The sum

[MathBuff]

hello, can somebody help me to solve this problem?

 

[MarkFL]

I would write the sum as:

[MATH]S_n=\sum_{k=1}^{n}\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}}[/MATH]
Let's rationalize the general summand:

[MATH]\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}}\cdot\frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{(k+1)\sqrt{k}-k\sqrt{k+1}}=\frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{(k+1)^2k-k^2(k+1)}=\frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{k(k+1)}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}[/MATH]
Hence:

[MATH]S_n=\sum_{k=1}^{n}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)[/MATH]
Now you should be able to get the sum to telescope. Can you proceed?

 

[MathBuff]

Really thank you for the answer but i still dont know what to do next

 

[MarkFL]

I would next write:

[MATH]S_n=\sum_{k=1}^{n}\frac{1}{\sqrt{k}}-\sum_{k=1}^{n}\frac{1}{\sqrt{k+1}}[/MATH]
Re-index the second sum:

[MATH]S_n=\sum_{k=1}^{n}\frac{1}{\sqrt{k}}-\sum_{k=2}^{n+1}\frac{1}{\sqrt{k}}[/MATH]
Now strip off the first term of the first sum and the last term of the second sum:

[MATH]S_n=1+\sum_{k=2}^{n}\frac{1}{\sqrt{k}}-\sum_{k=2}^{n}\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{n+1}}[/MATH]
Now the sums add to zero, and you may conclude what?

 

[Steven G]

Really thank you for the answer but i still dont know what to do next

You need to do some investigating on your own. Try to write out the first 8 terms and you will see something special happening.

 

[MarkFL]

To follow up:

[MATH]S_n=1-\frac{1}{\sqrt{n+1}}[/MATH]