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}. $