Find the value of $A$ where $$A=\sum_{n=1}^{\infty}\left(n\sin\left(\frac{\pi n}{2}\right)\left(e^x-1-\frac{x}{1!}-\frac{x^2}{2!}-\cdots-\frac{x^n}{n!}\right)\right)$$
By using the expansion of $e^x$, I have simplified the expression into $$A=\sum_{n=1}^{\infty}\left(n\sin\left(\frac{\pi n}{2}\right)\left(\sum_{i=n+1}^{\infty}\frac{x^i}{i!}\right)\right)$$
After this I am having a hard time in exploring further into this problem.
Any help is greatly appreciated.
First of all, for all fixed $x$, $\sum_{i \geqslant n + 1} \left|\frac{x^i}{i!}\right| = \mathrm{O}\left(\frac{1}{n!}\right)$ by Taylor inequality hence the sum is absolutely convergent. Then, notice that $\sin\left(\frac{\pi n}{2}\right) = 1$ if $n \equiv 1 \ [4]$, $-1$ if $n \equiv 3\ [4]$, $0$ if $n \equiv 0\ [2]$ so the sum simplifies as, $$ A = \sum_{k \geqslant 0} (-1)^k(2k + 1)\sum_{i \geqslant 2k + 2} \frac{x^i}{i!} = \sum_{i \geqslant 2} \frac{x^i}{i!}\sum_{k = 0}^{\left\lfloor \frac{i - 2}{2} \right\rfloor} (-1)^k(2k + 1). $$ We can exchange the sums by absolute convergence. Then, I let you check by induction that for all integer $j$, $$ \sum_{k = 0}^{j - 1} (-1)^k(2k + 1) = -(-1)^jj. $$ Therefore, $$ A = -\sum_{i \geqslant 2} \frac{x^i}{i!}(-1)^{\left\lfloor \frac{i}{2} \right\rfloor}\left\lfloor \frac{i}{2} \right\rfloor = A_{\mathrm{even}} + A_{\mathrm{odd}}, $$ where $A_{\mathrm{even}}$ (resp. $A_{\mathrm{odd}}$) are computing by summing on all the $i \geqslant 2$ even (resp. odd). We have, \begin{align*} A_{\mathrm{even}} & = -\sum_{i \geqslant 2|i \textrm{ even}} \frac{x^i}{i!}(-1)^{\left\lfloor \frac{i}{2} \right\rfloor}\left\lfloor \frac{i}{2} \right\rfloor\\ & = -\sum_{j \geqslant 0} \frac{x^{2j + 2}}{(2j + 2)!}(-1)^{j + 1}(j + 1)\\ & = \frac{x}{2}\sum_{j \geqslant 0} (-1)^j\frac{x^{2j + 1}}{(2j + 1)!}\\ & = \frac{x}{2}\sin(x). \end{align*} I let you compute $A_{\mathrm{odd}}$, it is the same idea. You have to make appear the series of $\sin$ and $\cos$. You should also verify my computations, it is easy to make mistakes with this kind of calculus.
Some thoughts. (We need to prove that the function is differentiable etc.)
We have $$A'(x) = A(x) + \sum_{n=1}^{\infty}\left(n\sin\left(\frac{\pi n}{2}\right)\cdot \frac{x^n}{n!}\right) = A(x) + x\cos x.$$ Solve the ODE to obtain $$A(x) = - \frac12 x\cos x + \frac12 \sin x + \frac12 x\sin x + c\mathrm{e}^x.$$
Since $A(0) = 0$, we have $c = 0$. Thus, $$A(x) = - \frac12 x\cos x + \frac12 \sin x + \frac12 x\sin x.$$
Another way to solve the problem is to use:
or the regularized gamma function to find:
$$\sum_{j=n+1}^\infty\frac{x^j}{j!}=\frac{e^x}{n!}\int_0^x e^{-t}t^ndt$$
This gives:
$$A(x)=\int_0^xe^{x-t}\sum_{n=1}^\infty \frac{t^n}{(n-1)!}\sin\left(\frac{\pi n}2\right)dt$$
The inner sum can be interpreted as $e^x$’s Maclaurin series if writing $\sin(\frac{\pi n}2)$ in complex form, so we are left with:
$$A(x)=\int_0^xe^{x-t} t\cos(t) dt=\frac12x\sin(x)-\frac12x\cos(x)+\frac12\sin(x)$$
