Evaluate the series $\sum _{n=1}^{\infty} \frac{n}{5^n}$ $$\sum _{n=1}^{\infty}\frac{n}{5^n}$$
I tried to plug in $n=1,2,3,4,...$ but I can't use common ratio to solve problem.
I think there is another way like using differentiation or integral but I don't no exactly what to do.
 A: As the comments suggest consider the sum (for $|x| < 1$) $$\sum_{n \geq 0} x^n = \frac{1}{1-x} \implies \sum_{n \geq 1}nx^{n-1} = \frac{1}{(1-x)^2} \implies \sum_{n \geq 1}nx^n = \frac{x}{(1-x)^2}.$$
Now set $x = 1/5$ to get $$\sum_{n\geq 1}\frac{n}{5^n} = \color{green}{\frac{5}{16}}.$$
A: $$S=\sum_{n\geq 1}\frac{n}{5^n}$$
is clearly absolutely convergent. We have
$$\begin{eqnarray*} 4S = 5S-S &=& \sum_{n\geq 1}\frac{n}{5^{n-1}}-\sum_{n\geq 1}\frac{n}{5^n}\\&=&1+\sum_{n\geq 1}\frac{n+1}{5^n}-\sum_{n\geq 1}\frac{n}{5^n}\\&=&1+\sum_{n\geq 1}\frac{1}{5^n}\\&=&1+\frac{\frac{1}{5}}{1-\frac{1}{5}}=1+\frac{1}{4}=\frac{5}{4} \end{eqnarray*}$$
hence it follows that $\color{red}{\large S=\frac{5}{16}}$.
A: I am not using calculus let the sum be $S$ thus $S=\frac{1}{5}+\frac{2}{5^2}$...now $\frac{S}{5}=\frac{1}{5^2}+\frac{2}{5^3}+...$ subtracting two we have $\frac{4S}{5}=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...$ which is GP with common ratio less than 1 using formula that sum of such gp is $\frac{a}{1-r}$ we have  $\frac{4S}{5}=\frac{1}{4}$ thus $S=\frac{5}{16}$
