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 $\infty$?

  • $\begingroup$ This integral does have a solution but it's a rather long one. Are you su're you aren't missing anything ? $\endgroup$ – 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.

Can you take it from here?

  • $\begingroup$ Yes, and it is $\int_{0}^{1}\frac{1}{1+x}dx=ln2$, no? $\endgroup$ – user651692 Mar 9 at 8:14
  • $\begingroup$ @JacobDenicula Yes, you are correct! $\endgroup$ – Robert Z Mar 9 at 8:18

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