Need explain how to find sum of series

Question

Can someone explain me how to find sum of next series:

$\sum_{n=1}^\infty n^4 \tan^{n-1}(x)$

Thanks for answers in advance.

Answer 1

We can write $n^4$ as a sum of combinatorial polynomials of degree $4$ or less: $$ n^4=24\binom{n}{4}+36\binom{n}{3}+14\binom{n}{2}+\binom{n}{1}\tag{1} $$ Next, for $|x|\lt1$, we can use the Generalized Binomial Theorem and negative binomial coefficients to show $$ \begin{align} \sum_{n=k}^\infty\binom{n}{k}x^{n-1} &=\sum_{n=k}^\infty\binom{n}{n-k}x^{n-1}\\ &=\sum_{n=k}^\infty\binom{-k-1}{n-k}(-1)^{n-k}x^{n-1}\\ &=\sum_{n=0}^\infty\binom{-k-1}{n}(-1)^{n}x^{n+k-1}\\ &=\frac{x^{k-1}}{(1-x)^{k+1}}\tag{2} \end{align} $$ Therefore, using $(1)$ and $(2)$, we get $$ \begin{align} \sum_{n=1}^\infty n^4x^{n-1} &=\frac{24x^3}{(1-x)^5}+\frac{36x^2}{(1-x)^4}+\frac{14x}{(1-x)^3}+\frac1{(1-x)^2}\\ &=\frac{1+11x+11x^2+x^3}{(1-x)^5}\tag{3} \end{align} $$ Now, just substitute $x\mapsto\tan(x)$.

Answer 2

Hint: $~\displaystyle\sum_{n=0}^\infty t^n~=~?~$ Now, differentiate both sides with respect to t, and notice what happens... :-)