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Here's a problem that I think I've managed to solve: $$\lim_{n\rightarrow\infty} \sqrt{2^n+n^2}-\sqrt{2^n+1}$$

Here's how I did it:

$(\sqrt{2^n+n^2}-\sqrt{2^n+1})\frac{\sqrt{2^n+n^2}+\sqrt{2^n+1}}{\sqrt{2^n+n^2}+\sqrt{2^n+1}}=\frac{n^2-1}{\sqrt{2^n+n^2}+\sqrt{2^n+1}}$

And now you can see that (for sufficiently large $n$):

$0\le\frac{n^2-1}{\sqrt{2^n+n^2}+\sqrt{2^n+1}}\le\frac{n^2}{\sqrt{2^n}}$

Limit of $0$ clearly is $0$, and this is also true for $\frac{n^2}{\sqrt{2^n}}$, so that means that: $$\lim_{n\rightarrow\infty} \sqrt{2^n+n^2}-\sqrt{2^n+1}=0$$

The thing is I got kind of lucky I've solved it and I'm interested if there are any other ways to solve this limit (not involving any derivatives nor integrals). Does anybody see any clever solution other than mine? And also - is my solution correct?

qiubit
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    Yes, your solution is correct. Another way to find the limit in similar situations is to use $$1 + \frac{x}{2} - \frac{x^2}{8} \leqslant \sqrt{1+x} \leqslant 1 + \frac{x}{2}$$ for $0 \leqslant x \leqslant 1$, and write the roots as $\sqrt{2^n}\cdot \sqrt{1 + a\cdot 2^{-n}}$ with $a = n^2$ resp $a = 1$. – Daniel Fischer Nov 02 '14 at 13:10
  • Your idea seems to be supported by the Squeeze Theorem.

    The other way would be by taking the ratio between the two functions.

     – 0 kelvin Nov 02 '14 at 13:32
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    @0kelvin: the ratio won't allow you to conclude. take $a_n=n^2+n$, $b_n=n^2$: the ratio goes to $1$, the difference to infinity. – Clement C. Nov 02 '14 at 13:38
  • Let $$f(n)=\sqrt{2^n+n^2}-\sqrt{2^n+1},,,,n\in\mathbb{N}.$$ Then we want to find the limit $$\lim_{n\to\infty}f(n).$$ $f(1)=0$ and it is not hard to see that $f(n)\ge0,,,,\forall n\ge1.$
    I have note that $f$ has its maximum at $n=7$ and it is strictly decreasing for $n\ge8.$ Sometimes this may leads you to a new proof of your result.     – Bumblebee Nov 02 '14 at 14:19

3 Answers3

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$$\sqrt{2^n+n^2}-\sqrt{2^n+1}=2^{n/2} (\sqrt{1+n^22^{-n}}-\sqrt{1+2^{-n}})=2^{n/2}\left(1+\frac12n^22^{-n}+o(n^22^{-n})-1-\frac122^{-n}-o(2^{-n})\right)=\frac{n^2-1}22^{-n/2}+o(n^22^{-n/2}).$$ The expression converges to $0$.

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If you know about Taylor expansions, you can do the following: as long as $\frac{a_n}{2^n}\xrightarrow[n\to\infty]{}0$, $$ \sqrt{2^n + a_n} = \sqrt{2^n}\sqrt{1 + \frac{a_n}{2^n}} = \sqrt{2^n}\left(1+\frac{a_n}{2\cdot2^n}+o\!\left(\frac{a_n}{2^n}\right)\right) = 2^{\frac{n}{2}}+\frac{a_n}{2^{\frac{n}{2}+1}}+o\!\left(\frac{a_n}{2^n}\right) $$ Applying this to your case with first $a_n=n^2$, then $a_n=1$, you get that the difference is $$ 2^{\frac{n}{2}}+\frac{n^2}{2^{\frac{n}{2}+1}}+o\!\left(\frac{n^2}{2^n}\right) - 2^{\frac{n}{2}} $$ as $\frac{1}{2^n} = o\!\left(\frac{n^2}{2^n}\right)$. I.e., the difference is equivalent to $\frac{n^2}{2^{\frac{n}{2}+1}}$, which goes to 0.

Clement C.
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You can rewrite your expression as $$ \frac{n^2-1}{2^{n/2}}\frac{1}{\sqrt{1+n^22^{-n}}+\sqrt{1+2^{-n}}} $$ and the second fraction has limit $1/2$. The first fraction has limit $0$.

For both results you just need to know that $$ \lim_{n\to\infty}\frac{P(n)}{k^n}=0 $$ for $k>1$ and $P$ a polynomial. This can be done in several ways, the easiest one is induction using l'Hôpital's theorem.

egreg
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