How to prove that $(1+\frac{1}{n})^{n+\frac{1}{2}}>e$ for all positive integer $n$?

Question

It has already been proved that $$ \left(1+\dfrac{1}{n}\right)^n<\mathrm{e}<\left(1+\dfrac{1}{n}\right)^{n+1} $$ The problem is to prove that the geometric average between $\left(1+\dfrac{1}{n}\right)^n$ and $\left(1+\dfrac{1}{n}\right)^{n+1}$ is still bigger than $\mathrm{e}$: $$ \left(1+\dfrac{1}{n}\right)^{n+\frac{1}{2}}>\mathrm{e} $$ One possible approach is to prove that $\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}$ strictly declines with $n$. Then, consider the limit of $\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}(n\to\infty)$ equals to $e$. But how to prove it? I need help.

Answer 1

In comments, @BenjaminWang asked an interesting question :

"What is the maximum $f(n)$ such that $(1+\frac1n)^{n+f(n)}<e$"

The equality is given by $$f(n)=\frac{1}{\log \left(1+\frac{1}{n}\right)}-n$$

Expanding as a series, $$f=\sum_{k=0}^\infty (-1)^k \,\frac {a_k}{b_k\,n^k}$$ where the $a_k$ and $b_k$ form sequences $A002206$ and $A002207$ in $OEIS$ (Gregory coefficients).

This gives as bounds

$$(1+\frac1n)^{n+f(n)}>e -\sum_{k=0}^{2p} (-1)^k \,\frac {a_k}{b_k\,n^k}=e\left(1-\frac{c_p}{n^{2p+2} }\right)$$

$$(1+\frac1n)^{n+f(n)}<e -\sum_{k=0}^{2p+1} (-1)^k \,\frac {a_k}{b_k\,n^k}=e\left(1+\frac{d_p}{n^{2p+3} }\right)$$ Moreover, coefficients $c_p$ and $d_p$ are not only small but also decreasing functions of $p$

Answer 2

Let $$f(x)=\left(1+\dfrac{1}{x}\right)^{x+\dfrac{1}{2}}, x\ge 1.$$ Then $$ f'(x)=\left(1+\dfrac{1}{x}\right)^{x+\dfrac{1}{2}}\bigg[\bigg(x+\dfrac{1}{2}\bigg)\ln\left(1+\dfrac{1}{x}\right)\bigg]'=\left(1+\dfrac{1}{n}\right)^{n+\dfrac{1}{2}}\bigg[\ln\left(1+\dfrac{1}{x}\right)-\dfrac{1+2x}{2x+2x^2}\bigg]. $$ Let $$ g(x)=\ln\left(1+\dfrac{1}{x}\right)-\dfrac{1+2x}{2x+2x^2} $$ and then $$ g'(x)=\frac{1}{2x^2(1+x)^2}>0. $$ So $g(x)$ is strictly increasing and hence for $1\le x<\infty$, $g(x)<g(\infty)=0$. This implies $f'(x)<0$ for $x\ge1$ or $f(x)$ is strictly decreasing. Thus for $n<\infty$, $f(n)>f(\infty)=e$.

Answer 3

It is equivalent to prove that $\left(n+\dfrac12\right)\ln\left(1+\dfrac1n\right)\gt1$.

The function $f(x)=\left(x+\dfrac12\right)\ln\left(1+\dfrac1x\right)$ has derivative $f'(x)=\ln\left(1+\dfrac1x\right)-\dfrac{2x+1}{2x^2+2x}$ so it is decreasing with $f(1)\approx1.03972077$ and $$\lim_{x\to \infty}f(x)=\lim_{x\to \infty}\frac{\ln\left(1+\dfrac1x\right)}{\frac{1}{x+\frac12}}=\lim_{x\to \infty}\dfrac{\frac{-1}{x^2+x}}{\frac{-4}{(2x+1)^2}}=1$$ Therefore always $f(x)$ is greater than $1$ so $\left(n+\dfrac12\right)\ln\left(1+\dfrac1n\right)\gt1$ and $\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}>e$. We are done.

Answer 4

Hi Oliver I shall propose some hint to solve the problem :

Writing the function as :

$$g(x)=\left(1+1/x\right)^{x+1/2}=(1+1/x)^{x(1-a)+1/2}(1+a),a=0$$

Invert the variable we need to study :

$$f(x)=(1+1/a)^{a(1-x)+1/2}(1+x),x\simeq 0$$

Introduce the logarithm :

$$(\ln(f(x)))'=\frac{1}{x+1}-a\ln\left(1+1/a\right)$$

Setting $x=0$ in the derivative (we can use the Lambert's function to invert the function but it's another story) we have that the derivative (already shown) is decreasing and positive so the function in $x$ is increasing around $x=0$.

The function in logarithmic form is also plus a minus almost completely monotone on $(0,\infty]$

$$h\left(0\right)+h'\left(0\right)x+\frac{h''\left(0\right)}{2}x^{2}+\frac{h'''\left(0\right)}{6}x^{3}+\cdot\cdot\cdot+\frac{h^{\left(2n\right)}\left(0\right)}{\left(2n\right)!}x^{\left(2n\right)}\leq h(x)=\ln(f(x))\leq h\left(0\right)+h'\left(0\right)x+\frac{h''\left(0\right)}{2}x^{2}+\frac{h'''\left(0\right)}{6}x^{3}+\cdot\cdot\cdot+\frac{h^{\left(2n+1\right)}\left(0\right)}{\left(2n+1\right)!}x^{\left(2n+1\right)}$$

So as the function is increasing $h'(x_{max})=0,h(x_{max})\geq h(0)$

$$ h(x_{max})=\ln(f(x_{max}))\leq h\left(0\right)+h'\left(0\right)x_{max}+\frac{h''\left(0\right)}{2}x_{max}^{2}+\frac{h'''\left(0\right)}{6}x_{max}^{3}+\cdot\cdot\cdot+\frac{h^{\left(2n+1\right)}\left(0\right)}{\left(2n+1\right)!}x_{max}^{\left(2n+1\right)}$$

Or :

$$h(x_{max})-(h'\left(0\right)x_{max}+\frac{h''\left(0\right)}{2}x_{max}^{2}+\frac{h'''\left(0\right)}{6}x_{max}^{3}+\cdot\cdot\cdot+\frac{h^{\left(2n+1\right)}\left(0\right)}{\left(2n+1\right)!}x_{max}^{\left(2n+1\right)})\leq h\left(0\right)$$

So :

$$h(x_{max})-(h'\left(0\right)x_{max}+\frac{h''\left(0\right)}{2}x_{max}^{2}+\frac{h'''\left(0\right)}{6}x_{max}^{3})\leq h(0)$$

One can show that the function where $x=x_{max}$ is decreasing in $a$ using the fact $p(x)=(1+1/x)^{1+x}$ is decreasing as :

$$P(x)=\left(1+\frac{2}{a}\right)^{1+a\left(1-x\right)}\left(1+x\right)^{2}$$

$P(x)$ is decreasing in $a$ as mentionned by the OP's post but as one can show :

$$\sqrt{P(x_{max})P(0)}-(P'\left(0\right)x_{max}+\frac{P''\left(0\right)}{2}x_{max}^{2})<\sqrt{2}\sqrt{P(x_{max})}$$

And by the mean value theorem $\exists k\in[-1/2,0]$ such that :

$$2<P(k)=\sqrt{P(x_{max})P(0)}-(P'\left(0\right)x_{max}+\frac{P''\left(0\right)}{2}x_{max}^{2})$$

So by a continuity argument $\sqrt{P(0)}$ is decreasing in $a$