Calculus – How to Prove $(1+\frac{1}{n})^{n+\frac{1}{2}}>e$ for All Positive Integers?

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 $p$