# Evaluating $\sum_{n=1}^{\infty}\frac{n}{16^n}$ [duplicate]

I'm trying to evaluate the sum of the following infinite series: $$\sum_{n=1}^{\infty}\frac{n}{16^n}$$

I know it converges to $\frac{16}{225}$, but I don't know how to reach this solution. It's not a geometric series or a telescoping sum, and I haven't found any way to relate it to a Taylor or Maclaurin series. How should I approach this problem?

## marked as duplicate by Hans Lundmark, Arnaud Mortier, Namaste calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 5 '18 at 0:04

Hint:

What is the derivative of $\;\sum_{n=1}^{\infty} x^n$?

• @Dibbs. When you have taken the derivative, what series do you get when you set $x=1/16$? Also, how can you rewrite $\sum_{n=1}^{\infty} x^n$ in a closed form (not as a series), and what is it's derivative at $x=1/16$? – md2perpe Jul 28 '18 at 22:44
• It took me a second to wrap my head around it, but wow, it's incredible how that works out. Thanks so much! – Dibbs Jul 29 '18 at 1:01

Since $$\sqrt[n]{\dfrac{n}{16^n}}=\dfrac{\sqrt[n]{n}}{16} \to \frac{1}{16}<1,~~~~(n \to \infty)$$then by radical test, $\sum\limits_{n=1}^{\infty}\dfrac{n}{16^n}$ is convergent. Thus, we may denote

$$S=\sum_{n=1}^{\infty}\frac{n}{16^n}=\frac{1}{16}+\frac{2}{16^2}+\frac{3}{16^3}+\cdots.\tag1$$

Then $$16S=1+\frac{2}{16}+\frac{3}{16^2}+\cdots\tag2.$$

By $(2)-(1)$, we obtain $$15S=1+\frac{1}{16}+\frac{1}{16^2}+\cdots=\dfrac{1}{1-\dfrac{1}{16}}=\frac{16}{15}.$$

As a result, $$S=\frac{16}{225}.$$

• This helped a lot, thanks. I'm wondering, this should work for all infinite summations of the form $\sum_{n=1}^{\infty}\frac{n}{a^n}$ where $a$ is greater than $1$, right? And the answer will always be $\frac{a}{(a-1)^2}$ too, right? – Dibbs Jul 29 '18 at 1:07

$$S=\frac { 1 }{ 16 } +\frac { 2 }{ { 16 }^{ 2 } } +\frac { 3 }{ { 16 }^{ 3 } } +\frac { 4 }{ { 16 }^{ 4 } } +...\\ \frac { S }{ 16 } =\frac { 1 }{ { 16 }^{ 2 } } +\frac { 2 }{ { 16 }^{ 3 } } +\frac { 3 }{ { 16 }^{ 4 } } +\frac { 4 }{ { 16 }^{ 5 } } +...\\ S-\frac { S }{ 16 } =\frac { 1 }{ 16 } +\left( \frac { 2 }{ { 16 }^{ 2 } } -\frac { 1 }{ { 16 }^{ 2 } } \right) +\left( \frac { 3 }{ { 16 }^{ 3 } } -\frac { 2 }{ { 16 }^{ 3 } } \right) +\left( \frac { 4 }{ { 16 }^{ 4 } } -\frac { 3 }{ { 16 }^{ 4 } } \right) +...\\ \frac { 15 }{ 16 } S=\frac { 1 }{ 16 } +\frac { 1 }{ { 16 }^{ 2 } } +\frac { 1 }{ { 16 }^{ 3 } } +\frac { 1 }{ { 16 }^{ 4 } } +...\\ \frac { 15 }{ 16 } S=\frac { \frac { 1 }{ 16 } }{ 1-\frac { 1 }{ 16 } } =\frac { 1 }{ 15 } \\ S=\frac { 16 }{ 255 } \\ \\$$