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My teacher posed an infinite series question to me today and I'm not quite sure how to start to go about it. $$\sum_{r=1}^{\infty} \dfrac{2^{-r}}{r^{2}}$$ Any hints would be much appreciated.

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$r$ from $1$ and not $0$, I would assume. –  user17762 May 10 '13 at 22:34
    
Haha yes, from r = 1, thanks xD –  TI82 May 10 '13 at 22:34
    
What is the Question? If convergence, what theorems do you know, e.g. the Ratio Test? –  hardmath May 10 '13 at 22:36
    
Just to evaluate the infinite sum. And no, I don't :S I'm working on elementary knowledge here, I'm no undergrad :L –  TI82 May 10 '13 at 22:36

3 Answers 3

Claim: $$\color{blue}{\boxed{\displaystyle \sum_{k=1}^{\infty} \dfrac1{2^k k^2} = \dfrac{\pi^2}{12} - \dfrac{\log^2(2)}2}}$$


Proof

What you want to evaluate is the PolyLogarithm function, $\text{Li}_2(x)$, at $x=1/2$. The PolyLogarithm function $\text{Li}_s(x)$ is defined as $$\text{Li}_s(x) = \sum_{k=1}^{\infty} \dfrac{x^k}{k^s}$$

We are lucky here to evaluate this sum, since in general, we cannot obtain such a nice answer for $\text{Li}_s(x)$. For instance, there is no nice answer to $$\text{Li}_{2}(1/3) = \sum_{k=1}^{\infty} \dfrac1{3^k k^2}$$

Below is the procedure on how to obtain the solution $\displaystyle \sum_{k=1}^{\infty} \dfrac1{2^k k^2} = \dfrac{\pi^2}{12} - \dfrac{\log^2(2)}2$. From geometric series, we have $$\sum_{r=1}^{\infty} x^{r-1} = \dfrac1{1-x}$$ Integrating from 0 to $x$, we get that $$\sum_{r=1}^{\infty} \dfrac{x^r}r = -\log(1-x) \implies \sum_{r=1}^{\infty} \dfrac{x^{r-1}}r = -\dfrac{\log(1-x)}x$$ Again integrating from $0$ to $1/2$, we get that $$S = \sum_{r=1}^{\infty} \dfrac1{2^rr^2} = - \int_0^{1/2} \dfrac{\log(1-t)}t dt = -\int_1^{1/2} \dfrac{\log(t)}{1-t}(-dt) =-\int_{1/2}^1 \dfrac{\log(t)}{1-t}dt$$ Now we have \begin{align} \int_{1/2}^1 \dfrac{\log(t)}{1-t}dt & = \int_{1/2}^1 \sum_{k=0}^{\infty} t^k \log(t) dt = \sum_{k=0}^{\infty} \int_{1/2}^1 t^k \log(t) dt\\ & = \sum_{k=0}^{\infty}\dfrac{-1 + 2^{-(1+k)} + (k+1) \log(2)2^{-(1+k)}}{(1+k)^2}\\ & = - \dfrac{\pi^2}6 + S + \log(2) \cdot \overbrace{\sum_{k=0}^{\infty} \dfrac{2^{-(k+1)}}{k+1}}^{-\log(1-1/2) = \log(2)} \end{align} Hence, we get that $$S = -\left(-\dfrac{\pi^2}6 + S + \log^2(2)\right) \implies 2S = \dfrac{\pi^2}6 - \log^2(2) \implies S = \dfrac{\pi^2}{12} - \dfrac{\log^2(2)}2$$ Hence, $$\color{red}{\boxed{\displaystyle \sum_{k=1}^{\infty} \dfrac1{2^k k^2} = \dfrac{\pi^2}{12} - \dfrac{\log^2(2)}2}}$$

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Wow! :O Haha, thanks very much! I only asked for a hint but you did that in what, 15 minutes? You're a tank xD –  TI82 May 10 '13 at 23:01
    
This approach adapts a general identity for the dilogarithm, $$Li_2(z) + Li_2(1-z) = \frac{\pi^2}{6} -\log z \cdot \log(1-z)$$ to the case $z = 1-z = 1/2$. Given OP's unfamiliarity with the Ratio Test, the Question may have been posed at a pre-college level. If so, a numerical evaluation of the series may have been intended. –  hardmath May 11 '13 at 1:27

Consider $f(x)=\sum\limits_{r=1}^\infty \dfrac{x^r}{r^2}$. Try differentiating, multiplying by $x$, and then differentiating again. This function you should recognize. Now try to solve for $f(x)$ and then set $x=1/2$.

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The problem is that the solution of the resulting DE is I believe non-elementary. –  André Nicolas May 10 '13 at 22:47
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@Ted Ha ha. Easier said than done. –  user17762 May 10 '13 at 22:53
    
I said "try." I don't believe in giving complete solutions here ... certainly not without a lot of effort from the OP. Admittedly, this problem is hard :) –  Ted Shifrin May 10 '13 at 23:18

Hint:

$$\frac1{2^rr^2}\le\frac1{2^r}=\left(\frac12\right)^r$$

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3  
I assume the OP wants to evaluate it. –  user17762 May 10 '13 at 22:35
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He said nothing, just asked for a hint so I gave him one for convergence... –  DonAntonio May 10 '13 at 22:35
    
Your observation can be useful in estimating the error of a partial sum; truncating the series after $r=N$ removes a tail whose combined terms are less than $2^{-N}$. –  hardmath May 11 '13 at 1:33

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