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I am trying to calculate $\sum_{n=0}^\infty(n+2)x^n$.

I was thinking it is like the second derivative of $x^{n+2}/(n+1)$ but I am not sure how to go about calculating it. Any hints?

Martin Sleziak
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Ron
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  • Shouldn't it be i – Archis Welankar May 12 '16 at 05:12
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    If it is $\sum_{n=0}^\infty (n+2)x^n$, it may be easiest to write it as $\sum nx^n+\sum 2x^n$. – André Nicolas May 12 '16 at 05:14
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    With a bit of searching you should be able to find many very similar questions on this site. Just a few: 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 – Martin Sleziak May 12 '16 at 09:20

3 Answers3

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Hint. You are on the right track, one may recall that $$ \sum_{n=0}^\infty\frac{x^{n+2}}{n+1}=-x\ln(1-x),\quad |x|<1. $$ Then differentiate termwise twice.

Olivier Oloa
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$$\sum_{n=0}^{\infty}(n+2)x^n=\sum_{n=0}^{\infty}nx^n+2\sum_{n=0}^{\infty}x^n$$

we know $$\sum_{n=0}^Nnx^n=-\frac{x(x^N-1)}{(x-1)^2}+\frac{Nx^{N+1}}{x-1}$$ and $$\sum_{n=0}^Nx^n=\frac {1-x^{N+1}}{1-x}$$

When $x\lt 1$ and $N \to \infty$ $$\sum_{n=0}^{\infty}(n+2)x^n=\frac{x}{(x-1)^2}+2\frac {1}{1-x}=\frac{2-x}{(x-1)^2}$$

user115350
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Hint: It is sometimes convenient to work with operators. Since \begin{align*} \left(x\frac{d}{dx}\right)\sum_{n=0}^\infty a_nx^n=x\sum_{n=0}^\infty na_nx^{n-1}=\sum_{n=0}^\infty na_nx^{n} \end{align*}

we consider $xD:=\left(x\frac{d}{dx}\right)$ as operator and obtain for $k\geq 0$ \begin{align*} \left(xD\right)^k\sum_{n=0}^{\infty}a_nx^n=\sum_{n=0}^\infty n^ka_nx^{n} \end{align*}

We conclude \begin{align*} \sum_{n=0}^{\infty}(n+2)x^n&=\left(xD+2\right)\sum_{n=0}^{\infty}x^n\\ &=\left(xD+2\right)\frac{1}{1-x}\\ &=x\left(\frac{1}{1-x}\right)^\prime+\frac{2}{1-x}\\ &=\ldots \end{align*}

Markus Scheuer
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