<s>[INST] 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 $\frac{2}{5}$
 [/INST] 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. 
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