Closed form of $\sum_{n=0}^\infty \frac{x^{(2n+1)k}}{((2n+1)k)!}$

Question

I encountered the following series and I have been unable to find a closed form, except for the values $k=1$ and $k=2$ $$\sum_{n=0}^\infty \frac{x^{(2n+1)k}}{((2n+1)k)!}$$ Is there a general form of writing this for any $k$ in terms of elementary or special functions? (being $k$ a positive integer).

Thanks in advance

Answer 1

$\frac{\tanh^{-1}(x^k)}{k!}$ is one way to write the closed form of the infinite sum with a trigonometric function.

Note: $\tanh^{-1}=\frac{e^{2z}-1}{e^{2z}+1}$.

Answer 2

These are generalized hypergeometric functions.

$$f_k(x)=\sum_{n=0}^\infty \frac{x^{(2n+1)k}}{((2n+1)k)!}$$

I do not simplify in order you see the patterns $$f_3(x)=\frac{x^3}{3!} \, _0F_5\left(;\frac{4}{6},\frac{5}{6},\frac{7}{6},\frac{8}{6},\frac{9}{6};\left (\frac{x}{6}\right)^6\right)$$ $$f_4(x)=\frac{x^4}{4!} \, _0F_7\left(;\frac{5}{8},\frac{6}{8},\frac{7}{8},\frac{9}{8},\frac{10}{8},\frac {11}{8},\frac{12}{8};\left (\frac{x}{8}\right)^8\right)$$ $$f_5(x)=\frac{x^5}{5!} \, _0F_9\left(;\frac{6}{10},\frac{7}{10},\frac{8}{10},\frac{9}{10},\frac{11}{10}, \frac{12}{10},\frac{13}{10},\frac{14}{10},\frac{15}{10};\left (\frac{x}{10}\right)^{10}\right)$$

For sure, you could write $$f_3(x)=\frac{\sinh (x)}{3}-\frac{2}{3} \cos \left(\frac{\sqrt{3} x}{2}\right) \sinh \left(\frac{x}{2}\right)$$ $$f_4(x)=\frac{1}{4} (\cos (x)+\cosh (x))-\frac{1}{2} \cos \left(\frac{x}{\sqrt{2}}\right) \cosh \left(\frac{x}{\sqrt{2}}\right)$$ I prefer not to type the next one.

The advantage with the hyeprgeomatric function is that there are very efficient codes for their computations.

Answer 3

Here is the general way of obtaining a closed formula for $\sum_{n\ge 0}a_{nk}x^{nk}$ from $f(x)=\sum_{n\ge 0}a_{n}x^{n}$.

We will make use of the following fact. Let $\omega_k$ be any $k$th primitive root of unity, say $\omega_k=e^{2\pi i/k}=\cos(2\pi/k)+i\sin(2\pi/k)$, and let $n$ be any integer. Then $$ \frac{1}{k}\sum_{j=0}^{k-1}\omega_k^{nj}= \begin{cases} 1, & \text{if} \ k\mid n,\\ 0, & \text{if} \ k\not\mid n.\\ \end{cases} $$ Obviously, this is true in the first case, because each summand is $1$. In the second case, this is true because $$ \sum_{j=0}^{k-1}\omega_j^{nj}=\frac{\omega_k^{kn}-1}{\omega_k^n-1}=\frac{1-1}{\omega_k^n-1}=0, $$ since $\omega_k^n\ne 1$ when $k\not\mid n$. Therefore, $$ \sum_{n\ge 0}a_{nk}x^{nk}=\sum_{n\ge 0}\left(\frac{1}{k}\sum_{j=0}^{k-1}\omega_k^{nj}\right)a_{n}x^{n}=\frac{1}{k}\sum_{j=0}^{k-1}\left(\sum_{n\ge 0}a_n\omega_k^{nj}x^n\right)=\frac{1}{k}\sum_{j=0}^{k-1}f(\omega_k^jx) $$ Now we just need to find the closed formula for $f(x)=\sum_{n\ge 0}\frac{x^{2n+1}}{(2n+1)!}$, but by essentially the same logic, this is the odd part of $e^x$, i.e. $$ f(x)=\frac{e^x-e^{-x}}{2}=\sinh x. $$