What is $\underset{n\to \mathrm{\infty}}{lim}n(\sum _{k=1}^{n}\frac{1}{\sqrt{{n}^{2}+k}}-1)$ ?

Janet Forbes
2022-07-12
Answered

What is $\underset{n\to \mathrm{\infty}}{lim}n(\sum _{k=1}^{n}\frac{1}{\sqrt{{n}^{2}+k}}-1)$ ?

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Ashley Parks

Answered 2022-07-13
Author has **11** answers

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

$\u27f6-\frac{1}{2}{\int}_{0}^{1}x\phantom{\rule{mediummathspace}{0ex}}dx=-\frac{1}{4}$

Thus the limit is $\overline{){\displaystyle -\frac{1}{4}}}$ by squeeze theorem.

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

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

$\u27f6-\frac{1}{2}{\int}_{0}^{1}x\phantom{\rule{mediummathspace}{0ex}}dx=-\frac{1}{4}$

Thus the limit is $\overline{){\displaystyle -\frac{1}{4}}}$ by squeeze theorem.

malalawak44

Answered 2022-07-14
Author has **4** answers

We have

$n(\sum _{k=1}^{n}\frac{1}{\sqrt{{n}^{2}+k}}-1)=n\sum _{k=1}^{n}(\frac{1}{\sqrt{{n}^{2}+k}}-\frac{1}{n})=n\sum _{k=1}^{n}\frac{n-\sqrt{{n}^{2}+k}}{n\sqrt{{n}^{2}+k}}\phantom{\rule{0ex}{0ex}}=\sum _{k=1}^{n}\frac{-k}{\sqrt{{n}^{2}+k}(n+\sqrt{{n}^{2}+k})}$

and, for $1\le k\le n$

$\frac{1}{(n+1)(2n+1)}\le \frac{1}{\sqrt{{n}^{2}+k}(n+\sqrt{{n}^{2}+k})}\le \frac{1}{n(2n)}$

while $\sum _{k=1}^{n}k=n(n+1)/2$. Thus (momentarily dropping the minus sign in the numerator $-k$) we have

$\frac{n}{2(n+1)(2n+1)}\le \sum _{k=1}^{n}\frac{k}{\sqrt{{n}^{2}+k}(n+\sqrt{{n}^{2}+k})}\le \frac{(n+1)}{4n}$

so by the squeeze theorem (and putting the minus sign back in), we have

$\underset{n\to \mathrm{\infty}}{lim}n(\sum _{k=1}^{n}\frac{1}{\sqrt{{n}^{2}+k}}-1)=-\frac{1}{4}$

$n(\sum _{k=1}^{n}\frac{1}{\sqrt{{n}^{2}+k}}-1)=n\sum _{k=1}^{n}(\frac{1}{\sqrt{{n}^{2}+k}}-\frac{1}{n})=n\sum _{k=1}^{n}\frac{n-\sqrt{{n}^{2}+k}}{n\sqrt{{n}^{2}+k}}\phantom{\rule{0ex}{0ex}}=\sum _{k=1}^{n}\frac{-k}{\sqrt{{n}^{2}+k}(n+\sqrt{{n}^{2}+k})}$

and, for $1\le k\le n$

$\frac{1}{(n+1)(2n+1)}\le \frac{1}{\sqrt{{n}^{2}+k}(n+\sqrt{{n}^{2}+k})}\le \frac{1}{n(2n)}$

while $\sum _{k=1}^{n}k=n(n+1)/2$. Thus (momentarily dropping the minus sign in the numerator $-k$) we have

$\frac{n}{2(n+1)(2n+1)}\le \sum _{k=1}^{n}\frac{k}{\sqrt{{n}^{2}+k}(n+\sqrt{{n}^{2}+k})}\le \frac{(n+1)}{4n}$

so by the squeeze theorem (and putting the minus sign back in), we have

$\underset{n\to \mathrm{\infty}}{lim}n(\sum _{k=1}^{n}\frac{1}{\sqrt{{n}^{2}+k}}-1)=-\frac{1}{4}$

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