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Can I compute the sum $$ \sum_{n\geq 0}n\frac{1}{4^n} $$ by use of some trick?

First I thought of a geometrical series?

mathfemi
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  • See also some of these questions: http://math.stackexchange.com/questions/87030/proof-by-induction-sum-limits-i-0n-i-2i-1-n1-2n-1 http://math.stackexchange.com/questions/11464/how-to-compute-the-formula-sum-r-1d-r-cdot-2r http://math.stackexchange.com/questions/90637/what-is-the-limit-of-sum-limits-n-1-inftyn2-3n http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-infty-n1xn (You could probably find several very similar questions on this site.) – Martin Sleziak Jun 16 '15 at 22:09

5 Answers5

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This is very similar to the geometric series, infact $$\sum_{n=0}^\infty n x^n = x \cdot \frac{d}{dx} \sum_{n=0}^\infty x^n = x \cdot \frac{d}{dx} \frac{1}{1-x} = \frac{x}{(1-x)^2}$$

Plugging in $x=1/4$ will provide you with an answer.

Joel
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  • +1, but let me just point out that, in general, one must be very careful making sure that the operations of differentiation and infinite series commute. – triple_sec Jun 16 '15 at 21:37
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    This is true. We are saved here since a power series converges uniformly on closed intervals of inside of its radius of convergence as does the series derivative. Since we are evaluating the function at $x=1/4$, this method is valid. @triple_sec – Joel Jun 16 '15 at 21:59
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Suppose $\sum_{n\ge0}n\dfrac1{4^n}=S$, then $4S=S+\sum_{n\ge0}\dfrac1{4^n}$.

$\Rightarrow3S=\dfrac43$

$\Rightarrow S=\dfrac49$

I think this might be the solution.

Amir Naseri
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The sum can be expanded as follows: \begin{align*} \begin{array}{l|cccc} \text{once}&1/4^1&&&\\ \text{twice}&1/4^2&1/4^2&&\\ \text{3 times}&1/4^3&1/4^3&1/4^3&&\\ \vdots\\ \text{$n$ times}&1/4^n&1/4^n&\cdots&1/4^n\\ \vdots \end{array} \end{align*} Now, instead of summing row-by-row, try summing column-by-column: \begin{align*} \left(\frac{1}{4^1}+\frac{1}{4^2}+\frac{1}{4^3}+\ldots\right)+\left(\frac{1}{4^2}+\frac{1}{4^3}+\ldots\right)+\left(\frac{1}{4^3}+\ldots\right)+\ldots \end{align*} Each summand is of the form $$\sum_{j=k}^{\infty}\frac{1}{4^j}=\frac{1}{4^k}\sum_{j=0}^{\infty}\frac{1}{4^j}=\frac{1}{4^k}\times\frac{1}{1-1/4}=\frac{1}{4^k}\times\frac{4}{3}$$ for $k\in\{1,2,\ldots\}$. Now sum over $k$ to get $$\sum_{k=1}^{\infty}\frac{1}{4^k}\times \frac{4}{3}=\frac{1/4}{1-1/4}\times\frac{4}{3}=\frac{1}{3}\times\frac{4}{3}=\frac{4}{9}.$$

triple_sec
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The idea is to take the derivative of the geometric series. For $|x|<1$, $$\frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}$$

So, $$\frac x{(1-x)^2}=\sum_{n=1}^\infty nx^n.$$

Tim Raczkowski
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Yes, it's just a geometric series of geometric series: $$ \sum_{n=1}^{\infty}nx^{-n}=\sum_{k=1}^{\infty}\sum_{n=k}^{\infty}x^{-n}=\sum_{k=1}^{\infty}\frac{x^{-k}}{1-1/x}=\frac{x^{-1}}{(1-1/x)^2}=\frac{x}{(x-1)^2}. $$ So your sum is $4 / 9$.

mjqxxxx
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