# What is <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeXAtom-ORD"

What is $\underset{n\to \mathrm{\infty }}{lim}n\left(\sum _{k=1}^{n}\frac{1}{\sqrt{{n}^{2}+k}}-1\right)$ ?
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Ashley Parks
We can rewrite the limit as
$\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n}\frac{n}{\sqrt{{n}^{2}+k}}-n=\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n}\frac{n-\sqrt{{n}^{2}+k}}{\sqrt{{n}^{2}+k}}=\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n}\frac{1-\sqrt{1+\frac{k}{{n}^{2}}}}{\sqrt{1+\frac{k}{{n}^{2}}}}$
Since the numerator approaches zero we will rationalize and continue to simplify
$\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n}\frac{-\frac{k}{{n}^{2}}}{\sqrt{1+\frac{k}{{n}^{2}}}+1+\frac{k}{{n}^{2}}}=\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n}\frac{-\frac{k}{n}}{\sqrt{1+\frac{k}{{n}^{2}}}+1+\frac{k}{{n}^{2}}}\cdot \frac{1}{n}$
We can sandwich this complicated expression with two other limits
$\underset{n\to \mathrm{\infty }}{lim}-\frac{1}{2}\sum _{k=1}^{n}\frac{k}{n}\cdot \frac{1}{n}<\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n}\frac{-\frac{k}{n}}{\sqrt{1+\frac{k}{{n}^{2}}}+1+\frac{k}{{n}^{2}}}\cdot \frac{1}{n}<\underset{n\to \mathrm{\infty }}{lim}\frac{-1}{\sqrt{1+\frac{1}{n}}+1+\frac{1}{n}}\sum _{k=1}^{n}\frac{k}{n}\cdot \frac{1}{n}$
Both of which approach the Riemann sum
$⟶-\frac{1}{2}{\int }_{0}^{1}x\phantom{\rule{mediummathspace}{0ex}}dx=-\frac{1}{4}$
Thus the limit is $\overline{)-\frac{1}{4}}$ by squeeze theorem.
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malalawak44
We have
$n\left(\sum _{k=1}^{n}\frac{1}{\sqrt{{n}^{2}+k}}-1\right)=n\sum _{k=1}^{n}\left(\frac{1}{\sqrt{{n}^{2}+k}}-\frac{1}{n}\right)=n\sum _{k=1}^{n}\frac{n-\sqrt{{n}^{2}+k}}{n\sqrt{{n}^{2}+k}}\phantom{\rule{0ex}{0ex}}=\sum _{k=1}^{n}\frac{-k}{\sqrt{{n}^{2}+k}\left(n+\sqrt{{n}^{2}+k}\right)}$
and, for $1\le k\le n$
$\frac{1}{\left(n+1\right)\left(2n+1\right)}\le \frac{1}{\sqrt{{n}^{2}+k}\left(n+\sqrt{{n}^{2}+k}\right)}\le \frac{1}{n\left(2n\right)}$
while $\sum _{k=1}^{n}k=n\left(n+1\right)/2$. Thus (momentarily dropping the minus sign in the numerator $-k$) we have
$\frac{n}{2\left(n+1\right)\left(2n+1\right)}\le \sum _{k=1}^{n}\frac{k}{\sqrt{{n}^{2}+k}\left(n+\sqrt{{n}^{2}+k}\right)}\le \frac{\left(n+1\right)}{4n}$
so by the squeeze theorem (and putting the minus sign back in), we have
$\underset{n\to \mathrm{\infty }}{lim}n\left(\sum _{k=1}^{n}\frac{1}{\sqrt{{n}^{2}+k}}-1\right)=-\frac{1}{4}$