Question

Evaluate the infinite sum

Answer 1

The sum converges to 1000.

The $n$-th partial sum of the series is

$S_n=\displaystyle\sum_{i=1}^n100\left(\dfrac9{10}\right)^{i-1}=100\left(1+\dfrac9{10}+\left(\dfrac9{10}\right)^2+\cdots+\left(\dfrac9{10}\right)^{n-1}\right)$

Then

$\dfrac9{10}S_n=100\left(\dfrac9{10}+\left(\dfrac9{10}\right)^2+\left(\dfrac9{10}\right)^3+\cdots+\left(\dfrac9{10}\right)^n\right)$

so that

$S_n-\dfrac9{10}S_n=\dfrac1{10}S_n=100\left(1-\left(\dfrac9{10}\right)^n\right)$

$\implies S_n=1000\left(1-\left(\dfrac9{10}\right)^n\right)$

As $n\to\infty$, $\left(\dfrac9{10}\right)^n\to0$, so we're left with

$\displaystyle\sum_{i=1}^\infty100\left(\dfrac9{10}\right)^{i-1}=\lim_{n\to\infty}S_n=1000$