Find the sum of $\sum_{n = 1}^{+\infty}\frac{n(n+1)}{2^n}$

Question

Given the series $$\sum_{n = 1}^{+\infty}\frac{n(n+1)}{2^n} = \alpha$$ find $\alpha$.

What technique am I supposed to use here?

Thanks in advance.

Answer 1

If$$f(z)=\sum_{n=1}^\infty n(n+1)z^n$$and if$$F(z)=\sum_{n=1}^\infty nz^{n+1},$$then $F'(z)=f(z)$ and the sum that you're after is $f\left(\frac12\right)$. Now,$$F(z)=z^2\sum_{n=1}^\infty nz^{n-1}=z^2\left(\sum_{n=1}^\infty z^n\right)'.$$Can you take it from here?

Answer 2

Note that

$$\sum_{n = 1}^{\infty}\frac{n(n+1)}{2^n}=\sum_{n = 1}^{\infty}n(n+1)\left(\frac{1}{2}\right)^n$$

then recall that for $|x|<1$

$$f(x)=\sum_{k=0}^\infty x^k=\frac{1}{1-x}$$

$$f'(x)=\sum_{k=1}^\infty kx^{k-1}=\frac{1}{(1-x)^2}$$

$$f''(x)=\sum_{k=2}^\infty k(k-1)x^{k-2}=\frac{2}{(1-x)^3}$$

therefore

$$\sum_{k=2}^\infty k(k-1)x^{k-2}=\sum_{k=1}^\infty k(k+1)x^{k-1}=\frac{2}{(1-x)^3}\implies\sum_{k=1}^\infty k(k+1)x^{k}=\frac{2x}{(1-x)^3}$$

and then

$$\sum_{n = 1}^{\infty}n(n+1)\left(\frac{1}{2}\right)^n=\frac{2\frac12}{(1-\frac12)^3}$$

Answer 3

$$f(x) = \frac{1}{1-x} =\sum_{n=0}^\infty x^n \Rightarrow xf''(x) = \frac{2x}{(1-x)^3}= \sum_{n=1}^\infty n(n+1)x^n$$ It follows $$ \sum_{n=1}^\infty n(n+1)\frac{1}{2^n} =\frac{2\cdot\frac{1}{2}}{(1-\frac{1}{2})^3}= 8$$