I am trying to calculate the sum of an infinite geometric series. The problem is that in this series, '$i$' is part of the ratio.

The equation is as follows, as best as I can produce it here:

$$\sum_\limits{i=0}^{\infty} \frac{i}{4^i}$$

The part I am confused about is the fact that i itself is part of the ratio. Because it is included in the ratio, $S = \displaystyle \frac{a_1}{1-R}$, the equation for calculating the sum, makes no sense.

Thank you in advance for your help.


  • 2
    $\begingroup$ Are you asking about $\sum_\limits{i=0}^{\infty} \frac{i}{4^i}$? $\endgroup$ – zz20s Jan 25 '16 at 0:00

I'd guess $i$, here, means the index, not the square root of $-1$.

Let $S$ be the sum. Then

$$S=\displaystyle\sum_{i=0}^\infty \frac{i}{4^i}$$ $$S= 0 + \frac{1}{4} + \frac{2}{16}+\frac{3}{64}+\cdots $$ $$4S = 1 + \frac{2}{4} + \frac{3}{16} +\frac{4}{64}+\cdots$$

If we subtract $S$ from $4S$, we get

$$3S = 1 + \frac{1}{4}+\frac{1}{16}+\frac{1}{64} + \cdots $$

But the right hand side is an infinite geometric series, with first term $1$ and constant ratio $1/4$. This means

$$3S = \frac{4}{3}$$

$$ S = \boxed{\frac{4}{9}}$$

Your original sum is an example of an arithmetico-geometric series.

  • 1
    $\begingroup$ One must be careful with manipulations like this because you must be sure that the infinite series converges. Still though, +1 for clever manipulation. $\endgroup$ – zz20s Jan 25 '16 at 0:15

First note that the sum converges by the ratio test.

To determine the exact value of the sum, write $f(x)=\frac{1}{1-x}= \sum_\limits{i=0}^{\infty} x^i$ for $x<|1|$.

Differentiate both sides to obtain $\frac{1}{(1-x)^2}= \sum_\limits{i=0}^{\infty} ix^{i-1}=\sum_\limits{i=0}^{\infty} \frac{ix^i}{x}$

Now, let $x=\frac{1}{4} \Rightarrow \frac{1}{(1-(1/4))^2}= \sum_\limits{i=0}^{\infty} i(1/4)^{i-1}=\sum_\limits{i=0}^{\infty} \frac{i(1/4)^i}{1/4}$

$$\frac{1/4}{(1-(1/4))^2}=\frac{4}{9}=\sum_\limits{i=0}^{\infty} \frac{i}{4^i}$$


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