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How to prove that

$$\sum_{k\geq 1} \frac{1}{2^k k^2}=\frac{\pi^2}{12}-\frac{1}{2}\log(2)^2$$

without using the well-known $\operatorname{Li}_2\left( \frac{1}{2} \right)$ ?

Edited : Thanks for L.F , but I should have made it clear that I want an answer using only series manipulations .

Zaid Alyafeai
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    I don't think this is going to be an easy one; the problem easily generalises to $\sum \frac{x^k}{k^2}$ (so setting $x=\frac{1}{2}$ gives your problem back), so if there existed a nice way to do this problem, I would imagine the generalisation wouldn't be so far behind, and the Basel problem would be less infamous! – FireGarden Aug 13 '13 at 22:21
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    @FireGarden, I believe there is a way to solve the problem , and generalization would be impossible since we only know the values listed for the dilogarithm . I think it wouldn't be impossible to prove that the sum is $\frac{\pi^2}{12}-\frac{1}{2} \log^2(2)$ , knowing that the first term is the alternating zeta function . – Zaid Alyafeai Aug 13 '13 at 22:38
  • I was hoping for an answer without using integration . I meant by pure series manipulation . – Zaid Alyafeai Aug 13 '13 at 22:58
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    I don't understand why this question was closed. The answer below as well as the answer to the "duplicate" question are just disguised derivations of the identity $$\mathrm{Li}_2 (z) +\mathrm{Li}_2 (1-z) = \zeta(2)- \ln z \ln (1-z) $$ without calling dilogarithms "dilogarithms". This is clearly not the answer OP was looking for. Vote to reopen. – Start wearing purple Aug 13 '13 at 23:53
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    @O.L , my initial intention was solving the series without integration ! – Zaid Alyafeai Aug 13 '13 at 23:58

3 Answers3

6

$$ \sum_{k\geq 1}x^{k-1}=\frac{1}{1-x}\Rightarrow \sum_{k\geq 1}\frac{1}{2^kk^2}=-\int_0^{\frac{1}{2}} \frac{\ln (1-x)}{x}\,dx$$

Notice that:

$$\begin{aligned}\int_0^{\frac{1}{2}} \frac{\ln (1-x)}{x}\,dx\overset{x\to \frac{x}{2}}=\int_0^1 \frac{\ln (1-\frac{x}{2})}{x}\,dx \overset{ \text{IBP}}=\int_0^1 \frac{\ln x}{2-x}\,dt \overset{x\to 1-x}=\int_0^1 \frac{\ln (1-x)}{1+x}\,dx\end{aligned}$$

Now consider:

$$f(t)=\int_0^1 \frac{\ln (1-tx)}{1+x}\,dx$$ We want $-f(1)$. Differentiate: $$\begin{aligned} f'(t) &= \int_0^1 \frac{x\,dx}{(tx-1)(1+x)} \\&= \int_0^1 \frac{dx}{tx-1}+\int_0^1 \frac{dx}{(t+1)(1+x)}-\int_0^1 \frac{t\,dx}{(tx-1)(t+1)}\\& =\frac{\ln (1-t)}{t}+\frac{\ln 2}{t+1} -\frac{\ln (1-t)}{1+t}\end{aligned}$$

Hence:

$$f(1) =\int_0^1 \frac{\ln (1-t)}{t}\,dt+\int_0^1\frac{\ln 2}{t+1}\,dt-\underbrace{\int_0^1 \frac{\ln (1-t)}{1+t}\,dt}_{f(1)}$$

$$\Rightarrow -f(1) =\frac{1}{2}\left(-\int_0^1 \frac{\ln (1-t)}{t}\,dt-\int_0^1\frac{\ln 2}{t+1}\right)\,dt=\frac{\pi^2}{12}-\frac{\ln^2 2}{2}$$

L. F.
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6

Apply Euler Series Transformation to the series $$ \frac{\pi^2}{12}=\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)^2}\tag{1} $$ which yields $$ \begin{align} \frac{\pi^2}{12} &=\sum_{n=0}^\infty2^{-n-1}\sum_{k=0}^n\binom{n}{k}\frac{(-1)^k}{(k+1)^2}\tag{2}\\ &=\sum_{n=0}^\infty2^{-n-1}\frac1{n+1}\sum_{k=0}^n\binom{n+1}{k+1}\frac{(-1)^k}{k+1}\tag{3}\\ &=\sum_{n=0}^\infty2^{-n-1}\frac1{n+1}\sum_{k=0}^n\sum_{j=0}^n\binom{j}{k}\frac{(-1)^k}{k+1}\tag{4}\\ &=\sum_{n=0}^\infty2^{-n-1}\frac1{n+1}\sum_{j=0}^n\sum_{k=0}^n\binom{j+1}{k+1}\frac{(-1)^k}{j+1}\tag{5}\\ &=\sum_{n=0}^\infty2^{-n-1}\frac1{n+1}\sum_{j=0}^n\frac1{j+1}\tag{6}\\ &=\sum_{n=1}^\infty2^{-n}\frac1{n}\sum_{j=1}^n\frac1j\tag{7}\\ &=\sum_{n=1}^\infty\frac{H_n}{n2^n}\tag{8} \end{align} $$ Use the series for $\log(2)=-\log\left(1-\frac12\right)$ $$ \log(2)=\sum_{m=1}^\infty\frac1{k2^k}\tag{9} $$ and square to get $$ \begin{align} \log(2)^2 &=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{mn2^{m+n}}\tag{10}\\ &=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m(m+n)2^{m+n}}+\frac1{n(m+n)2^{m+n}}\tag{11}\\ &=2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m(m+n)2^{m+n}}\tag{12}\\ &=2\sum_{m=1}^\infty\sum_{n=m+1}^\infty\frac1{mn2^n}\tag{13}\\ &=2\sum_{m=1}^\infty\sum_{n=m}^\infty\frac1{mn2^n}-2\sum_{m=1}^\infty\frac1{m^22^m}\tag{14}\\ &=2\sum_{n=1}^\infty\sum_{m=1}^n\frac1{mn2^n}-2\sum_{m=1}^\infty\frac1{m^22^m}\tag{15}\\ &=2\sum_{n=1}^\infty\frac{H_n}{n2^n}-2\sum_{m=1}^\infty\frac1{m^22^m}\tag{16}\\ &=\frac{\pi^2}{6}-2\sum_{m=1}^\infty\frac1{m^22^m}\tag{17} \end{align} $$ Therefore, $$ \sum_{m=1}^\infty\frac1{m^22^m}=\frac{\pi^2}{12}-\frac12\log(2)^2\tag{18} $$

Explanation

$\hphantom{0}(2)$ Euler Series Transformation
$\hphantom{0}(3)$ $\binom{n+1}{k+1}=\frac{n+1}{k+1}\binom{n}{k}$
$\hphantom{0}(4)$ $\binom{n+1}{k+1}=\sum_{j=0}^n\binom{j}{k}$
$\hphantom{0}(5)$ $\binom{j+1}{k+1}=\frac{j+1}{k+1}\binom{j}{k}$
$\hphantom{0}(6)$ $\sum_{k=0}^j(-1)^k\binom{j+1}{k+1}=1$
$\hphantom{0}(7)$ $j\mapsto j-1$ and $n\mapsto n-1$
$\hphantom{0}(8)$ collect $H_n$
$(10)$ multiply the series
$(11)$ $\frac1{mn}=\frac1{m(m+n)}+\frac1{n(m+n)}$
$(12)$ use symmetry
$(13)$ $n\mapsto n-m$
$(14)$ add and subtract the terms for $m=n$
$(15)$ change order of summation
$(16)$ collect $H_n$
$(17)$ use $(8)$

robjohn
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3

We can use the following identity

$$ \mathrm{Li}_2(1-x) = -\mathrm{Li}_2 \left( {1-\frac {1}{x}} \right)- \frac{1}{2}\, \ln^2 \left( \frac{1}{x} \right)-\zeta(2) ,$$

which relates two dilogarithm functions. Now, in your case, letting $x=\frac{1}{2}$, we have

$$ \mathrm{Li}_2\left(\frac{1}{2}\right) = -\mathrm{Li}_2 \left( {-1} \right)- \frac{1}{2}\, \ln^2 \left( 2 \right)-\zeta(2)$$

$$ = \frac{\zeta(2)}{2}- \frac{1}{2}\, \ln^2 \left( {2} \right)-\zeta(2)$$

$$\implies \mathrm{Li}_2\left(\frac{1}{2}\right)= \frac{1}{2}\ln^2\left( {2} \right)-\frac{\zeta(2)}{2}. $$

Note:

$$ \mathrm{Li}_2(-1)= \sum_{k=1}^{\infty}\frac{(-1)^k}{k^2}= -\frac{\zeta(2)}{2}. $$

Mhenni Benghorbal
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