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?

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.
