# Evaluate the series $\sum _{n=1}^{\infty} \frac{n}{5^n}$ [duplicate]

$$\sum _{n=1}^{\infty}\frac{n}{5^n}$$

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.

• This is a special case of the sum $\sum_{n=1}^\infty nx^n$ which converges for $|x|<1$. It can be derived from the formula for $\sum_{n=0}^\infty x^n$. Commented Jun 3, 2017 at 13:59
• Commented Jun 3, 2017 at 14:05
• @userSeventeen Oh, whoops, good catch. Fixed hint: $$nx^n=x(nx^{n-1})=x\frac{dx^n}{dx}$$ Commented Jun 3, 2017 at 14:06
• When you say "but I can't use common ratio to solve problem", do you mean that you are not allowed to use it, or that you don't see how you can apply it? Commented Jun 3, 2017 at 14:12
• I mean I don't know how to apply it! Thank you everybody, now I understand!!
– Area
Commented Jun 3, 2017 at 15:28

## 3 Answers

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 $$\sum_{n\geq 1}\frac{n}{5^n} = \color{green}{\frac{5}{16}}.$$

$$S=\sum_{n\geq 1}\frac{n}{5^n}$$ 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 $\color{red}{\large S=\frac{5}{16}}$.

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 $S=\frac{5}{16}$