What is the value of $1+\frac{1}{2^2}+ \frac{1}{4^2}+ \frac{1}{8^2}+... $?

Question

I need to find the value of the following series:

$$ 1+\frac{1}{2^2}+ \frac{1}{4^2}+ \frac{1}{8^2}+\cdots $$

That is $$ 1+ \frac{1}{2^2}+ \frac{1}{(2^2)^2} +\frac {1}{(2^3)^2}+ \cdots$$

It is the summation of a geometric progression where all the terms are squared. I am unable to go further with it.

Answer 1

You're right. You're summing:

$$\sum_{n=0}^\infty{\bigg[\color{green}1\cdot\color{red}{\bigg(\frac14\bigg)}^n\bigg]}=\frac{\color{green}1}{1-\color{red}{\frac14}}=\frac43$$

Answer 2

Why is it that the squares give you trouble? Your series is simply$$1+\frac1{2^2}+\frac1{(2^2)^2}+\frac1{(2^2)^3}+\cdots$$and therefore its sum is$$\frac1{1-\frac1{2^2}}=\frac43.$$

Answer 3

$$ \begin{align} 1+\frac{1}{2^2}+ \frac{1}{4^2}+ \frac{1}{8^2}+\cdots &=\frac{1}{(2^0)^2}+\frac{1}{(2^1)^2}+ \frac{1}{(2^2)^2}+ \frac{1}{(2^3)^2}+\cdots\\ &=\frac{1}{(2^2)^0}+\frac{1}{(2^2)^1}+ \frac{1}{(2^2)^2}+ \frac{1}{(2^2)^3}+\cdots\\ &=\sum\limits_{n=0}^{\infty}\frac{1}{\left(2^2\right)^n}\\ &=\sum\limits_{n=0}^{\infty}\left(\frac{1}{4}\right)^n. \end{align} $$

$\sum\limits_{n=0}^{\infty}\left(\frac{1}{4}\right)^n$ is a geometric series because it's an expression of the form $\sum\limits_{n=0}^{\infty}ar^n$ where $a=1$ and $r=\frac{1}{4}$. We also know that if $|r|<1$, a geometric series converges (otherwise it diverges which means that the sum does not approach a finite number, in other words, it blows up to either positive or negative infinity) and we can even find what it converges to using the formula $\sum\limits_{n=0}^{\infty}ar^n=\frac{a}{1-r}$. In our case, $\left|\frac{1}{4}\right|<1$. So, we know that our series converges and we can find what it converges to:

$$ \sum\limits_{n=0}^{\infty}\left(\frac{1}{4}\right)^n=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}=1\frac{1}{3}. $$