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

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.

Solution 1:

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 $n=143$.

Solution 2:

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.

Solution 3:

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 $n\in\mathbb{N}.$

Solution 4:

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} $

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