Evaluating $\sum_{n=1}^\infty \frac{2^{2n-1}}{5^{n+1}}$ Evaluating 
$$\sum_{n=1}^\infty \frac{2^{2n-1}}{5^{n+1}}$$
Its $\frac{\infty}{\infty}$ so I should use L Hospital rule, but the terms are exponential and differentiation won't do much good? I am thinking maybe I somehow use $\ln$ both sides? But how? Or perhaps I should do something else?
UPDATE
Following @Paul's answer: Since $|r| < 1$ so sequence is convergent. So I use the formula
$$\sum_{n=1}^\infty ar^{n-1} = \frac{a}{1-r}$$
$$\frac{1}{10} \cdot \frac{1}{1-\frac{4}{5}} = \frac{1}{10}\cdot 5 = \frac{1}{2}$$
But answer is $\frac{2}{5}$
 A: You can rewrite it as 
$$\sum_{n=1}^\infty \frac{2^{2n-1}}{5^{n+1}}=\sum_{n=1}^\infty \frac{\frac{1}{2}\cdot2^{2n}}{5\cdot 5^n}=\frac{1}{10}\sum_{n=1}^\infty \frac{2^{2n}}{5^n}=\frac{1}{10}\sum_{n=1}^\infty \frac{4^{n}}{5^n}=\frac{1}{10}\sum_{n=1}^\infty \left(\frac{4}{5}\right)^n$$
which is a geometric series with ratio $r=\displaystyle\frac{4}{5}$. I think you can finish it from here. 
A: This is a series, not a sequence. L'Hopital's rule is used for the latter, not the former. 
When working with series, it is a good idea to first determine if the the given series is of a certain known type. One type is a series of the form 
$$\sum\limits_{n=m}^\infty a r^{n+k} =ar^{m+k} + ar^{m+k+1}++ ar^{m+k+2}+\cdots.$$
These are called geometric series and the quantity  $r$ is called the ratio of the series. 
A geometric series converges if and only if $|r|<1$. When the series converges, it converges to to the first term of the series divided by the quantity $(1-r)$. The first term of the series above is obtained when you take $n=m$: $ar^{m+k}$.
I would not suggest you use a formula to find the sum of the series, but rather do the following:


*

*Identify the ratio $r$.

*If $|r|<1$, the series converges to $\text{the first term of the
    series}\over 1-r $. If $|r|\ge1$, the series diverges.


In your case
$$
\sum_{n=1}^\infty { {2^{2n-1}\over 5^{n+1}} }=\sum_{n=1}^\infty\textstyle {1\over 10}({4\over5})^n 
$$
This series is geometric with $r=4/5$.  The first term of the series is ${1\over10}\cdot{4\over5}$ (obtained by setting $n=1$). So, the series converges and
$$
\sum_{n=1}^\infty{\textstyle {1\over 10}({4\over5})^n} = {{1\over10}\cdot{4\over5}\over 1-{4\over5}}
={{1\over10}\cdot{4\over5}\over  {1\over5}}={2\over5}.
$$
Another example:
$$
\sum_{n=2}^\infty 4\cdot (-1/3)^{n+5}\quad \buildrel {r=-1/3}\over{ = }\quad
{4\cdot(-1/3)^7\over 1-(-1/3)}  ={-4/3^7\over 5/3}.

$$
Here, the first term of the series is obtained when $n=2$: $4\cdot(-1/3)^{2+5}=4\cdot(-1/3)^{7}$.
And one more:
$$
\sum_{n=0}^\infty  ( 1/3)^{n }\quad \buildrel {r= 1/3}\over{ = }\quad
{ ( 1/3)^0\over 1-(1/3)}  ={1\over 2/3}={3\over2}.

$$
