Closed-form term for $\sum_{i=1}^{\infty} i q^i$ 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 $x>1$. 
 A: We have
$$\sum_{i=1}^\infty i q^i=q\sum_{i=1}^\infty i q^{i-1}=q\frac{d}{dq}\left(\sum_{i=1}^\infty  q^i\right)=q\frac{d}{dq}\left(\frac{q}{1-q}\right)$$
A: $$\sum_{i=1}^{\infty} i q^i=q\sum_{i=1}^{\infty} i q^{i-1}=q\frac{d}{dq}\sum_{i=1}^{\infty} q^{i}=q\frac{d}{dq}\sum_{i=0}^{\infty} q^{i}=q\frac{d}{dq}\frac{1}{1-q}=\frac{q}{(1-q)^2}$$
$$\sum_{i=x}^{\infty} i q^i=q\sum_{i=x}^{\infty} i q^{i-1}=q\frac{d}{dq}\sum_{i=x}^{\infty} q^{i}=q\frac{d}{dq}\left[\sum_{i=0}^{\infty} q^{i}-\sum_{i=0}^{x-1} q^{i}\right]=$$$$=q\frac{d}{dq}\left[\frac{1}{1-q}-\frac{1-q^{x}}{1-q}\right]=q\frac{d}{dq}\frac{q^{x}}{1-q}=\frac{(1-x)q^{x+1}+xq^{x}}{(1-q)^2}$$
A: 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*}

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*}
