5
$\begingroup$

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

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.

$\endgroup$
5
  • 2
    $\begingroup$ 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 $\sum_{n=0}^\infty x^n$. $\endgroup$ Commented Jun 3, 2017 at 13:59
  • $\begingroup$ Similar : math.stackexchange.com/questions/757263/…. $\endgroup$ Commented Jun 3, 2017 at 14:05
  • 1
    $\begingroup$ @userSeventeen Oh, whoops, good catch. Fixed hint: $$nx^n=x(nx^{n-1})=x\frac{dx^n}{dx}$$ $\endgroup$ Commented Jun 3, 2017 at 14:06
  • $\begingroup$ 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? $\endgroup$ Commented Jun 3, 2017 at 14:12
  • $\begingroup$ I mean I don't know how to apply it! Thank you everybody, now I understand!! $\endgroup$
    – Area
    Commented Jun 3, 2017 at 15:28

3 Answers 3

8
$\begingroup$

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

$\endgroup$
5
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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

$\endgroup$
0

Not the answer you're looking for? Browse other questions tagged .