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I know how to derive the formula $\sum_{k=0}^\infty q^k = \frac{1}{1-q}$ for $|q| < 1$. I also recall having found the respective formulae for $\sum_{k=0}^\infty k q^k$ and $\sum_{k=0}^\infty k^2 q^k$ rewriting e.g. $$\sum_{k=0}^\infty k q^k = q \frac{\mathrm d}{\mathrm dq} \sum_{k=0}^\infty q^k.$$

Now I recall having proven the first formula (for $\sum_{k=0}^\infty q^k$) before having discussed functions (in particular derivatives and integrals), that is to say right when I started to learn about infinite series. However, I've only come across the second and third formula near the end of Calculus 1, thus only deriving them using derivatives.

My question is: Is it possible to derive the formulae with less knowledge, particularly not using any derivatives or integrals? If yes, does the same apply for higher powers ($\geq 3$) of $k$?

Thanks for your answers in advance.

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2 Answers 2

up vote 7 down vote accepted

Definitely, for instance

$$\begin{eqnarray}\sum_{k=0}^\infty k q^k & = & \sum_{k=1}^\infty (k-1) q^{k-1} \\ & = & \frac{1}{q}\sum_{k=1}^\infty k q^{k} - \sum_{k=1}^\infty q^{k-1} \end{eqnarray}$$

The last term is the one you know already, the first term contains the original sum again. From there you can solve this equation to get

$$\sum_{k=0}^\infty k q^k = \frac{q}{(1-q)^2} \; .$$

You can proceed in a similar way for the other series sum.

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I'm not sure this is what you mean but you can take the formula ${1 \over 1 - q} = \sum_{k\geq 0} q^k$ and square it. You get $$ {1 \over (1-q)^2} = \sum_{k \geq 0} q^k \sum_{\ell \geq 0} q^\ell = \sum_{k,\ell \geq 0} q^{\ell + k} = \sum_{n \geq 0} (n+1)q^n. $$ Thus ${q \over (1-q)^2} = \sum_{n\geq 0} (n+1)q^{n+1} = \sum_{n\geq 0} nq^n$. This is hardly more enlightening though because in fact knowing that $\sum nq^n = q d/dq (\sum n q^n)$ told me to find that $$ {q \over (1-q)^2} = q {d \over dq} {1 \over 1-q} $$

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