Is there a closed form of $(\frac{1}{3!}-\frac{1}{4!})+\frac{2}{6!}-(\frac{4}{7!}-\frac{4}{8!})-\frac{8}{10!}...$

Question

Is there a closed form of $$(\frac{1}{3!}-\frac{1}{4!})+\frac{2}{6!}-(\frac{4}{7!}-\frac{4}{8!})-\frac{8}{10!}+(\frac{16}{11!}-\frac{16}{12!})+\frac{32}{14!}-(\frac{64}{15!}-\frac{64}{16})-\frac{128}{18!}+(\frac{256}{19!}-\frac{256}{20!})+\frac{512}{22!}...?$$ I want to know how to find the closed form of the series if possible? thanks for any help

Answer 1

As far as I see, it can be written in the form: $$ S=\displaystyle\sum^{\infty}_{k=0}\frac{(-1)^k4^k}{(3+4k)!}-\displaystyle\sum^{\infty}_{k=0}\frac{(-1)^k4^k}{(4+4k)!}+2\displaystyle\sum^{\infty}_{k=0}\frac{(-1)^k4^k}{(6+4k)!} $$

Wolfram knows each of these sums: $$ \displaystyle\sum^{\infty}_{k=0}\frac{(-1)^k4^k}{(3+4k)!}=\frac{1}{4}(sin(1)cosh(1)-cos(1)sinh(1))\\ \displaystyle\sum^{\infty}_{k=0}\frac{(-1)^k4^k}{(4+4k)!}=\frac{1}{4}(1-cos(1)cosh(1))\\ \displaystyle\sum^{\infty}_{k=0}\frac{(-1)^k4^k}{(6+4k)!}=\frac{1}{8}(1-sin(1)sinh(1)) $$

Stack back: $$ S=\frac{1}{4}(sin(1)+cos(1))(cosh(1)-sinh(1))=\frac{1}{4e}(sin(1)+cos(1)) $$

Answer 2

With the same spirit of the Daniel Fischer's answer, take the fourth root of the unity $\zeta=e^{2\pi i/4} $. We have $$e^{\zeta z}=\sum_{k\geq0}\frac{\zeta^{k}z^{k}}{k!}=\sum_{k\geq0}\frac{z^{4k}}{\left(4k\right)!}+\zeta\sum_{k\geq0}\frac{z^{4k+1}}{\left(4k+1\right)!}+\zeta^{2}\sum_{k\geq0}\frac{z^{4k+2}}{\left(4k+2\right)!}+\zeta^{3}\sum_{k\geq0}\frac{z^{4k+3}}{\left(4k+3\right)!} $$ and $1+\zeta+\zeta^{2}+\zeta^{3}=0 $ so writing a suitable combination of $e^{\zeta^{m}z} $ we can find the value of each series written by Kostiantyn Lapchevskyi. For example we have that $$\sum_{k\geq0}\frac{z^{4k}}{\left(4k\right)!}=\frac{1}{4}\sum_{k=1}^{4}e^{\zeta^{k}z} $$ hence $$\sum_{k\geq0}\frac{\left(-1\right)^{k}4^{k}}{\left(4k+4\right)!}=-\frac{1}{4}\sum_{k\geq1}\frac{\left(-4\right)^{k}}{\left(4k\right)!}=-\frac{1}{4}\left(\frac{1}{4}\sum_{k=1}^{4}\exp\left(\zeta^{k}\left(-4\right)^{1/4}\right)-1\right) $$ $$=-\frac{\left(1+e^{2}\right)\cos\left(1\right)}{8e}+\frac{1}{4} $$ which is equivalent to your result. In the same page I linked you can find a very useful scheme made by Lucian.

Answer 3

It seems the $n$th summand of your series is $$a_n =\frac{(-1)^{n+1}\cdot 2^{\left\lfloor\frac23(n-1)\right\rfloor}}{(n+\left\lfloor\frac13n\right\rfloor +2)!} $$ Letting $b_n=a_{3n-2}+a_{3n-1}+a_{3n}$, we find $$\begin{align}b_n&=(-1)^{n+1}\cdot\left(\frac{ 2^{2n-2}}{(4n-1)!}-\frac{2^{2n-2}}{(4n)!}+\frac{ 2^{2n-1}}{(4n+2)!} \right)\\ &=\frac{(-1)^{n+1}4^{n-1}}{(4n+2)!}\cdot\bigl(4n(4n+1)(4n+2)-(4n+1)(4n+2)+2\bigr)\end{align} $$ To treat this, it may be most efficient to investigate the general case of $\sum_n\frac{c^nn^k}{(4n+r)!}$