Let be $$b_n := \sqrt{n+\sqrt{2n}}-\sqrt{n-\sqrt{2n}}, n\in\mathbb{N}$$ a sequence. I am to determine $\lim\limits_{n\to\infty}b_n$ which is obviously $\sqrt{2}$.

My first step was to transform the sequence to $-\sqrt{(-\sqrt{2}+\sqrt{n})\sqrt{n}}+\sqrt{(\sqrt{2}+\sqrt{n})\sqrt{n}}$ however this didn't helped me out. Some fellow students used some funky binomial formula to bound the limit to a specific range via the "sandwich rule" (I didn't found a suitable translation from german to english) which is defined by:

Definition. Let $a_n\leq a_n'\leq a_n''$ for nearly all $n\in\mathbb{N}$. Then one has $$\lim\limits_{n\to\infty}a_n = \lim\limits_{n\to\infty}a_n''=a\Rightarrow\lim\limits_{n\to\infty}a_n'=a.$$

I do understand the basic idea to find an expression similiar to $\sqrt{n\pm\sqrt{2n}}$ to be able to bound the limit to a range however it doesn't seem obvious to me what I can do here. My second idea was to show that $b_n$ is bounded by $[0;2)$ and then to conclude, that $\sqrt{2}$ is the proper limit.

Is anyone able to explain the idea with the binomial formula and the sandwich-rule?


My fellow students tried to prove $\lim\limits_{n\to\infty}\sqrt{n\pm\sqrt{2}}=\lim\limits_{n\to\infty}\sqrt{n\pm\sqrt{2}+0.5}=\lim\limits_{n\to\infty}\sqrt{(n\pm\sqrt{2}/2)^2}$ however I do not understand how to develop this idea. Their next step was to show $\lim\limits_{n\to\infty}b_n=\sqrt{2}$ via $\lim\limits_{n\to\infty}\left(\sqrt{n\pm\sqrt{2}}-(n\pm\sqrt{2}/2)\right)=0$. Can anyone tell me whether this is correct and explain what is the clue behind this?

  • $\begingroup$ Watch out: $b_1$ is not defined. $\endgroup$ – Did Nov 1 '11 at 11:46
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    $\begingroup$ Not a maths guy, but by no means incompetent. I have a hard time seeing the obvious about the $\sqrt{2}$ solution, a guesstimate maybe. Any clues as to why this is obvious? $\endgroup$ – Captain Giraffe Nov 1 '11 at 16:56
  • $\begingroup$ @CaptainGiraffe: By obvious I mean that if you think aboout it, there is something about this idea. Furthermore I checked the limit with Mathematica and know that $\sqrt{2}$ is the solution. $\endgroup$ – Christian Ivicevic Nov 1 '11 at 18:44

First method: use the conjugate quantity.

Multiplying and dividing $b_n$ by $\sqrt{n+\sqrt{2n}}+\sqrt{n-\sqrt{2n}}$ yields $$b_n := \frac{(n+\sqrt{2n})-(n-\sqrt{2n})}{\sqrt{n+\sqrt{2n}}+\sqrt{n-\sqrt{2n}}}=\frac{2\sqrt{2n}}{\sqrt{n+\sqrt{2n}}+\sqrt{n-\sqrt{2n}}}. $$ The denominator is equivalent to $2\sqrt{n}$ hence $b_n\sim2\sqrt{2n}\frac1{2\sqrt{n}}=\sqrt2$.

Second method: use integrals.

The derivative of $x\mapsto\sqrt{x}$ is the function $\varphi:x\mapsto1/(2\sqrt{x})$, hence $$ b_n=\int_{n-\sqrt{2n}}^{n+\sqrt{2n}}\varphi(x)\mathrm dx. $$ The function $\varphi$ is decreasing hence $\varphi(n+\sqrt{2n})\leqslant \varphi(x)\leqslant \varphi(n-\sqrt{2n})$ for every $x$ in the interval of integration. This interval has length $2\sqrt{2n}$ hence, integrating the double inequality for $\varphi(x)$, one gets $$ 2\sqrt{2n}\varphi(n+\sqrt{2n})\leqslant b_n\leqslant 2\sqrt{2n}\varphi(n-\sqrt{2n}). $$ Since(1) $\varphi(n-\sqrt{2n})\sim \varphi(n+\sqrt{2n})\sim 1/(2\sqrt{n})$, this proves that $b_n\to\sqrt2$.

Third method: use Taylor expansions (see @Phira's answer).

(1) This uses only the fact that $\sqrt{1+z}=1+o(1)$ when $z\to0$, that is, the continuity of the square root function at $1$. To wit, $1/\varphi(n\pm\sqrt{2n})=2\sqrt{n}\sqrt{1+z_n}$ with $z_n=\pm\sqrt{2/n}\to0$ hence $1/\varphi(n\pm\sqrt{2n})\sim2\sqrt{n}$.

  • $\begingroup$ You do not explain what you mean by some funky binomial formula. The second method above is based on what you call the sandwich-rule. $\endgroup$ – Did Nov 1 '11 at 11:33
  • $\begingroup$ Although I never heard of either these methods, the first one seems the most convenient and human-readable one which solves my problem here! After reading the Wikipedia article this will be the solution I will use. Thanks! $\endgroup$ – Christian Ivicevic Nov 1 '11 at 12:44

Why not just multiply/divide by the conjugate, then divide numerator/denominator by $\sqrt{n}$? $$\circ=\frac{2\sqrt{2n}}{\sqrt{n+\sqrt{2n}}+\sqrt{n-\sqrt{2n}}}=\frac{2\sqrt{2}}{\sqrt{1+\sqrt{2/n}}+\sqrt{1-\sqrt{2/n}}}\to\frac{2\sqrt{2}}{1+1}=\sqrt{2}.$$


This can be more or less considered as a variation on Phira's and Didier's answers.

You can write $$b_n = \sqrt{n+\sqrt{2n}}-\sqrt{n-\sqrt{2n}}=\sqrt{n}\left(\sqrt{1+\sqrt{\frac2n}}-\sqrt{1-\sqrt{\frac2n}}\right).$$

Then we can use the Mean Value Theorem for the function $f(x)=\sqrt{x}$ on the interval with the endpoints $1-\sqrt{\frac2n}$ and $1+\sqrt{\frac2n}$ which shows that $$b_n=\sqrt n \cdot 2\sqrt{\frac2n} \cdot \frac1{2\xi_n},$$ where $\xi_n\in(1-\sqrt{\frac2n},1+\sqrt{\frac2n})$. We see that $\xi_n\to 1$ and $$\lim\limits_{n\to\infty} b_n = \sqrt2.$$

EDIT: You description of an alternative method (suggested by your fellow students) is not entirely clear to me, but I understood it as follows:

They noticed that $n\pm\frac{\sqrt2}2=\sqrt{\left(n\pm\frac{\sqrt2}2\right)^2}=\sqrt{n\pm\sqrt{2n}+\frac12}$

In case you succeed in proving $$\lim\limits_{n\to\infty} (\sqrt{n+\sqrt{2n}}-\sqrt{n+\sqrt{2n}+\frac{\sqrt{2}}2}) = \lim\limits_{n\to\infty} (\sqrt{n-\sqrt{2n}}-\sqrt{n-\sqrt{2n}+\frac{\sqrt{2}}2})=0 \qquad (*)$$ then you can use this to show that $\lim\limits_{n\to\infty} b_n = \lim\limits_{n\to\infty} (\sqrt{n+\sqrt{2n}}-(n+\frac{\sqrt2}2) ) - $ $\lim\limits_{n\to\infty} (\sqrt{n-\sqrt{2n}}-(n-\frac{\sqrt2}2) ) + $ $\lim\limits_{n\to\infty} ((n+\frac{\sqrt2}2) - (n-\frac{\sqrt2}2) ) = \sqrt{2}$

So it remains only question is how to prove $(*)$. If you show it by some of the methods suggested for your original problem then you could have the used this method for the original problem as well. (Und dann ist es also alles umsonnts gewesen.)
So I think that if you cannot find some new method to prove $(*)$, this approach did not bring too much. It only added an intermediary step which may or may not help you make things clearer - I believe this depends on your personal taste. Another possible advantage of this approach could be, that if you solved $(*)$ in a similar manner as suggested in the answers, you would probably work with slightly simpler expressions.

  • $\begingroup$ +1 for the explanation of the method my fellow students used! However, as you already mentioned, I sticked to the conjugate quantity method which was much easier ;) $\endgroup$ – Christian Ivicevic Nov 1 '11 at 16:07

The following is in response to the edited question. It may be interesting to complete the square, in order to bring out the symmetry. Note that $$n+\sqrt{2n}=\left(\sqrt{n}+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2} \qquad\text{and}\qquad n-\sqrt{2n}=\left(\sqrt{n}-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}.$$ Maybe completing the square qualifies as a "funky binomial formula."

To save some typing, we write $x$ for $\sqrt{n}+\frac{1}{\sqrt{2}}$ and $y$ for $\sqrt{n}-\frac{1}{\sqrt{2}}$. Note that $x-y=\sqrt{2}$, and therefore $$\sqrt{x^2+1/2}-\sqrt{y^2+1/2}=\left(\sqrt{x^2+1/2}-x\right)-\left(\sqrt{y^2+1/2}-y\right)+\sqrt{2}.$$ As $n$ grows without bound, so do $x$ and $y$. Thus it will be enough to show that $\displaystyle\lim_{u\to\infty}\left(\sqrt{u^2+1/2}-u\right)=0$.

This can be proved using any of the several ideas described in the other answers. For example, if $u$ is positive, then $$u^2+\frac{1}{2}<\left(u+\frac{1}{4u}\right)^2,$$ (just expand the right-hand side) and therefore $$0<\sqrt{u^2+1/2}-u <u+\frac{1}{4u}-u=\frac{1}{4u}.$$ Thus, by "squeezing," $\displaystyle\lim_{u\to\infty}\left(\sqrt{u^2+1/2}-u\right)=0$.


If you know Taylor expansions, you can use the series for $(1+x)^{\frac12}$ which is probably the binomial formula you are referring to.

We know $(1+x)^{\frac12}= 1+ \frac12 x + O(x^2)$ for small $x$, so $\sqrt{n\pm\sqrt{2n}}= \sqrt n \cdot \sqrt{1\pm\sqrt{\frac2n}}=\sqrt n \cdot (1\pm \frac 12 \sqrt{\frac 2n}+O(\frac 1n))$. The difference of the expression for the two signs gives the desired limits.

No sandwiches needed for this approach, your fellow students might have referred to the Bernoulli inequality or something similar.

I want to add that I don't see anything "obvious" about the limit.

  • $\begingroup$ Damn, I was in the middle of typing up an answer using Newton's binomial series. O well, here's a useful link: en.wikipedia.org/wiki/Binomial_series $\endgroup$ – Ragib Zaman Nov 1 '11 at 11:40
  • $\begingroup$ I think the "obvious" part, was that if you evaluate the expression with a sequence of reasonably large values of N (say N~[1000, 10000, 100000]) you can see that it approaches sqrt(2) rapidly (e.g., differ from sqrt(2) by ~4x10^-4, 4x10^-5 and 4x10^-6 respectively). $\endgroup$ – dr jimbob Nov 1 '11 at 17:26
  • $\begingroup$ @drjimbob Obvious; According to me, is that I can do the evaluation in my head; by experience and/or intuition. Evaluation for N@1000/10000/100000 is not something my mind is built for :/ $\endgroup$ – Captain Giraffe Nov 1 '11 at 17:42

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