Limit of $\frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$ What is the limit of this sequence $\frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$?
Where $a$ is a constant and $n \to \infty$.
If answered  with proofs, it will be best. 
 A: With $S_n = \frac{1}{a} + \frac{2}{a^2} + \frac{3}{a^3} + \cdots \frac{n}{a^n}$ and using the closed form of geometric sums,
$$
\begin{align}
\lim_{n \to \infty} S_n &= \frac{1}{a} + \frac{2}{a^2} + \frac{3}{a^3} + \cdots \\
&= \frac{1}{a} + \frac{1}{a^2} + \frac{1}{a^3} + \cdots \\
& \phantom{=\frac{1}{a}} + \frac{1}{a^2} + \frac{1}{a^3} + \cdots \\
& \phantom{=\frac{1}{a} + \frac{1}{a^2}} + \frac{1}{a^3} + \cdots \\
& \phantom{=\frac{1}{a} + \frac{1}{a^2} + \frac{1}{a^3}} \ddots \\
&= \frac{1}{a - 1} + \frac{1}{a} \cdot \frac{1}{a - 1} + \frac{1}{a^2} \cdot \frac{1}{a - 1} + \cdots \\
&= \frac{1}{a - 1} \left( 1 + \frac{1}{a} + \frac{1}{a^2} + \cdots \right) \\
&= \frac{1}{a - 1} \cdot \frac{a}{a - 1} \\
&= \frac{a}{(a - 1)^2}
\end{align}
$$
A: Let $f(x) = 1+x + x^2+x^3+ \cdots$. Then the radius of convergence of $f$ is $1$, and inside this disc we have $f(x) = \frac{1}{1-x}$, and $f'(x) = 1+2x+3x^2+\cdots = \frac{1}{(1-x)^2}$.
Suppose $|x|<1$, then we have $xf'(x) = x+2x^2+3x^3+\cdots = \frac{x}{(1-x)^2}$. 
If we choose $|a| >1$, then letting $x = \frac{1}{a}$ we have $\frac{1}{a} + \frac{2}{a^2}+ \frac{3}{a^3}+ \cdots = \frac{\frac{1}{a}}{(1-\frac{1}{a})^2}= \frac{a}{(a-1)^2}$.
A: Hint:
$\frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$ = $(\frac{1}{a} + \frac{1}{a^2} + \cdots + \frac{1}{a^n}) + (\frac{1}{a^2} + \frac{1}{a^3} + \cdots + \frac{1}{a^n}) + ...$ 
Simplify each sum, factor out common terms and it will become more clear to solve. :)
A: The limit does not exist if $|a|\le 1$, and does if $|a|\gt 1$. The fact that it exists if $|a|\gt 1$ can be shown using the Ratio Test.
So we concentrate on the value of the limit, when it exists.
Let $S_n$ be our sum, and let $S$ be its limit.  Then 
$$aS_{n+1}-S_n=1+\frac{1}{a}+\frac{1}{a^2}+\cdots+\frac{1}{a^n}.$$
Let $n\to\infty$. The left-hand side approaches $(a-1)S$, while the right-hand side approaches the sum of an infinite geometric series. That sum is $\frac{a}{a-1}$. Thus $S=\frac{a}{(a-1)^2}$. 
