Prove that $2(\sqrt{n+1}-\sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1})$ if $n \ge 1$ using induction.

Can someone help me with this problem please. Base case is easily shown, and for the inductive step I know I have to use the inequality for $n\gt1$ to show that it is true for $n+1$, but I am not quite sure how to.


  • $\begingroup$ I would try to make 2 inductions: left to middle, and middle to right $\endgroup$
    – ulead86
    Dec 21, 2012 at 9:15
  • $\begingroup$ This inequality is a hint for a problem in the book of Richard A. Silverman. Page 61. $\endgroup$
    – Mikasa
    Dec 21, 2012 at 9:37
  • $\begingroup$ Related question: math.stackexchange.com/questions/1139563/… $\endgroup$ Jul 15, 2016 at 17:53

3 Answers 3


Does it have to be full induction? Because it seems to be way simpler:

$$(1)\;\;\;\;\;\;2\left(\sqrt{n+1}-\sqrt n\right)=\frac{2}{\sqrt{n+1}+\sqrt n}\leq\frac{2}{\sqrt n+\sqrt n}=\frac{1}{\sqrt n}$$

$$(2)\;\;\;\;\;\;2\left(\sqrt n-\sqrt{n-1}\right)=\frac{2}{\sqrt n+\sqrt{n-1}}\geq\frac{2}{\sqrt n+\sqrt n}=\frac{1}{\sqrt n}$$

  • $\begingroup$ Well, the question was in exercises for a section on induction on the textbook I am working on. So I assumed it has to be through induction (silly me!). But I like this better, its simpler. Thanks $\endgroup$
    – user53876
    Dec 21, 2012 at 11:10
  • $\begingroup$ I see. Nevertheless it would be worth for you to try the induction thing as well, though in this case it seems to be a rather cumbersome and unnecessary. $\endgroup$
    – DonAntonio
    Dec 21, 2012 at 14:39

Let's only concentrate on the middle to right inequality.

This can be easily proven without induction. Indeed,

$$\frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1})\iff \frac12<n-\sqrt{n^2-n}\iff \\ \sqrt{n^2-n}<n-\frac12\iff n^2-n<n^2-n+\frac14\iff 0<\frac14$$ which holds. Induction would complicate things a lot as you will see:

Assume $$\frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1})$$ for some $n\ge 1$. We must prove that $$\frac{1}{\sqrt{n+1}} < 2(\sqrt{n+1} - \sqrt{n})$$ or equivalently that $$\frac{1}{\sqrt{n}} < 2\frac{(\sqrt{n+1} - \sqrt{n})\sqrt{n+1}}{\sqrt{n}}$$ In order to use our assumption, it seems natural to try and show that $$2(\sqrt{n}-\sqrt{n-1}) < 2\frac{(\sqrt{n+1} - \sqrt{n})\sqrt{n+1}}{\sqrt{n}}$$

This is only sufficient and not neccessary. Unfortunately however, not only does this not hold, but proving that it is wrong will be harder than showing the whole thing without induction!:

$$\sqrt{n}-\sqrt{n-1} < \frac{(\sqrt{n+1} - \sqrt{n})\sqrt{n+1}}{\sqrt{n}}\iff n-\sqrt{n^2-n}<n+1-\sqrt{n^2+n}\iff\\ \sqrt{n^2+n}<1+\sqrt{n^2-n}\iff n^2+n<1+2\sqrt{n^2-n}+n^2-n\iff \\ 2n-1<2\sqrt{n^2-n}\iff 4n^2-4n+1<4n^2-4n\iff 1<0$$ which is absurd. So, unless I have done a mistake in my calculations (very likely) , induction isn't a shortcut to solve this problem (unless of course you use the assumption in a different and smarter way)


Let's consider the first inequality, and let's multiply both sides by $\frac{\sqrt{n}}2$: $$ \sqrt{n(n+1)}-n < \frac{1}{2} $$ Since $\lim_{n\to\infty}a_n=\frac{1}{2}$, it suffices to show that the sequence $a_n=\sqrt{n(n+1)}-n$ is strictly increasing, i.e. $a_n> a_{n-1}$ for $n>1$, so that $a_1< a_2 \ldots < a_n < \lim_{n\to\infty}a_n = \frac{1}{2}$ holds by induction. \begin{equation} a_n > a_{n-1} \quad\Leftrightarrow\quad \sqrt{n(n+1)}-n > \sqrt{n(n-1)}-n+1 \quad\Leftrightarrow\quad\\ n\left( \sqrt{1+\frac{1}{n}} - \sqrt{1-\frac{1}{n}} \right) > 1 \end{equation} Let's raise both sides to the second power: $$ \quad\Leftrightarrow\quad 2n^2\left( 1 - \sqrt{1-\frac{1}{n^2}} \right) > 1 \quad\Leftrightarrow\quad 2n^2 - 1 > 2n\sqrt{n^2-1} \quad\Leftrightarrow\quad\\ 4n^4 + 1 - 4n^2 > 4n^4 - 4n^2 \quad\Leftrightarrow\quad 1>0 $$ so the first inequality is proven.

Concerning the second one, for the same reasoning you show that the sequence $b_n=n-\sqrt{n(n-1)}$ is strictly decreasing (for simpler calculations, I'd suggest to show that $b_{n+1}>b_n$ rather than $b_n> b_{n-1}$); actually you end up with showing that $\sqrt{n(n+1)}-\sqrt{n(n-1)}>1$, which we proved above.

  • $\begingroup$ DonAntonio is right, you don't need induction... $\endgroup$
    – AndreasT
    Dec 21, 2012 at 10:21

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