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Please explain different methods to calculate the sum of infinite series with $\dfrac{n^2}{2^n}$ as it's general term i.e. Calculate

$$\sum_{n=0}^\infty \dfrac {n^2}{2^n}$$

Please avoid the method used for summation of arithmetic geometric series. It is very tedious approach. Does any other simpler approach exist?

Martin Sleziak
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divban
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  • If this isn't a polylogarithm, then I don't know what is... – Lucian Oct 02 '14 at 18:54
  • @Lucian As usual, it is much preferable to call it a rational function evaluated at $\frac12$. You seem to be missing the point that every polylogarithm of nonpositive integer order trivializes into a rational function (with denominator a power of $1-z$). To react to every question about $\sum\limits_nz^n$ and its derivatives by saying that they are polylogarithms is distracting at best. – Did Oct 07 '14 at 17:49

2 Answers2

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For $|x|<1,$ $$\sum_{r=0}^\infty x^r=\frac1{1-x}$$

Differentiate wrt $x,$ $$\sum_{r=0}^\infty rx^{r-1}=\frac1{(1-x)^2}$$

Multiply by $x$

Differentiate wrt $x$

Multiply by $x$

Can you recognize $x$ here?

lab bhattacharjee
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The answer below is (perhaps) not given with perfect seriousness, since in a sense it is on the complicated side. However, if one thinks in probabilistic terms, it is not completely unreasonable.

Imagine tossing a fair coin until we obtain a head. Let $X$ be the number of tosses. Then $$E(X^2)=\sum_1^\infty \frac{n^2}{2^n}.$$ Condition on the result of the first toss. The conditional expectation of $X^2$, given that we got a head on the first toss, is $1$.

The conditional expectation of $X^2$, given that we got a tail on the first toss, is $E(1+X)^2$, which by the linearity of expectation is equal to $1+2E(X)+E(X^2)$. It follows that $$E(X^2)=\frac{1}{2}\cdot 1+\frac{1}{2}(1+2E(X)+E(X^2)).$$ We conclude that $$E(X^2)=2+2E(X).$$ It remains to find $E(X)$. By a conditioning argument similar to the one above, but quite a bit simpler, we get that $E(X)=2$, giving $E(X^2)=6$.

André Nicolas
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