How do you solve this summation: $\displaystyle\sum_{n=1}^{\infty} \dfrac{n^3}{2^n}$?

Question

How would you solve $\displaystyle\sum_{n=1}^{\infty} \dfrac{n^3}{2^n}$?

I stumbled upon it online and it has been giving me a very difficult time despite how simple it looks.

I would love it if someone could explain it to me.

Answer 1

By this post, the generating function for $n^3$ is $$ \sum_{n\ge 0}n^3x^n = \frac{x(1+4x+x^2)}{(1-x)^4}. $$ This was obtained by the standard trick of applying $(xD)^3$ (differentiating and multiplying $x$) to the generating function $\sum_{n\geq0}x^n = \frac1{1-x}$. Then, we can just evaluate this at $1/2$.

Answer 2

It's not much different from the other answers in terms of the process, but here is a probabilistic interpretation of the sum.

Consider the Geometric Random Variable $X$ with $p=\frac{1}{2}.$ We see that $$\mathbb{E}[X^3]=\sum_{n=1}^{\infty} n^3 \left( \frac{1}{2} \right)^n,$$ the third moment of $X.$

Let us consider the moment generating function

\begin{align}M_{X}(t)&=\mathbb{E}[e^{tX}]\\ & =\sum_{n=1}^{\infty}e^{tn} \left(\frac{1}{2}\right)^n =\frac{e^t}{2-e^t} \end{align}

We differentiate $M_{X}(t)$ thrice at $t=0.$ On one hand, we recover $\mathbb{E}[X^3]$ from the series representation, and on the other hand,

$$\frac{d^3}{dt^3} M_X(t)|_{t=0}= \frac{2e^t(4+8e^t+e^{2t})}{(2-e^t)^4}|_{t=0}=26$$

Answer 3

Start with $$f(z) = \sum_{n=0}^{\infty} z^n$$

Differentiate once, multiply by $z$, differentiate again, multiply by $z$ again and differentiate and multiply.

You will get

$$ \sum_{n=0}^{\infty} n^3z^n$$

Now set $z = \frac{1}{2}$.

Now use $$f(z) = \frac{1}{1-z}$$ to do the actually algebra and find out the value.

Answer 4

As Lord Shark the Unknown commented consider $$A=\sum_{n=1}^\infty n^3x^n$$ Now, the trick is to rewrite $$n^3=n(n-1)(n-2)+an(n-1)+b n$$ By identification, you should find $a=3$ and $b=1$. So $$A=\sum_{n=1}^\infty n(n-1)(n-2)x^n+3\sum_{n=1}^\infty n(n-1)x^n+\sum_{n=1}^\infty n x^n$$ that is to say $$A=x^3\sum_{n=1}^\infty n(n-1)(n-2)x^{n-3}+3x^2\sum_{n=1}^\infty n(n-1)x^{n-2}+x\sum_{n=1}^\infty n x^{n-1}$$ that is to say $$A=x^3 \left(\sum_{n=1}^\infty x^{n} \right)'''+3x^2\left(\sum_{n=1}^\infty x^{n} \right)''+x\left(\sum_{n=1}^\infty x^{n} \right)'$$