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?

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 $S(1), S(2),S(3), S(4),,...$
