Can I change this summation to a sum of other summations? The form of the summation I have is 
$$\sum _{ x=0 }^{ \infty  }{ x{ a }^{ x } } $$
I need to somehow remove the $x$ from the original summation in order to achieve the geometric series in each  other summation. For instance, $$\sum _{ x=? }^{ \infty  }{ { a }^{ x }\quad +\quad  } \sum _{ x=? }^{ \infty  }{ { a }^{ x }\quad +\quad  } \sum _{ x=? }^{ \infty  }{ { a }^{ x }\quad +\quad  } ...$$ I have seen this done before, but forget how to work with the bounds of each new summation . It would be greatly appreciated if anyone had a clue what I was talking about.
 A: You should know that $\sum_{n=0}^\infty a^n = \frac{1}{1-a}$.  
If we were to derive both sides with respect to $a$, then we have 
$\frac{d}{da}[\sum_{n=0}^\infty a^n] = \sum_{n=0}^\infty \frac{d}{da}[a^n]=\sum_{n=0}^\infty na^{n-1} = \frac{d}{da}[\frac{1}{1-a}]=\frac{1}{(1-a)^2}$.  
Multiplying both sides by $a$ gives us the desired result:
$$\sum_{n=0}^\infty na^n = \frac{a}{(1-a)^2}$$
A: Here is the way forward that aligns with approach suggested in the OP.
$$\begin{align}
\sum_{x=0}^\infty xa^x&=\sum_{x=1}^\infty a^x\sum_{y=1}^x(1)\\\\
&=\sum_{y=1}^\infty\sum_{x=y}^{\infty}a^x\\\\
& =\sum_{y=1}^\infty\frac{a^y}{1-a}\\\\
&=\frac{a}{(1-a)^2}
\end{align}$$
A: $a^x=e^{x\log a}$ and then apply the Taylor Expansion for $e^{x\log a}$. That should clean things up and make it a better form to solve the problem.
A: This is one of my favourite tricks. Multiplying a series by a carefully chosen term. It is especially useful in dealing with arithmetic-geometric series. 
$$\begin{align}
 S &= a + 2a^2 + 3a^3 + \dots \\
 aS &= a^2 + 2a^3 + 3 a^4 + \dots \\
S - aS &= a + a^2 + a^3 + \dots \\
S &= \frac {a + a^2 + a^3 + \dots}{1-a}\\
S &= \frac{a}{(1-a)^2}\\
\end{align}$$
The last step assumes $\left\lvert a \right\rvert \lt 1$. Otherwise, the infinite series does not converge. 
