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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.

xiaohan2012
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    Could you do $\sum \frac{1}{a^i}$? If so, what does it look like one has to do to that sum to make it look like the one in question? Look at it term-by-term. – zibadawa timmy Nov 04 '13 at 07:23

4 Answers4

10

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,

$$ \begin{align} \lim_{n \to \infty} S_n &= \frac{1}{a} + \frac{2}{a^2} + \frac{3}{a^3} + \cdots \\ &= \frac{1}{a} + \frac{1}{a^2} + \frac{1}{a^3} + \cdots \\ & \phantom{=\frac{1}{a}} + \frac{1}{a^2} + \frac{1}{a^3} + \cdots \\ & \phantom{=\frac{1}{a} + \frac{1}{a^2}} + \frac{1}{a^3} + \cdots \\ & \phantom{=\frac{1}{a} + \frac{1}{a^2} + \frac{1}{a^3}} \ddots \\ &= \frac{1}{a - 1} + \frac{1}{a} \cdot \frac{1}{a - 1} + \frac{1}{a^2} \cdot \frac{1}{a - 1} + \cdots \\ &= \frac{1}{a - 1} \left( 1 + \frac{1}{a} + \frac{1}{a^2} + \cdots \right) \\ &= \frac{1}{a - 1} \cdot \frac{a}{a - 1} \\ &= \frac{a}{(a - 1)^2} \end{align} $$

Sammy Black
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5

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 $\frac{1}{a} + \frac{2}{a^2}+ \frac{3}{a^3}+ \cdots = \frac{\frac{1}{a}}{(1-\frac{1}{a})^2}= \frac{a}{(a-1)^2}$.

copper.hat
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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. :)

aaa
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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 $S=\frac{a}{(a-1)^2}$.

André Nicolas
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