What is the closed form for

$$\sum_{n=0}^{\infty}{(-1)^n\left({{\pi\over 2}}\right)^{2n}\over (2n+k)!}=F(k)?$$

My try:

I have found a few values of $F(k)$, but was unable to find a closed form for it.

$F(0)=0$

$F(1)={2\over \pi}$

$F(2)=\left({2\over \pi}\right)^2$

$F(3)=\left({2\over \pi}\right)^2-\left({2\over \pi}\right)^3$

$F(4)={1\over 2}\left({2\over \pi}\right)^2-\left({2\over \pi}\right)^4$

$F(5)={1\over 6}\left({2\over \pi}\right)^2-\left({2\over \pi}\right)^4+\left({2\over \pi}\right)^5$

  • I included the closed form of the sum. Indeed the hypergoemetric function can be simplified. – Zaid Alyafeai Jan 3 '17 at 5:42
up vote 8 down vote accepted

Note that

$$\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)! } = \cos(x)$$

You can reach the result by integrating $k$-times.

$$\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n+k)! } = \frac{1}{x^{k}}\int^{x}_0 \mathrm{d}t_{k-1}\int^{t_{k-1}}_0 \mathrm{d}t_{k-2} \cdots\int^{t_1}_0\mathrm{d}t_0\cos(t_0) $$

For example when $k=1$

$$\sum_{n=0}^\infty \frac{(-1)^n(\pi/2)^{2n}}{(2n+1)! } = \frac{2}{\pi}\int^{\pi/2}_0 \mathrm{d}t_0 \cos(t_0)\,= \frac{2}{\pi}$$

For $k=2$

\begin{align} \sum_{n=0}^\infty \frac{(-1)^n(\pi/2)^{2n}}{(2n+2)! } &= \left( \frac{2}{\pi}\right)^2\int^{\pi/2}_0 \mathrm{d}t_{1}\int^{t_1}_0\mathrm{d}t_0\cos(t_0) \\ &= \left( \frac{2}{\pi}\right)^2\int^{\pi/2}_0 \mathrm{d}t_{1}\sin(t_1)\\ & = \left( \frac{2}{\pi}\right)^2 \end{align}

For

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

We can define recursively

$$f_0(x) = \cos(x)$$

$$f_k(x) =\int^x_0 f_{k-1}(t)\,dt$$

By solving the recursive formula

$$\sum_{n=0}^\infty \frac{(-1)^n x^{2n+k}}{(2n+k)!} = \begin{cases} \sum_{n=0}^{\lceil k/2 \rceil-2}\frac{x^{2n+1}(-1)^{\lceil k/2 \rceil+n}}{(2n+1)!}-(-1)^{\lceil k/2 \rceil}\sin(x) & \text{If $k$ is odd} \\ \sum_{n=0}^{k/2-1}\frac{x^{2n}(-1)^{n+k/2}}{(2n)!}-(-1)^{k/2}\cos(x) & \text{If $k$ is even}\end{cases}$$

Finally we have the closed form

$$F(k) = \begin{cases}\left( \frac{2}{\pi}\right)^k \sum_{n=0}^{\lceil k/2 \rceil-2}\frac{(\pi/2)^{2n+1}(-1)^{\lceil k/2 \rceil+n}}{(2n+1)!}-\left( \frac{2}{\pi}\right)^k(-1)^{\lceil k/2 \rceil} & \text{If $k$ is odd} \\ \left( \frac{2}{\pi}\right)^k\sum_{n=0}^{k/2-1}\frac{(\pi/2)^{2n}(-1)^{n+k/2}}{(2n)!}& \text{If $k$ is even}\end{cases}$$

To check the correctness of the formula

For $k=5$ we have

$$F(5)=\left(\frac{2}{\pi}\right)^5 \sum_{n=0}^{1}\frac{(\pi/2)^{2n+1}(-1)^{3+n}}{(2n+1)!}-\left( \frac{2}{\pi}\right)^5(-1)^{3} = {1\over 6}\left({2\over \pi}\right)^2-\left({2\over \pi}\right)^4+\left({2\over \pi}\right)^5$$

  • 1
    interesting derivation, but seems that there is something missing in the recursive formula at the end. – G Cab Jan 2 '17 at 17:15
  • @GCab, I made an edit. – Zaid Alyafeai Jan 2 '17 at 17:48
  • does not look sufficient, as it is it will cycle around $cos(x)$, $sin(x)$. Shouldn't there be a $1/x$ in front of the integral ? – G Cab Jan 2 '17 at 23:29
  • @GCab, no it will not cycle. $\int^x_0 \sin (t)dt =1-\cos (x)$ . – Zaid Alyafeai Jan 3 '17 at 4:21
  • @GCab, I included the closed form of the formula. – Zaid Alyafeai Jan 3 '17 at 5:41

Your sum is the special value at $x=(-(\pi/2)^2)$ of $$\sum_{n=0}^\infty \frac{x^n}{(2n + k)!}=\frac{\, _1F_2\left(1;\frac{k}{2}+\frac{1}{2},\frac{k}{2}+1;\frac{x}{4}\right)}{k!}$$

The sum can also be expressed in terms of the Regularized Incomplete Gamma function ($Q(a,z)$).
In fact, premised that for a general function of a integer $f(k)$ we have $$ \begin{gathered} \frac{{\left( {i^{\,k} + \left( { - i} \right)^{\,k} } \right)}} {2}f(k) = \left( {\frac{{e^{\,i\,k\frac{\pi } {2}} + e^{\, - \,i\,k\frac{\pi } {2}} }} {2}} \right)f(k) = \cos \left( {k\frac{\pi } {2}} \right)f(k) = \hfill \\ = \left[ {k = 2j} \right]\left( { - 1} \right)^{\,j} f(2j)\quad \left| {\;k,j\; \in \;\;\mathbb{Z}\,} \right. \hfill \\ \end{gathered} $$ and that the Lower Incomplete Gamma function can be expressed as: $$ \begin{gathered} \gamma (s,z) = \int_{\,0\,}^{\,z\,} {t^{\,s - 1} \;e^{\, - \,t} dt} = \hfill \\ = z^{\,s} \;e^{\,\, - z} \;\Gamma (s)\;\sum\limits_{0\, \leqslant \,k} {\frac{{z^{\,k} }} {{\Gamma (s + k + 1)}}} = \frac{{z^{\,s} \;e^{\,\, - z} }} {s}\sum\limits_{0\, \leqslant \,k} {\frac{{z^{\,k} }} {{\left( {s + 1} \right)^{\,\overline {\,k\,} } \,}}} = \hfill \\ = z^{\,s} \;e^{\,\, - z} \sum\limits_{0\, \leqslant \,k} {\frac{{z^{\,k} }} {{s^{\,\overline {\,k + 1\,} } \,}}} = z^{\,s} \sum\limits_{0\, \leqslant \,j} {\frac{{\left( { - 1} \right)^{\,j} }} {{\,\left( {s + j} \right)}}\frac{{z^{\,j} }} {{j!}}} \hfill \\ \end{gathered} $$ we can then write $$ \begin{gathered} F(x,m) = \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{x^{\,2k} }} {{\left( {2k + m} \right)!}}} = \hfill \\ = \frac{1} {2}\left( {\sum\limits_{0\, \leqslant \,k} {\frac{{\left( {i\,x} \right)^{\,k} }} {{\left( {k + m} \right)!}}} + \sum\limits_{0\, \leqslant \,k} {\frac{{\left( { - \,i\,x} \right)^{\,k} }} {{\left( {k + m} \right)!}}} } \right) = \hfill \\ = \frac{1} {{2\,\Gamma (m)}}\left( {\frac{{e^{\,\,i\,x} \,\gamma (m,i\,x)}} {{\left( {i\,x} \right)^{\,m} }} + \frac{{e^{\, - \,i\,x} \,\gamma (m, - i\,x)}} {{\left( { - i\,x} \right)^{\,m} }}} \right) = \hfill \\ = \frac{1} {{\left| x \right|^{\,\,m} }}\,\operatorname{Re} \left( {e^{\,\,i\,x} e^{\,\, - \,i\,m\,\left( {sign(x)\pi /2} \right)} \,\frac{{\gamma (m,i\,x)}} {{\Gamma (m)}}} \right) = \hfill \\ = \frac{1} {{\left| x \right|^{\,\,m} }}\,\operatorname{Re} \left( {e^{\,\,i\,x} e^{\,\, - \,i\,m\,\left( {sign(x)\pi /2} \right)} \,\left( {1 - Q(m,i\,x)} \right)} \right) \hfill \\ \end{gathered} $$

So, for $x=\pi /2$ we get $$ \begin{gathered} F(\pi /2,m) = \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\left( {\pi /2} \right)^{\,2k} }} {{\left( {2k + m} \right)!}}} = \hfill \\ = \left( {\frac{2} {\pi }} \right)^{\,\,m} \,\operatorname{Re} \left( {e^{\,\, - \,i\,\left( {m - 1} \right)\,\left( {\pi /2} \right)} \,\frac{{\gamma (m,i\,\pi /2)}} {{\Gamma (m)}}} \right) = \hfill \\ = \left( {\frac{2} {\pi }} \right)^{\,\,m} \,\operatorname{Re} \left( {e^{\,\, - \,i\,\left( {m - 1} \right)\,\left( {\pi /2} \right)} \,\left( {1 - Q(m,i\,\pi /2)} \right)} \right) \hfill \\ \end{gathered} $$ example with $m=5$ $$ \begin{gathered} F(\pi /2,5) = \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\left( {\pi /2} \right)^{\,2k} }} {{\left( {2k + 5} \right)!}}} = \hfill \\ = \left( {\frac{2} {\pi }} \right)^{\,\,5} \,\operatorname{Re} \left( {e^{\,\, - \,i\,4\,\left( {\pi /2} \right)} \,\left( {1 - Q(5,i\,\pi /2)} \right)} \right) = \hfill \\ = \left( {\frac{2} {\pi }} \right)^{\,\,5} \,\operatorname{Re} \,\left( {1 - \frac{1} {{24}}\left( {12\,\pi - \frac{1} {2}\pi ^{\,3} + i\left( {3\pi ^{\,2} - \frac{1} {{16}}\pi ^{\,4} - 24} \right)} \right)} \right) = \hfill \\ = \left( {\frac{2} {\pi }} \right)^{\,\,5} \left( {1 - \,\frac{\pi } {2} + \frac{1} {{48}}\pi ^{\,3} } \right) = \left( {\frac{2} {\pi }} \right)^{\,\,5} - \,\left( {\frac{2} {\pi }} \right)^{\,\,4} + \frac{1} {6}\left( {\frac{2} {\pi }} \right)^{\,2} \hfill \\ \end{gathered} $$ which matches with the value given by Zaid Alyafeai, as well as with the fomula indicated by Igor Rivin $$ \begin{gathered} F(\pi /2,5) = \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\left( {\pi /2} \right)^{\,2k} }} {{\left( {2k + 5} \right)!}}} = \left( {\frac{2} {\pi }} \right)^{\,\,5} - \,\left( {\frac{2} {\pi }} \right)^{\,\,4} + \frac{1} {6}\left( {\frac{2} {\pi }} \right)^{\,2} = \hfill \\ = 0.007860176 \cdots = \hfill \\ = \frac{1} {{5!}}{}_1F_2 \left( {1\;;\;\frac{5} {2} + \frac{1} {2},\;\frac{5} {2} + 1\;;\; - \frac{1} {4}\left( {\frac{\pi } {2}} \right)^{\,2} } \right) \hfill \\ \end{gathered} $$ and concerning the latter, a computer calculation over multiple values of $m$ and $x$ shows a full match.

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