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$$\lim_{n\to\infty}\left(\sqrt{n^2+n+1}-\left\lfloor\sqrt{n^2+n+1}\right\rfloor\right)\;=\;?\quad(n\in I) \\ \text{where $\lfloor\cdot\rfloor$ is the greatest integer function.}$$


This is what I did:

Since $[x] = x - \{x\}$ we get our limit equal to $$\lim_{n\to\infty}\left\{\sqrt{n^2+n+1}\right\}$$ Moving the limit inside the fractional part function and replacing $n=\frac 1h \; \text {where } h\to0^+$ we get $$\left\{\lim_{h\to0^+} \frac{\sqrt{h^2+h+1}}h\right\}$$

Applying L'Hospital Rule, we get our limit equal to $\left\{\frac 12\right\}$ which is $0$.


The problem:

The answer in the answer key is $\frac12$. So here, the only problem I seem to find in my solution is that $n\in I$ and simply assuming $n = \frac 1h$ doesn't ensure our $n$ to be an integer.

Can anyone provide a way to either correctly assume a new value for $n$ or any alternate way to solve this?

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  • $\begingroup$ i think the limit is zero $\endgroup$ – Bhaskara-III Feb 8 '16 at 12:47
  • $\begingroup$ You can't just put the limit inside the $\{\cdot\}$ function. $n-1/n$ does not have a limit, but $\{n-1/n\}$ does have a limit. $\endgroup$ – Thomas Andrews Feb 8 '16 at 12:50
  • $\begingroup$ But more importantly, $\{\cdot\}$ is not continuous, so $a_n\to a$ does not mean $\{a_n\}\to \{a\}$. $\endgroup$ – Thomas Andrews Feb 8 '16 at 12:50
  • $\begingroup$ You also can't use L'Hopital, beucase that limit is of the form $\frac{1}{0}$, and you'd need an indeterminate form to apply L'Hopital. $\endgroup$ – Thomas Andrews Feb 8 '16 at 12:52
  • $\begingroup$ @ThomasAndrews $\{\cdot\}$ is continuous at $\frac12$ so moving the limit inside is not wrong. $\endgroup$ – Pratyush Yadav Feb 8 '16 at 13:09
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For every natural number, we have $$\lfloor \sqrt{n^2+n+1} \rfloor =n$$ because $n^2\leq n^2+n+1< n^2+2n+1=(n+1)^2$. So we get $$\begin{align}\lim_{n\to\infty} (\sqrt{n^2+n+1}- \lfloor \sqrt{n^2+n+1} \rfloor)&=\lim_{n\to\infty}(\sqrt{n^2+n+1}-n)\\ &=\lim_{n\to\infty}\frac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n}\\ &=\lim_{n\to\infty}\frac{n+1}{\sqrt{n^2+n+1}+n}\\ &=\lim_{n\to\infty}\frac{1+\frac{1}{n}}{\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}+1}\\ &=\frac{1}{2} \end{align}$$

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  • $\begingroup$ Wow. Didn't think it would be this easy. Can you give some elaboration on why the floor function becomes equal to n ? It is intuitively obvious but i don't know how to prove it. $\endgroup$ – user230452 Feb 8 '16 at 22:42
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    $\begingroup$ @user230452 didn't he give the elaboration on the next line? In the inequality, since n is positive. So $n^2+n+1$ lies between $n^2$ and $n^2 + 2n + 1$. And since $\lfloor x\rfloor=$ the greatest integer $\leq x$, the answer is n. $\endgroup$ – Max Payne Feb 9 '16 at 12:05
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Alternative idea for proving the essential facts: writing $$\sqrt{n^2+n+1} = n\sqrt{1+1/n+1/n^2}$$ and using the Taylor series of $\sqrt{1+x}$: $$1+{\frac{x}2}-{\frac{x^2}8}+O(x^3)$$ we have $$\sqrt{n^2+n+1} = n+\frac12+\frac3{8n}+O(1/n^2)$$ and $$\lfloor\sqrt{n^2+n+1}\rfloor = n.$$

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  • $\begingroup$ Ah, thanks for contributing, but I am in high school and Taylor Series is not in my syllabus, so I have very little idea about it. I'm sure someone else would find it useful $\endgroup$ – Pratyush Yadav Feb 10 '16 at 16:55
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The crux of this is that $\sqrt{n^2+n+1}\approx n+\frac{1}{2}$ for large $n$.

Note that $$n^2+n+1=\left(n+\frac{1}{2}\right)^2+\frac{3}{4}\implies\sqrt{n^2+n+1}-\left(n+\frac{1}{2}\right)=\frac{\frac{3}{4}}{n+\frac{1}{2}+\sqrt{n^2+n+1}}$$

One can thus quickly see that for large enough $n$, $n<\sqrt{n^2+n+1}<n+1$, so $\left\lfloor\sqrt{n^2+n+1}\right\rfloor=n$ eventually.

Thus we can see that \begin{align}\sqrt{n^2+n+1}-\lfloor \sqrt{n^2+n+1}\rfloor&=\sqrt{n^2+n+1}-n\\ &=\sqrt{n^2+n+1}-\left(n+\frac{1}{2}\right)+\frac{1}{2}\\ &=\frac{\frac{3}{4}}{n+\frac{1}{2}+\sqrt{n^2+n+1}}+\frac{1}{2}\\ &\to\frac{1}{2}\end{align}

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