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Need to show that

$$ \sum_{r=1}^n \operatorname{sech}(r) < \int_0^n \operatorname{sech}(x) \mathrm{d} x $$ which is done and got to

$$ \sum_{r=1}^n \operatorname{sech}(r) < \tan^{-1} \left( \sinh(n) \right) $$

Then we need to find an upper bound for the summation. and that's what I got stuck.

I can only get the "sense" that the graph of $\sinh$ goes to $\infty$, so that the $\tan^{-1}$ is kinda of at the aysmpote, which is $\pi/2$.

But is there a formal way to see this?

CasperYC
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1 Answers1

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For $\cosh$ is strictly increasing (and positive) over $[0,+\infty)$, $\mathrm{sech} = \frac{1}{\cosh}$ is strictly decreasing over $[0,+\infty)$. Hence, $$ \forall k \geqslant 0 \text{ integer}, \quad \int_k^{k+1}\mathrm{sech}(t)dt > \mathrm{sech}(k+1) \cdot (k+1-k) = \mathrm{sech}(k+1). $$ To conclude, sum over $k\in [\![0,\ldots,n-1]\!]$.

Didier
  • 19,132