2
$\begingroup$

I am trying to attempt a homework problem but this summation pops up:

$$\sum_{m=0}^n \left(\frac{(-1)^m}{m!(m+1/2)} \times \frac{1}{(n-m)!}\right)$$

According to Mathematica,

$$\sum_{m=0}^n \left(\frac{(-1)^m}{m!(m+1/2)} \times \frac{1}{(n-m)!}\right) = \frac{\sqrt{\pi}}{(\Gamma(n+1/2))\times (n+1/2)}$$

Is there a simple way (ie by the use of some identities etc.) for me to obtain the expression on the right? I have tried to view the summation as some sort of a binomial expansion but to no avail. My guess that that is this identity would not be "standard" in the sense that the right-hand side involves some sort of $\Gamma(n+1/2)$ and $\pi$. Thanks in advance!

$\endgroup$
1
  • $\begingroup$ Mathematica uses Gamma where semi-factorials could do. $\endgroup$
    – user65203
    Commented Mar 10, 2020 at 8:24

1 Answer 1

2
$\begingroup$

Your expression is :

$$ \frac{1}{n!}\sum_0^n \binom {n} {m}\int_0^1 (-1)^m t^{m-1/2}dt=\frac{1}{n!}\int_0^1(1-t)^n t^{-1/2}dt$$

Now use

$$B(x,y)=\int_{0}^{1}t^{x-1}(1-t)^{y-1}dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

$\endgroup$
2
  • $\begingroup$ Thank you very much! On a side note, is there any good references that I can use for techniques to evaluate expressions related to functions like the gamma functions etc? $\endgroup$
    – HK Tan
    Commented Mar 10, 2020 at 11:12
  • $\begingroup$ You can see the page of wikipedia on gamma function, etc... $\endgroup$
    – Kelenner
    Commented Mar 10, 2020 at 12:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .