For which value of r does the following equality hold: $\sum_{n=0}^{\infty}r^n = \sum_{k=1}^{\infty}180/3^{4k}$ 
For which value of r does the following equality hold: $$\sum_{n=0}^{\infty}r^n = \sum_{k=1}^{\infty}180/3^{4k}$$

I'm just not making any progress with this. I know from memorization that $\sum_{n=0}^{\infty}r^n = 1/(1-r)$, but that's about it. I guess that means $\sum_{k=1}^{\infty}180/3^{4k}$ is also? 
 A: Assuming you know that,
$$\sum\limits_{n = 0}^\infty a r^n = \frac{a}{1 - r}$$
$$\sum\limits_{k=1}^{\infty}\frac{180}{3^{4k}}=180\sum\limits_{k=1}^{\infty}\frac{1}{3^{4k}}$$
The $R.H.S$ is a $G.P$ with $$a=\frac{1}{3^{4}}$$ and $$r=\frac{1}{3^{4}}$$ which can be evaluated to be  equal to 
$$\frac{\frac{1}{3^{4}}}{1-\frac{1}{3^{4}}}=\frac{1}{80}$$.
Hence the R.H.S evaluates to $$180*\frac{1}{80}=\frac{9}4$$
Which should be equal to $$L.H.S=\frac{1}{1-r}$$ as evaluated by you.
Hence the answer  $$r = \frac{5}9$$.
A: Hint: The closed form of a geometric series is
$$\sum\limits_{n = 0}^\infty a r^n = \frac{a}{1 - r}$$
Alternatively,
$$\sum\limits_{n = 1}^\infty a r^n = \frac{a}{1 - r} - a = \frac{ra}{1-r}$$
So the right hand side is simply given by
\begin{align}
\sum\limits_{k = 1}^\infty \frac{180}{3^{4k}} &= \sum\limits_{k = 1}^\infty \frac{180}{(3^4)^k}  \\
&= \sum\limits_{k = 1}^\infty 180 \left(\frac{1}{3^4}\right)^k
\end{align}
So put this into a closed form, put the left side into a closed form, and solve for $r$.
