# Sum with binomial coefficient

• Aug 10th 2011, 01:23 PM
RSF
Sum with binomial coefficient
Trying to solve this sum, wich involves binomial coefficient. But it looks very difficult:

$\sum_{n=1}^{M-1}\left(\begin{array}{cc}M\\n\end{array}\right)$ $\sqrt{\frac{M-n}{n}}$

It is the same as:

$\sum_{n=1}^{M-1}\left(\begin{array}{cc}M\\n\end{array}\right)$ $\sqrt{\frac{n}{M-n}}$

Thanks
• Aug 10th 2011, 02:52 PM
TKHunny
Re: Sum with binomial coefficient
Do you believe it to be bounded? Either of them?
• Aug 11th 2011, 07:59 AM
RSF
Re: Sum with binomial coefficient
Comparing with:
$\sum_{n=1}^{M-1}\left(\begin{array}{cc}M\\n\end{array}\right)(M-n)=M2^{(M-1)}-2$

one finds:

$S=\sum_{n=1}^{M-1}\left(\begin{array}{cc}M\\n\end{array}\right)$ $\sqrt{\frac{M-n}{n}}$ $

if $M>2$

and for $M>4$ it looks like this sum is

$S>2^M$

so

$2^M

for $M>4$
• Aug 11th 2011, 08:29 AM
Plato
Re: Sum with binomial coefficient
When I first read the post, I did not think it was true.
After using a computer algebra to check many cases, I convinced myself it is true. I think the should be done by induction. But I have not found the identity needed to make it work.
• Aug 11th 2011, 11:00 AM
RSF
Re: Sum with binomial coefficient
Thanks TKHunny for your question, it has been usefull to establish limits to the possible outcome of the sum.
Thanks Plato for getting intrested.
Both sums are exactly the same, it requires some algebra to show this:

$\sum_{n=1}^{M-1}\left(\begin{array}{cc}M\\n\end{array}\right)$ $\sqrt{\frac{n}{M-n}}$ $=$ $\sum_{n=1}^{M-1}\left(\begin{array}{cc}M\\n\end{array}\right)$ $\sqrt{\frac{M-n}{n}}$

The question I cannot answer is expresing the sum (either of them, as they are the same) as a more compact function of M, for example:

$\sum_{n=0}^{M }\left(\begin{array}{cc}M\\n\end{array}\right)$ $=2^M$