Closed-form term for $\sum_{i=1}^{\infty} i q^i$

Question

I am interested in the following sum $$\sum_{i=1}^{\infty} i q^i$$ for some $q<1$.

Is there a closed-form-term for this? If yes, how does one derive this?

I am also interested in

$\sum_{i=x}^{\infty} i q^i$

for some $x>1$.

Answer 1

We have

$$\sum_{i=1}^\infty i q^i=q\sum_{i=1}^\infty i q^{i-1}=q\frac{d}{dq}\left(\sum_{i=1}^\infty q^i\right)=q\frac{d}{dq}\left(\frac{q}{1-q}\right)$$

Answer 2

$$\sum_{i=1}^{\infty} i q^i=q\sum_{i=1}^{\infty} i q^{i-1}=q\frac{d}{dq}\sum_{i=1}^{\infty} q^{i}=q\frac{d}{dq}\sum_{i=0}^{\infty} q^{i}=q\frac{d}{dq}\frac{1}{1-q}=\frac{q}{(1-q)^2}$$

$$\sum_{i=x}^{\infty} i q^i=q\sum_{i=x}^{\infty} i q^{i-1}=q\frac{d}{dq}\sum_{i=x}^{\infty} q^{i}=q\frac{d}{dq}\left[\sum_{i=0}^{\infty} q^{i}-\sum_{i=0}^{x-1} q^{i}\right]=$$$$=q\frac{d}{dq}\left[\frac{1}{1-q}-\frac{1-q^{x}}{1-q}\right]=q\frac{d}{dq}\frac{q^{x}}{1-q}=\frac{(1-x)q^{x+1}+xq^{x}}{(1-q)^2}$$

Answer 3

We have \begin{align*} \sum_{i=1}^\infty iq^i &= \sum_{i=1}^\infty \sum_{k=1}^i q^i\\ &= \sum_{k=1}^\infty \sum_{i=k}^\infty q^i\\ &= \sum_{k=1}^\infty \frac{q^k}{1-q}\\ &= \frac{q}{(1-q)^2} \end{align*}

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Addendum: For another start we have \begin{align*} \sum_{i=x}^\infty iq^i &= \sum_{i=x}^\infty \sum_{k=1}^{i} q^i\\ &= \sum_{k=1}^\infty \sum_{i=\max\{x,k\}}^\infty q^i\\ &= \sum_{k=1}^\infty \frac{q^{\max\{x,k\}}}{1-q}\\ &= \frac{(x-1)q^x}{1-q} + \sum_{k=x}^\infty \frac{q^k}{1-q}\\ &= \frac{(x-1)q^x}{1-q} + \frac{q^x}{(1-q)^2}\\ &= \frac{(x-1)(1-q)q^x + q^x}{(1-q)^2}\\ &= \frac{xq^x - (x-1)q^{x+1} }{(1-q)^2} \end{align*}