3
$\begingroup$

I'm stuck on computing the sum of

\begin{align*} \sum\limits_{k=0}^n \frac{(-1)^k}{k! (2k+1)} \frac{1}{(n-k)!} \end{align*}

I tried some manipulations which include

\begin{align*} \frac{1}{n!} \binom{n}{k} = \frac{1}{k! (n-k)!} \end{align*}

but still that $2k+1$ at the denominator complicates things. By the way, wolframalpha says that

\begin{align*} \sum\limits_{k=0}^n \frac{(-1)^k}{k! (2k+1)} \frac{1}{(n-k)!} = \frac{\sqrt{\pi}}{2(n+\frac{1}{2})!} \end{align*}

for $n\geq 1$.

Can anyone help me?

$\endgroup$
2
$\begingroup$

Start with the binomial theorem: $$\sum_{k=0}^n \binom{n}{k}x^n=(x+1)^n$$ Substitute $x=y^2$: $$\sum_{k=0}^n \binom{n}{k}y^{2k}=(y^2+1)^n$$ Integrate both sides: $$\sum_{k=0}^n \binom{n}{k}\frac{y^{2k+1}}{2k+1}=\int_0^y(t^2+1)^ndt$$ Divide across by $n!$: $$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{y^{2k+1}}{2k+1}=\frac{1}{n!}\int_0^y(t^2+1)^ndt$$ Let $y=i$: $$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{i(-1)^k}{2k+1}=\frac{1}{n!}\int_0^i(t^2+1)^ndt$$ $$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{i(-1)^k}{2k+1}=\frac{1}{n!}\int_0^1 i(1-t^2)^ndt$$ $$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{(-1)^k}{2k+1}=\frac{1}{2 n!}\int_0^1 t^{-1/2}(1-t)^ndt$$ Use Euler's Beta Function: $$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{(-1)^k}{2k+1}=\frac{1}{2 n!}\frac{\Gamma(n+1)\Gamma(1/2)}{\Gamma(n+3/2)}$$ $$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{(-1)^k}{2k+1}=\frac{1}{2 n!}\frac{n!2^{n+1}}{(2n+1)!!}$$ $$\color{green}{\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{(-1)^k}{2k+1}=\frac{2^{n}}{(2n+1)!!}}$$

$\endgroup$
  • $\begingroup$ Shouldn't it be $\sum_{k=0}^n \binom{n}{k}x^k=(x+1)^n$ instead of $\sum_{k=0}^n \binom{n}{k}x^n=(x+1)^n$? Then all the $n$'s in the formulae that lead to $(-1)^n/(2n+1)$ should be $k$ to have $(-1)^k/(2k+1)$... $\endgroup$ – the_eraser Dec 23 '17 at 16:56
  • $\begingroup$ @the_eraser Yes, that's right, sorry for the typo! $\endgroup$ – Frpzzd Dec 23 '17 at 17:13
2
$\begingroup$

Use \begin{eqnarray*} \int_{0}^{1} x^{2k} dx =\frac{1}{2k+1}. \end{eqnarray*} interchange the order the integral and plum \begin{eqnarray*} \sum_{k=0}^{n} \frac{(-1)^k}{(2k+1) k! (n-k)!} &=& \frac{1}{n!} \int_{0}^{1} \sum_{k=0}^{n}x^{2k} \binom{n}{k} \\ &=& \frac{1}{n!} \int_{0}^{1} (1-x^2)^n dx \\ \end{eqnarray*} It is well known that \begin{eqnarray*} \int_{0}^{1} (1-x^{2})^n dx =\frac{(2n)!!}{(2n+1)!!}. \end{eqnarray*} Substituting this gives \begin{eqnarray*} \sum_{k=0}^{n} \frac{(-1)^k}{(2k+1) k! (n-k)!} = \color{red}{\frac{2^{2n} n! }{(2n+1)!}}. \end{eqnarray*}

$\endgroup$
  • $\begingroup$ if I'm not wrong $\frac{1}{n!} \int_{0}^{1} \sum_{k=0}^{n}x^{2k} \binom{n}{k}$ should be $\frac{1}{i n!} \int_{0}^{i} \sum_{k=0}^{n}x^{2k} \binom{n}{k}$... $\endgroup$ – the_eraser Dec 23 '17 at 17:46
0
$\begingroup$

You can consider the generating function $\displaystyle f(x)=e^{x^2}\int^x_0e^{-t^2}\,dt$ (that was your start, wasn't it?) and derive the differential equation $$f'(x)=2\,x\,f(x)+1.$$ The ansatz $$f(x)=\sum^\infty_{n=0}a_n\,x^n$$ gives $a_0=f(0)=0$, $a_1=f'(0)=1$ and $$n\,a_n=2\,a_{n-2}$$ for $n\ge2$. This means $a_{2k}=0$ and $\displaystyle a_{2k+1}=\frac{2^k}{(2k+1)!!}$ for $k\ge0$.

$\endgroup$
  • $\begingroup$ I do not understand how you deduce that $a_{2k}=0$ and $\displaystyle a_{2k+1}=\frac{2^k}{(2k+1)!!}$ for $k\geq 0$... anyway you reverse-engineered my mind :D $\endgroup$ – the_eraser Dec 23 '17 at 18:01
  • $\begingroup$ ok don't mind, got it! $\endgroup$ – the_eraser Dec 23 '17 at 18:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.