~park/Fall2011/sequences/seriesfunc.pdf $\endgroup$ – onurctirtir Mar 21 '16 at 18:53
  • $\begingroup$ If you correct it, I will assign it as answer $\endgroup$ – onurctirtir Mar 21 '16 at 18:53
  • $\begingroup$ And you have another fail that $$ \int \frac{1}{1-x} dx = -ln(1-x)$$ $\endgroup$ – onurctirtir Mar 21 '16 at 19:00
  • 1
    $\begingroup$ I don't see why shifting indexes is necessary : $f'(x) = (x+ \frac{x^2}{2^2}+ \frac{x^3}{3^2} + \cdots )' = 1 + \frac{x}{2} + \frac{x^2}{3} + \cdots = \sum_{n=1}^\infty \frac{x^{n-1}}{n}$. Then if you multiply by $x$, you get $\sum_{n=1}^\infty \frac{x^n}{n}$. And I don't see what's this "another fail" you're talking about $\endgroup$ – Tryss Mar 21 '16 at 20:32

I think you should integrate $$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$$ a couple times. We see $$-\log(1-x) =\sum^\infty_{n=0} \frac{x^{n+1}}{n+1}.$$ Then dividing by $x$, we see $$-\frac{\log(1-x)}{x} = \sum^\infty_{n=0} \frac{x^n}{n+1}.$$ Then integrating from $0$ to $1/2$ gives $$\sum^\infty_{n=0} \frac{1}{(n+1)^22^{n+1}} = - \int^{1/2}_{0} \frac{\log(1-x)}{x} dx = \frac{1}{12}(\pi^2 - 6\log(2)^2).$$ I'm not exactly sure how to perform that integral; I used WolframAlpha.

  • $\begingroup$ This is about the same as plugging the original sum into WA, really, seeing as the original sum is the series expansion for $Li_2(1/2)$ and what you wrote is the integral for $Li_2(1/2)$. But $Li_2$ is not an elementary function, so there is some nontrivial trickery in explicitly computing $Li_2(1/2)$. $\endgroup$ – Ian Mar 21 '16 at 3:01

Note that we can write

$$\begin{align} \sum_{n=1}^N\frac{x^n}{n^2}&=\sum_{n=1}^N \int_0^x s^{n-1}\,ds \int_0^1 t^{n-1}\,dt\\\\ &=\int_0^1 \int_0^x \frac{1-(st)^N}{1-(st)}\,ds\,dt \tag 1\\\\ \end{align}$$

For $x<1$, using the Dominated Convergence Theorem to evaluate the limit of $(1)$ as $N\to \infty$, we can write

$$\begin{align} \sum_{n=1}^\infty \frac{x^n}{n^2}&=\int_0^1 \int_0^x \frac{1}{1-(st)}\,ds\,dt\\\\ &=-\int_0^1 \frac{\log(1-xt)}{t}\,dt\\\\ &=-\int_0^x \frac{\log(1-u)}{u}\,du\\\\ &=\text{Li}_2(x) \tag 2 \end{align}$$

where in $(2)$ $\text{Li}_2(x)$ is the dilogarithm function. Therefore, for $x=1/2$ we have

$$\sum_{n=1}^\infty \frac{x^n}{n^2}=\text{Li}_2(1/2) \tag 3$$

Note that $\text{Li}_2(1)=\frac{\pi^2}{6}$ See Basel Problem. Furthermore, it can be shown (see the NOTE at the end of this solution) that the dilogarithm function satisfies the reflection identity

$$\begin{align} \text{Li}_2(x)+\text{Li}_2(1-x)&=\text{Li}_2(1)-\log(x)\log(1-x)\\\\ &=\frac{\pi^2}{6}-\log(x)\log(1-x) \tag 4 \end{align}$$

Using $(4)$ with $x=1/2$ in $(3)$ reveals

$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{x^n}{n^2}=\frac{\pi^2}{12}-\frac12 \log^2(1/2)}$$

And we are done!


Here, we prove the reflection identity given in $(4)$. Observe that we can write

$$\begin{align} \text{Li}_2(x)+\text{Li}_2(1-x)&=-\int_0^x \frac{\log(1-u)}{u}\,du-\int_0^{1-x}\frac{\log(1-u)}{u}\,du\\\\ &=-\int_0^x \frac{\log(1-u)}{u}\,du-\int_0^{1}\frac{\log(1-u)}{u}\,du-\int_1^{1-x} \frac{\log(1-u)}{u}\,du\\\\ &=\frac{\pi^2}{6}-\int_0^x \frac{\log(1-u)}{u}\,du-\int_1^{1-x} \frac{\log(1-u)}{u}\,du \\\\ &=\frac{\pi^2}{6}-\int_0^x \frac{\log(1-u)}{u}\,du+\int_0^{x} \frac{\log(u)}{1-u}\,du \\\\ &=\frac{\pi^2}{6}-\int_0^x \frac{\log(1-u)}{u}\,du+\left.\left(-\log(u)\log(1-u)\right)\right|_0^x +\int_0^x \frac{\log(1-u)}{u}\,du\\\\ &=\frac{\pi^2}{6}-\log(x)\log(1-x) \end{align}$$

as was to be shown!


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