<s>[INST] $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-2)!}(2n+1)x^{2n}$ to function I want to calculate $3-\frac{5}{2}+\frac{7}{24}-\frac{9}{720}+...$, while using $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-2)!}(2n+1)x^{2n}$.
Here is my attempt:
Firs I proved that the radius of converges for this series is $R= \infty$.
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-2)!}(2n+1)x^{2n} =~_{k=n-1} ~ = \sum_{k=0}^\infty \frac{(-1)^{k+2}}{(2k)!}(2k+3)x^{2k+2} = \sum_{k=0}^\infty \frac{(-1)^{k}}{(2k)!}(x^{2k+3})'$$
$$= (\sum_{k=0}^\infty \frac{(-1)^{k}}{(2k)!}x^{2k+3})'= (x^3 \sum_{k=0}^\infty \frac{(-1)^{k}}{(2k)!}x^{2k})'=(x^3 \cos(x))'=3x^2 \cos(x)-x^3 \sin(x)$$
So $f(x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-2)!}(2n+1)x^{2n}=3x^2 \cos(x)-x^3 \sin(x)$.
If I did everything correct, than I should find a value for $x$ which will help me to calculate the initial sum.
Did I do it ok?
Thanks a lot!
 [/INST] $$S=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-2)!}(2n+1)x^{2n}$$
Let $k=n-1$, then
$$S=\sum_{k=0}^{\infty} (-1)^k (2k+3)\frac{x^{2k+2}}{(2k)!}$$
We konw that $$x^3\cos x=\sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+3}}{(2k)!}$$
D.w.r.t. $x$, then
$$-x^3\sin x+3x^2 \cos x=S$$
OP's summation is verified. The sum of the numerical series is nothing but
$$3\cos 1-\sin 1$$ which is when $x=1$.
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