<s>[INST] What is the value of $1+\frac{1}{2^2}+ \frac{1}{4^2}+ \frac{1}{8^2}+... $? 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.  
 [/INST] $$
\begin{align}
1+\frac{1}{2^2}+ \frac{1}{4^2}+ \frac{1}{8^2}+\cdots
&=\frac{1}{(2^0)^2}+\frac{1}{(2^1)^2}+ \frac{1}{(2^2)^2}+ \frac{1}{(2^3)^2}+\cdots\\
&=\frac{1}{(2^2)^0}+\frac{1}{(2^2)^1}+ \frac{1}{(2^2)^2}+ \frac{1}{(2^2)^3}+\cdots\\
&=\sum\limits_{n=0}^{\infty}\frac{1}{\left(2^2\right)^n}\\
&=\sum\limits_{n=0}^{\infty}\left(\frac{1}{4}\right)^n.
\end{align}
$$
$\sum\limits_{n=0}^{\infty}\left(\frac{1}{4}\right)^n$ 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:
$$
\sum\limits_{n=0}^{\infty}\left(\frac{1}{4}\right)^n=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}=1\frac{1}{3}.
$$
</s>