How can I prove that $\sum_{n=1}^\infty \dfrac{n^4}{2^n} = 150 $? I can easily prove that this series converges, but I can't imagine a way to prove the statement above. I tried with the techniques for finding the sum of $\sum_{n=1}^\infty \frac{n}{2^n}$, but I didn't find an useful connection with this case. Could someone help me?
 A: Here is a quite general process to the problems of this kind:
$$\begin{align}
S(x) &= \sum_{n\ge1}x^n \\ 
S'(x) &= \sum_{n\ge 1} nx^{n-1} \\
xS'(x) &= \sum_{n\ge 1} nx^{n} \\
(xS'(x))' &= \sum_{n\ge1} n^2 x^{n-1} \\
x(xS'(x))' &= \sum_{n\ge1} n^2 x^{n} \\
(x(xS'(x))')' &= \sum_{n\ge1} n^3 x^{n-1} \\
x(x(xS'(x))')' &= \sum_{n\ge1} n^3 x^{n} \\
(x(x(xS'(x))')')' &= \sum_{n\ge1} n^4 x^{n-1} \\
x(x(x(xS'(x))')')' &= \sum_{n\ge1} n^4 x^{n} \tag{$*$}\end{align}
$$
and note that $\displaystyle S(x) = \frac{x}{1-x}$ for $|x|<1$.
For you problem, you should compute $(*)$ at $x = 1/2$.
A: We have that
$$\sum_{n\ge 1} \frac{1}{2^n} n^4
= \sum_{n\ge 1} \frac{1}{2^n}
\sum_{m=1}^4 n^{\underline{m}} {4\brace m}
= \sum_{m=1}^4 {4\brace m} m!
\sum_{n\ge 1} \frac{1}{2^n} {n\choose m}.$$
The inner sum is
$$\sum_{n\ge m} \frac{1}{2^n} {n\choose m}
= \frac{1}{2^m} \sum_{n\ge 0} \frac{1}{2^n} {n+m\choose m}
= \frac{1}{2^m} \frac{1}{(1-1/2)^{m+1}} = 2.$$
We thus obtain evaluating the Stirling numbers combinatorially (e.g.
${4\brace 2} = {4\choose 1} + \frac{1}{2} {4\choose 2} = 7$)
$$2\sum_{m=1}^4 {4\brace m} m!
\\ = 2 \times 1 \times 1 + 2 \times 7 \times 2
+ 2 \times 6 \times 6 + 2 \times 1 \times 24
= 150$$
as claimed.
A: Consider
$$S=\sum_{n=1}^\infty n^4 x^n$$ and now, use the trick
$$n^4=n(n-1)(n-2)(n-3)+6 n(n-1)(n-2)+7 n(n-1)+n$$ which makes
$$S=x^4\left(\sum_{n=1}^\infty x^n\right)''''+6x^3\left(\sum_{n=1}^\infty x^n\right)'''+7x^2\left(\sum_{n=1}^\infty x^n\right)''+x \left(\sum_{n=1}^\infty x^n\right)'$$
A: Let $$S(k)=\sum_{n=1}^\infty \frac{n^k}{2^n}$$
You know that $S(0)=\frac{1}{2}$.
Next, note that
$$
S(k)=\sum_{n=1}^\infty \frac{n^k}{2^n}=\frac{1}{2} \sum_{n=1}^\infty \frac{n^k}{2^{n-1}}=\frac{1}{2}\left(1+\sum_{n=2}^\infty \frac{n^k}{2^{n-1}}\right)\\
=\frac{1}{2}\left(1+\sum_{n=1}^\infty \frac{(n+1)^k}{2^{n}}\right)\\
=\frac{1}{2}+\frac{1}{2}S(k)+\frac{1}{2}\binom{k}{1}S(k-1)+\frac{1}{2}\binom{k}{2}S(k-2)+...+\frac{1}{2}\binom{k}{k}S(k-k)
$$
Baring a stupid mistake, we get the recurrence
$$
S(k)=1+\binom{k}{1}S(k-1)+\binom{k}{2}S(k-2)+...+\binom{k}{k}S(k-k)
$$
which allows you calculate recursivelly $S(1), S(2),S(3), S(4),,...$
