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Show that for $N = 1,2,3,\dots$ we have $$\sum _{k=1} ^N \frac 1 {\sqrt {k^2 + 1} + k} > \frac 1 2 \ln \frac {2N+1} 3$$

I got this as a calculus homework. I am supposed to show this, but it doesn't say how. I have tried solving the inequality but that wasn't correct. I'm totally lost here...Can someone please explain how to compute this?

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    $\begingroup$ Please adjust your formula. $\endgroup$ – Aretino May 15 '16 at 10:47
  • $\begingroup$ Kindly edit the title $\endgroup$ – bulbasaur May 15 '16 at 10:48
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Use the fact that $$\left(\frac1{\sqrt{k^2+1}-k}\right)\left(\frac1{\sqrt{k^2+1}+k}\right)=\frac1{k^2+1-k^2}=1$$

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$$ \sum_{k=1} ^{N} \frac{1}{\sqrt{k^2+1} +k} \ge \sum_{k=1} ^{N} \frac{1}{2\sqrt{k^2+1}} \ge \sum_{k=1} ^{N} \frac{1}{2(k+\frac{1}{2})} $$ $$ \ge \frac{1}{2} \int_1 ^{N} ln(\frac{1}{2}+x)dx = \frac{1}{2}[ln(N+\frac{1}{2})-ln(\frac{3}{2})] =\frac{1}{2} ln\frac{2N+1}{3} $$ The second inequality is true for all $k\ge1 $, since $(k+\frac{1}{2})^2 = k^2 +k +\frac{1}{4} \ge k^2 +1 $

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  • $\begingroup$ The 3 is on the outside, not the inside. $\endgroup$ – Jack Lam May 15 '16 at 11:24

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