For an integer $n >2$ show that $$\frac{1}{2}\sqrt{1+\frac{1}{n}} < \sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}$$.

I tried to integrate the RHS and got $$2n\left(\sqrt{1+\frac{1}{n}}-1\right)$$ But do not know how to proceed further.

Can someone help?


We have: $RHS \geq \dfrac{n}{\sqrt{n^2+n}}= \dfrac{\sqrt{n}}{\sqrt{n+1}}> \dfrac{\sqrt{n+1}}{2\sqrt{n}}=LHS$ is quite clear since $2n > n+1$ is true.

  • $\begingroup$ How did you get $RHS >=\frac{n}{\sqrt{n^2+n}}$ $\endgroup$ – Satish Ramanathan May 15 '15 at 0:23
  • $\begingroup$ $RHS > \displaystyle \sum_{k=1}^n \dfrac{1}{\sqrt{n^2+n}}= \dfrac{n}{\sqrt{n^2+n}}$ $\endgroup$ – DeepSea May 15 '15 at 0:24

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