Evaluating $\sum_{n=1}^{\infty}\frac{n}{16^n}$

Question

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?

Answer 1

Hint:

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

Answer 2

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, $$S=\frac{16}{225}.$$

Answer 3

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