Sum series with square roots

[wolly]

I'm trying to prove that the LHS is equal or at least equal to the RHS and the problem is that if I try to add all the numbers from the left side I get the solution 1 and not what the right side represents.
I used a solution created by me but I'm not 100% sure it's the correct one.

 

[MarkFL]

I think you have the right idea by telescoping the series:

[MATH]S=\sum_{a=1}^{x}\left(\frac{1}{\sqrt{2a-1}}-\frac{1}{\sqrt{1a+1}}\right)=\sum_{a=1}^{x}\left(\frac{1}{\sqrt{2a-1}}\right)-\sum_{a=1}^{x}\left(\frac{1}{\sqrt{2a+1}}\right)[/MATH]
Re-index the first sum:

[MATH]S=\sum_{a=0}^{x-1}\left(\frac{1}{\sqrt{2a+1}}\right)-\sum_{a=1}^{x}\left(\frac{1}{\sqrt{2a+1}}\right)[/MATH]
Strip away the first term of the first sum and the last term of the second sum:

[MATH]S=1+\sum_{a=1}^{x-1}\left(\frac{1}{\sqrt{2a+1}}\right)-\sum_{a=1}^{x-1}\left(\frac{1}{\sqrt{2a+1}}\right)-\frac{1}{\sqrt{2x+1}}[/MATH]
The two sums add to zero, hence:

[MATH]S=1-\frac{1}{\sqrt{2x+1}}\quad\checkmark[/MATH]