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

Question

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.

Answer 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.

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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 $n=143$.

Answer 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.

Answer 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 $n\in\mathbb{N}.$

Answer 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} $

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

Answer 5

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

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

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

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

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

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

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.