Is $\left(1+\frac1n\right)^{n+1/2}$ decreasing?

Preliminaries: A couple of extensions to Bernoulli's Inequality.

Bernoulli's Inequality says that $(1+x)^n$ is at least as big as the first two terms of its binomial expansion. It turns out, at least for $n\in\mathbb{Z}$, that a sharper inequality can be obtained using any partial sum with an even number of terma.

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Theorem $\bf{1}$: for $m\ge1$, $n\ge0$, and $x\gt-1$, $$ (1+x)^n\ge\sum_{k=0}^{2m-1}\binom{n}{k}x^k\tag1 $$

Proof (Induction on $n$): $(1)$ is trivial for $n=0$. Assume $(1)$ is true for $n-1$, then $$ \begin{align} (1+x)^n &=(1+x)(1+x)^{n-1}\tag{1a}\\[9pt] &\ge(1+x)\sum_{k=0}^{2m-1}\binom{n-1}{k}x^k\tag{1b}\\ &=\sum_{k=0}^{2m-1}\left[\binom{n-1}{k}+\binom{n-1}{k-1}\right]x^k+\binom{n-1}{2m-1}x^{2m}\tag{1c}\\ &\ge\sum_{k=0}^{2m-1}\binom{n}{k}x^k\tag{1d} \end{align} $$ Explanation:
$\text{(1a)}$: factor
$\text{(1b)}$: assumption for $n-1$
$\text{(1c)}$: multiply sum by $1+x$
$\text{(1d)}$: Pascal's Rule

Thus, $(1)$ is true for $n$.
${\large\square}$

Theorem $\bf{2}$: for $m\ge1$, $n\ge0$, and $x\gt-1$, $$ (1+x)^{-n}\ge\sum_{k=0}^{2m-1}\binom{-n}{k}x^k\tag2 $$

Proof (Induction on $n$): Note that another way of writing $(2)$ is $$ (1+x)^n\sum_{k=0}^{2m-1}(-1)^k\binom{n+k-1}{k}x^k\le1\tag{2a} $$

$\text{(2a)}$ is trivial for $n=0$. Assume $\text{(2a)}$ is true for $n-1$, then $$ \begin{align} &(1+x)^n\sum_{k=0}^{2m-1}(-1)^k\binom{n+k-1}{k}x^k\\ &=(1+x)^{n-1}\sum_{k=0}^{2m-1}(-1)^k\binom{n+k-1}{k}x^k(1+x)\tag{2b}\\ &=(1+x)^{n-1}\sum_{k=0}^{2m-1}(-1)^k{\textstyle\left[\binom{n+k-1}{k}-\binom{n+k-2}{k-1}\right]}x^k-{\textstyle\binom{n+2m-2}{2m-1}}x^{2m}(1+x)^{n-1}\tag{2c}\\ &=(1+x)^{n-1}\sum_{k=0}^{2m-1}(-1)^k\binom{n+k-2}{k}x^k-\binom{n+2m-2}{2m-1}x^{2m}(1+x)^{n-1}\tag{2d}\\[9pt] &\le1\tag{2e} \end{align} $$ Explanation:
$\text{(2b)}$: factor
$\text{(2c)}$: multiply sum by $1+x$
$\text{(2d)}$: Pascal's Rule
$\text{(2e)}$: assumption for $n-1$

Thus, $\text{(2a)}$ is true for $n$.
${\large\square}$

Note that for positive integer exponents, Bernoulli's Inequality is the case $m=1$ of Theorem $1$, and for negative integer exponents, it is the case $m=1$ of Theorem $2$.

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Answer: Use the case $m=2$ of Theorem $1$: $$ \begin{align} &\frac{\left(1+\frac1{n-1}\right)^{2n-1}}{\left(1+\frac1n\right)^{2n+1}}\\ &=\left(1+\frac1{n^2-1}\right)^{2n}\frac{n-1}{n+1}\\ &\ge\left(1+\frac{2n}{n^2-1}+\frac{2n(2n-1)}{2\left(n^2-1\right)^2}+\frac{2n(2n-1)(2n-2)}{6\left(n^2-1\right)^3}\right)\frac{n-1}{n+1}\\ &=1+\frac{n^2+n+6}{3(n-1)(n+1)^4}\\[9pt] &\ge1 \end{align} $$ That is, $\left(1+\frac1n\right)^{n+1/2}$ is decreasing.

We'll prove that this sequence indeed decreases.

We need to prove that $$\left(1+\frac{1}{n+1}\right)^{n+\frac{3}{2}}<\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}.$$Let $$f(x)=\left(x+\frac{1}{2}\right)\ln\left(1+\frac{1}{x}\right)-\left(x+\frac{3}{2}\right)\ln\left(1+\frac{1}{1+x}\right)=$$ $$=(2x+2)\ln(1+x)-\left(x+\frac{1}{2}\right)\ln{x}-\left(x+\frac{3}{2}\right)\ln(x+2),$$ where $x>0$.

But $$f''(x)=\frac{2}{x^2(x+2)^2(x+1)}>0$$ and $$\lim_{x\rightarrow+\infty}f'(x)=\lim_{x\rightarrow+\infty}\left(\ln\frac{(x+1)^2}{x(x+2)}+2-\frac{x+\frac{1}{2}}{x}-\frac{x+\frac{3}{2}}{x+2}\right)=0,$$ which says that $f'(x)<0$, $f$ decreases and since it's obvious that $$\lim_{x\rightarrow+\infty}f(x)=0,$$ we obtain that $f(x)>0$ and our sequence indeed decreases.