My Math Forum approximation for the sum

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 December 26th, 2011, 05:05 PM #1 Member   Joined: Aug 2011 Posts: 71 Thanks: 0 approximation for the sum Find approximation for the folowing sum (not use Stirling's formula): $\sum_{k=1}^n(\frac{1}{35})^{k-1}\frac{\Gamma(n-k+1)\Gamma(k-\frac12)}{\Gamma(n+\frac12)}$
 December 26th, 2011, 07:11 PM #2 Member   Joined: Aug 2011 Posts: 71 Thanks: 0 Re: approximation for the sum Ok. I've tried with Stirling's formula as well. But then I have a problem to calculate the sum...
 December 27th, 2011, 01:29 PM #3 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 Re: approximation for the sum I suggest you try using ?(n+1/2)=(n-1/2)(n-3/2)...?(1/2) and similarly for ?(k-1/2). Also ?(n-k+1)=(n-k)!
 December 27th, 2011, 01:52 PM #4 Member   Joined: Aug 2011 Posts: 71 Thanks: 0 Re: approximation for the sum I have got: $\frac{(2n-1)}{4(n-\frac12)}\sum_{k=1}^n35^{1-k}(n-k-\frac12)!(k-2)!$ now using Stirling's approximation: $\sum_{k=1}^n35^{1-k}(n-k-\frac12)!(k-2)!=\frac{2\pi}{e^{n-2}}\sum_{k=1}^n35^{1-k}(n-k)^{n-k}(k-2)^{k-3/2}$
 December 27th, 2011, 01:53 PM #5 Member   Joined: Aug 2011 Posts: 71 Thanks: 0 Re: approximation for the sum But I still have no idea how to approximate the last sum...
 December 28th, 2011, 01:11 PM #6 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 Re: approximation for the sum I can't help you with your approach. I suggest you try using the information I gave you. With a little manipulation you will have no half integer gammas. You will have a lot of factorals and powers of 2. It may not work, but it will be clearer.

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### summation i^k stirling

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