# find the $\sum\limits_{k=0}^n \frac{(-1)^k}{k! (2k+1)} \frac{1}{(n-k)!}$

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?

## 3 Answers

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: $$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{(-1)^k}{2k+1}=\frac{1}{2 n!}\frac{\Gamma(n+1)\Gamma(1/2)}{\Gamma(n+3/2)}$$ $$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{(-1)^k}{2k+1}=\frac{1}{2 n!}\frac{n!2^{n+1}}{(2n+1)!!}$$ $$\color{green}{\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{(-1)^k}{2k+1}=\frac{2^{n}}{(2n+1)!!}}$$

• 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 $(-1)^k/(2k+1)$... – the_eraser Dec 23 '17 at 16:56
• @the_eraser Yes, that's right, sorry for the typo! – Frpzzd Dec 23 '17 at 17:13

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*}

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

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 $k\ge0$.

• 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 – the_eraser Dec 23 '17 at 18:01
• ok don't mind, got it! – the_eraser Dec 23 '17 at 18:04