# Find $\lim\limits_{n\rightarrow\infty}\sum\limits_{k=1}^{n}\frac{1}{n+\sqrt{(k^2-k+1)}}$

Find $$\lim\limits_{n\rightarrow\infty}\sum\limits_{k=1}^{n}\frac{1}{n+\sqrt{(k^2-k+1)}}$$.

I observed it is a Riemann integral and can be written as $$\frac{1}{n}\sum\limits_{k=1}^{n}\frac{1}{1+\sqrt{\left(\left(\frac{k}{n}\right)^2-\frac{k}{n^2}+\frac{1}{n^2}\right)}}$$, and for $$x_i=\frac{k}{n}$$ this is a Riemann sum. I have problems with passing to the limit as I obtain $$\int_{0}^{1}\frac{1}{1+\sqrt{x^2-\frac{x}{n}+\frac{1}{n^2}}}dx$$. can I apply this limit for the integral as to reduce the $$\frac{1}{n}$$ as n converges to $$\infty$$?

• This integral does have a solution but it's a rather long one. Are you su're you aren't missing anything ? – Rebellos Mar 8 at 19:43

If you evaluate a limit as $$n$$ goes to infinity, then the result should not depend on $$n$$.
Instead, note that $$\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}\leq \frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\sqrt{\frac{k^2}{n^2}-\frac{k-1}{n^2}}}\leq \frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\sqrt{\frac{(k-1)^2}{n^2}}}=\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{1+\frac{k}{n}}.$$ Now use the Riemann sum approach for the left-side and the right-side.
• Yes, and it is $\int_{0}^{1}\frac{1}{1+x}dx=ln2$, no? – user651692 Mar 9 at 8:14