Closed form for $ a_n = 1-2^4 +3^4 - 4^4 + ... \pm (-1)^n \cdot n^4 $ using Generating Functions Problem: Find closed form expression for $ a_n = 1-2^4 +3^4 - 4^4 + ... \pm (-1)^n \cdot n^4 $ using Generating Functions.
Hint: Calculate what function you need to multiply in order to get partial sums with alternating signs.
I don't really know... , I found that $ G(x) = \sum_{n=0} (-1)^n(n+1)^4 x^n =\sum_{n=0} (-1)^n \cos(\pi n) x^n  $. I get the same results when trying to solve the recurrence relation $ a_{n+1} = a_n + (-1)^n (n+1)^4 $ which describes the sequence above, so I get no progress. I don't really see how the hint helps me, do you have any ideas? Thanks in advance!
Note: I found Closed form for $1-2+3-4+\cdots(-1)^{n-1}n$ and some more questions that relate to the partial sum above, but I can't see how to relate the partial sum to generating functions.
 A: Introduce $A(x):=\sum_{n\geq 0}a_n x^{n+1}$ for
$$
a_n=1-2^4+3^4-...+(-1)^{n-1}n^4.
$$
Then observe that the relation
$$
a_{n+1}=a_{n}+(-1)^n(n+1)^4
$$
and multiply it by $x^n$
$$
a_{n+1}x^n=a_{n}x^n+(-1)^n(n+1)^4x^n
$$
and sum for $n\geq 0$ to get
$$
\frac{A(x)}x=A(x)+\frac{(1-x) (x^2-10 x+1)}{(x+1)^5}
$$
yielding
$$
A(x)=\frac{x(x^2-10 x+1)}{(x+1)^5}=x-15 x^2+66 x^3-190 x^4+435 x^5-861 x^6+1540 x^7+O(x^8)
$$
and expanding $A(x)$ in partial fractions
$$
A(x)=\frac{x^3}{(x+1)^5}-\frac{10 x^2}{(x+1)^5}+\frac{x}{(x+1)^5}
$$
should provide (after some work) the correct answer
$$
a_n=-\frac{1}{2} (-1)^n n (n+1) \left(n^2+n-1\right).
$$
EDIT: about the identity
$$
\sum_{n\geq 0}(-1)^n(n+1)^4x^n=\frac{(1-x) (x^2-10 x+1)}{(x+1)^5}
$$
we can reason as follows. First observe that for any formal power series $B(x)=\sum_{n\geq 0}b_nx^n$ we have $(x\partial_x+1)B(x)=\sum_{n\geq 0}(n+1)b_nx^n$. Hence
$$
\sum_{n\geq 0}(-1)^n(n+1)^4x^n=(x\partial_x+1)^4\sum_{n\geq 0}(-x)^n=(x\partial_x+1)^4\frac 1{1+x}
$$
and a (tedious) computation of the derivatives gives the answer.
A: [Too long for a comment.]
I would like to give an alternative method for finding decomposition
$$S:=\sum_{n\geq 0}(-1)^n(n+1)^4x^n=\frac{(1-x) (x^2-10 x+1)}{(x+1)^5}$$
in the excellent answer of @Giulio by using the ubiquitous $\color{red}{\text{Stirling numbers}}$ of the second kind allowing to convert powers into descending factorials like this here:
$$n^4=\color{red}{1}n+\color{red}{7}n(n-1)+\color{red}{6}n(n-1)(n-2)+\color{red}{1}n(n-1)(n-2)(n-3)\tag{1}$$
(see this nice article).
(1) gives, after a shift $n \to n+1$ and a multiplication by $(-1)^n x^n$:
$$S=\sum (-1)^n (n+1)^4 x^{n}=\begin{cases}\underbrace{\sum (-1)^n (n+1)x^n}_{f'(x)}+\\7\underbrace{\sum (-1)^n (n+1)n x^n}_{f''(x)}+\\6\underbrace{\sum (-1)^n (n+1)n(n-1)x^n}_{f'''(x)}+\\1 \underbrace{\sum (-1)^n (n+1)n(n-1)(n-2)x^n}_{f^{iv}(x)}\end{cases}$$
where we recognize the successive derivatives of the geometric series:
$$f(x)=1-x+x^2-x^3+...= \frac{1}{1+x}$$
This gives:
$$S=\frac{1}{(1+x)^2}-\frac{7.2}{(1+x)^3}+\frac{6.3.2}{(1+x)^4}-\frac{4.3.2}{(1+x)^5}$$
$$S=\frac{(1+x)^3-14(1+x)^2+36(1+x)-24}{(1+x)^5}$$
And from there, finally, the RHS of (1).
