haugmbd

2022-02-23

Evaluate:
$\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n}\frac{1}{n+\sqrt{{k}^{2}+n}}$

Ayda Cannon

$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\sum _{k=1}^{n}\frac{1}{1+\sqrt{{\left(\frac{k}{n}\right)}^{2}+\frac{1}{n}}}=\underset{n\to \mathrm{\infty }}{lim}\underset{m\to \mathrm{\infty }}{lim}\frac{1}{n}\sum _{k=1}^{n}\frac{1}{1+\sqrt{{\left(\frac{k}{n}\right)}^{2}+\frac{1}{m}}}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\sum _{k=1}^{n}\frac{1}{1+\left(\frac{k}{n}\right)}$
$={\int }_{0}^{1}\frac{dx}{1+x}$
$=\mathrm{log}2$
The relevant theorem is if ${x}_{nm}\to {y}_{n}$ unformly as $m\to \mathrm{\infty }$ and ${x}_{nm}$ converges as $n\to \mathrm{\infty }$, then the double limit exists and
$\underset{n\to \mathrm{\infty }}{lim}{x}_{\cap }=\underset{n\to \mathrm{\infty }}{lim}\underset{m\to \mathrm{\infty }}{lim}{x}_{nm}=\underset{m\to \mathrm{\infty }}{lim}\underset{n\to \mathrm{\infty }}{lim}{x}_{nm}$
To show uniform convergence in this case, note that
$|\frac{1}{n}\sum _{k=1}^{n}\frac{1}{1+\sqrt{{\left(\frac{k}{n}\right)}^{2}+\frac{1}{m}}}-\frac{1}{n}\sum _{k=1}^{n}\frac{1}{1+\frac{k}{n}}|\le \frac{1}{n}\sum _{k=1}^{n}\frac{\mid 1+\frac{k}{n}-\left(1+\sqrt{{\left(\frac{k}{n}\right)}^{2}+\frac{1}{m}}\right\}\left\{|1+\frac{k}{n}||1+\sqrt{{\left(\frac{k}{n}\right)}^{2}+\frac{1}{m}}|\right\}}{}$
$\le \frac{1}{n}\sum _{k=1}^{n}\left(\sqrt{{\left(\frac{k}{n}\right)}^{2}-\frac{1}{m}}-\frac{k}{n}\right)$

Randall Odom

Squeezing:
$\frac{1}{1+\sqrt{{\left(\frac{k}{n}\right)}^{2}+\frac{1}{n}}}<\frac{1}{1+\sqrt{{\left(\frac{k}{n}\right)}^{2}+0}}=\frac{1}{1+\frac{k}{n}}$
And it is obvious that:
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\sum _{k=1}^{n}\frac{1}{1+\frac{k}{n}}={\int }_{0}^{1}\frac{1}{1+x}dx=\mathrm{ln}\left(2\right)$
We then notice that:
$\frac{1}{1+\sqrt{{\left(\frac{k}{n}\right)}^{2}+\frac{1}{n}}}>\frac{1}{1+\frac{k}{n}+\frac{1}{2k}}>\frac{1-\frac{1}{2k}}{1+\frac{k}{n}}>\frac{1}{1+\frac{k}{n}}-\frac{1}{2k}$
And finally notice that:
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\sum _{k=1}^{n}\frac{1}{1+\frac{k}{n}}-\frac{1}{2k}\ge \mathrm{ln}\left(2\right)$
which follows easily since
$\frac{1}{n}\sum _{k=2}^{n}\frac{1}{k}<\frac{1}{n}{\int }_{1}^{n}\frac{1}{x}dx=\frac{\mathrm{ln}n}{n}$
and this limits may be taken care of by LHospitals rule.

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