Sum with binomial coefficient

Trying to solve this sum, wich involves binomial coefficient. But it looks very difficult:

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

It is the same as:

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

Thanks

Re: Sum with binomial coefficient

Do you believe it to be bounded? Either of them?

Re: Sum with binomial coefficient

Comparing with:

$\displaystyle \sum_{n=1}^{M-1}\left(\begin{array}{cc}M\\n\end{array}\right)(M-n)=M2^{(M-1)}-2$

one finds:

$\displaystyle S=\sum_{n=1}^{M-1}\left(\begin{array}{cc}M\\n\end{array}\right)$$\displaystyle \sqrt{\frac{M-n}{n}}$$\displaystyle <M2^{(M-1)}-2$

if $\displaystyle M>2$

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

$\displaystyle S>2^M$

so

$\displaystyle 2^M<S<M2^{(M-1)}-2$

for $\displaystyle M>4$

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.

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:

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

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