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I want to know how to prove the following inequality.

For $n = 1, 2, 3, \ldots $

$$ n < \left(1+\frac{1}{\sqrt{n}} \right)^n $$

I tried with math induction but I failed.

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    $\begingroup$ Have you tried putting continuous variable instead of $n$ and then doing some analysis? $\endgroup$
    – ploosu2
    Nov 25, 2014 at 22:54
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    $\begingroup$ Display the work that you did with induction and we will point you in the right direction for that proof. $\endgroup$ Nov 25, 2014 at 23:07
  • $\begingroup$ yes, I did with derivative but it's too complex $\endgroup$
    – Allen Math
    Nov 26, 2014 at 0:13

5 Answers 5

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We have $$\left(1+\dfrac1{\sqrt{n}}\right)^n = \sum_{k=0}^n \dbinom{n}k \dfrac1{n^{k/2}} \geq \underbrace{1 + \sqrt{n} + \dfrac{(n-1)}2 + \dfrac{n(n-1)(n-2)}{6} \dfrac1{n^{3/2}} > n}_{\text{For $n \geq 3$}}$$ Check for $n=1,2$ manually.


EDIT If you do not want to do the dirty calculus of showing $$1 + \sqrt{n} + \dfrac{(n-1)}2 + \dfrac{n(n-1)(n-2)}{6} \dfrac1{n^{3/2}} > n$$ by showing that the appropriate function is increasing, here is a more logical way. We have $$\underbrace{1 + \sqrt{n} + \dfrac{(n-1)}2 + \dfrac{n(n-1)(n-2)}{6n^{3/2}} > \dfrac{n+1}2 + \dfrac{n^{3/2}}{24}}_{\text{For }n \geq 3}$$ The above inequality comes from the fact that $\sqrt{n} > 0$, $n-1 \geq \dfrac{n}2$ and $n-2\geq \dfrac{n}2$ for $n\geq 3$. We have $\dfrac{n^{3/2}}{24} \geq \dfrac{n}2$ for $n \geq 144$. Put all this together to obtain the result for $n \geq 144$. Write a code to check if it is true for $n=1$ to $n=143$.

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  • $\begingroup$ You can lower that limit of 144 even further. Note that $$ 1 + \sqrt{n} + \dfrac{(n-1)}2 + \dfrac{n(n-1)(n-2)}{6} \dfrac1{n^{3/2}} = \frac{n^{3/2}}{6} + \frac{n}{2} + \frac{n^{1/2}}{2} + \frac{1}{3n^{1/2}} + \frac{1}{2}$$. The RHS is greater than (using the first three terms) $$\frac{n}{2} + \frac{n^{1/2}(n + 3)}{6}$$. It is quite easy to show that $$\frac{n^{1/2}(n + 3)}{6} > \frac{n}{2}$$. This gives you the result, without needing calculus or computing for particular values. $\endgroup$ Nov 26, 2014 at 8:21
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A general approach to study a sequence is first to force some continuous parameter into the problem, and then to use differentiability.

Taking logarithms, one sees that the required inequality holds for some given $n$ if and only if $u(\sqrt{n})\gt0$, where, for every $x\gt0$, $$u(x)=x^2\log(x+1)-(x^2+2)\log(x).$$ Equivalently, $$u(x)=x^2\log\left(1+\frac1x\right)+2\log\left(\frac1x\right)=x^4v(1/x),$$ where the function $v$ is defined by $$v(z)=z^2\log(1+z)+2z^4\log(z).$$ Thus, $$v'(z)=2z\log(1+z)+8z^3\log(z)+\frac{z^2}{1+z}+2z^3.$$ For every $z$ in $(0,1)$, $$z\log z\geqslant-\mathrm e^{-1},\qquad 2\log(1+z)\geqslant2z-z^2,$$ hence $$v'(z)\geqslant2z^2-z^3-8\mathrm e^{-1}z^2+\frac{z^2}{1+z}+2z^3=(3-8\mathrm e^{-1})z^2+\frac{z^4}{1+z}\gt0,$$ since $3\mathrm e\gt8$. Thus, $v$ is increasing and, since $v(0)=0$, $v(z)\gt0$ for every $z\gt0$, hence $u(x)\gt0$ for every $x\gt0$, which implies the result.

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  • $\begingroup$ How did you get the last equality ($3^{rd}$ last line, the middle part), I get; $2z^2-z^3-8e^{-1}z^2+\frac{z^2}{1+z}+2z^3\overset{3e\ge 8}\ge 2z^2-z^3-3z^2+\frac{z^2}{1+z}+2z^3=\frac{z^2(z^2-1)+z^2}{1+z}>0$ BUT in the beginning after equivalently, I have: $u(x)=x^2\log(1+\frac1x)-2\log(\color{red}{x})$ Just say, if I'm wrong $\endgroup$
    – OBDA
    Nov 26, 2014 at 11:15
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    $\begingroup$ @OBDA Well, $-2\log(x)=+2\log\left(\frac1x\right)$. $\endgroup$
    – Did
    Nov 26, 2014 at 12:41
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We have that

$$n < \left(1+\frac{1}{\sqrt{n}} \right)^n\iff \ln n< n\ln \left(1+\frac{1}{\sqrt{n}} \right)\iff \frac{\ln n}{n}<\ln \left(1+\frac{1}{\sqrt{n}} \right).$$

Now, we use the inequalities (see $(3)$ in http://ajmaa.org/RGMIA/papers/v7n2/pade.pdf)

$$\begin{equation}\frac{2x}{2+x}\le \ln (1+x)\le \frac{x}{2}\frac{2+x}{1+x}, \quad x>0.\tag{1}\end{equation}$$

In our case, we get from $(1),$ $$\ln \left(1+\frac{1}{\sqrt{n}} \right)\ge \frac{2}{\sqrt{n}+2}.$$

Thus we only have to show

$$\frac{2}{\sqrt{n}+2}>\frac{\ln n}{n}.$$ That is,

$$\ln n<\frac{2n}{\sqrt{n}+2}.$$ Since $\ln n=2\ln \sqrt{n}$ previous inequality is equivalent to

$$\ln \sqrt{n}<\frac{n}{\sqrt{n}+2}.$$ Now, using again $(1)$

$$\ln \sqrt{n}=\ln (1+\sqrt{n}-1)\le \frac{\sqrt{n}-1}{2}\frac{\sqrt{n}+1}{\sqrt{n}}=\frac{n-1}{2\sqrt{n}}.$$ Thus we have to show

$$\frac{n-1}{2\sqrt{n}}<\frac{n}{\sqrt{n}+2}.$$ But

$$\frac{n-1}{2\sqrt{n}}<\frac{n}{\sqrt{n}+2}\iff 2(n-1)<(n+1)\sqrt{n},$$ which holds for any $n\in\mathbb{N}.$

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  • $\begingroup$ @mfI How do you prove the last inequality? $\endgroup$ Aug 22, 2015 at 9:50
  • $\begingroup$ Also.. formula copied is incorrect. It's 2x/(2+x); and not 2x/(2x+1) right? @mfl $\endgroup$ Aug 22, 2015 at 10:12
  • $\begingroup$ @Kugelblitz You are right. Typo fixed. In order to prove the las inequality just note that $$\frac{n-1}{2\sqrt{n}}<\frac{n}{\sqrt{n}+2}\iff (\sqrt{n}+2)(n-1)<2n\sqrt{n}\iff 2(n-1)<(n+1)\sqrt{n}$$ $\endgroup$
    – mfl
    Sep 7, 2015 at 0:39
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Hint:

$U=\left\{\left(1+\dfrac{1}{x}\right)^x: x\geq1\right\} \implies \inf U=2<\left(1+\dfrac{1}{\sqrt{n}}\right)^\sqrt{n} $

$$\color{blue}{\boxed { n \leq 2^{\sqrt{n}} \implies {n<\left(1+\dfrac{1}{\sqrt{n}}\right)^n}\ \ \ \forall n\geq16}}$$

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  • $\begingroup$ "Hint"?? $ $ $ $ $\endgroup$
    – Did
    Nov 26, 2014 at 7:20
  • $\begingroup$ if n = 8, then the left side is no longer valid? $\endgroup$
    – Allen Math
    Nov 26, 2014 at 11:32
  • $\begingroup$ @AllenJung: Now the range of the values is included. $\endgroup$
    – user170039
    Nov 26, 2014 at 13:12
  • $\begingroup$ @Did: See the edit. $\endgroup$
    – user170039
    Nov 26, 2014 at 13:38
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$n<(1+\frac{1}{\sqrt{n}})^{n}$

$n^{1/n} < 1+1/\sqrt{n}$

$n^{1/n + 1/2} < \sqrt{n}+1$

$(n^{(2+n)/2n} -1) < \sqrt{n}$

$n^{(2+n)/n} - 2n^{(2+n)/2n} +1 < n$

$n*n^{2/n} - 2\sqrt{n}*n^{1/n} +1 < n$

For large possible values of "n", then we can use $$lim_{n->∞}n^{1/n} = 1$$ to approximate the original $$n*n^{2/n} - 2\sqrt{n}*n^{1/n} +1 < n$$ quite closely as $$n*1^2 - 2\sqrt{n}*1 +1 < n$$ or, more simply, $$n - 2\sqrt{n} +1 < n$$ $$1-2\sqrt{n}<0$$

This works for all possible values of "n" large enough for the previous approximation to work for, and if you are looking at "n" values that you feel are too small for the approximation to work, then you are looking at "n" values small enough to use trial and error.

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    $\begingroup$ Can you please add the number of cases you want to solve be trial and error to prove that the statement holds for n > 1. $\endgroup$ Nov 26, 2014 at 0:02

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