How to verify $(1+\frac{1}{n})^2(1-\frac{1}{n^2})^{n-1}\geq \exp(\frac{1}{n})$

Question

How to verify this inequality? Assuming that $n\in \mathbb{N}^+$, we have:

$$\left(1+\frac{1}{n}\right)^2\left(1-\frac{1}{n^2}\right)^{n-1}\geq \exp\left(\frac{1}{n}\right).$$

Answer 1

Consider $$A_n=\left(1+\frac{1}{n}\right)^2\times\left(1-\frac{1}{n^2}\right)^{n-1}$$ Take logarithms $$\log(A_n)=2\log\left(1+\frac{1}{n}\right)+(n-1)\log\left(1-\frac{1}{n^2}\right)$$ Now, use the Taylor series $$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+O\left(x^6\right)$$ and replace $x$ by $\frac 1n$ in the first term and by $-\frac 1{n^2}$ in the second term.

This should give $$\log(A_n)=\frac{1}{n}+\frac{1}{6 n^3}+\frac{1}{15 n^5}+O\left(\frac{1}{n^6}\right) >\frac 1n$$

You can check, using limits, that $A_1=4 > e$.

Edit

Taking more terms, the expansion of $\log(A_n)$ does not seem to contain any negative coefficient (I did stop at $O\left(\frac{1}{n^{1000}}\right)$, using, for sure, a CAS for this last part).

A closer look at the coefficients reveals that $$\log(A_n)=\sum_{k=1}^\infty \frac{1}{k (2 k-1)}\frac 1 {n^{2 k-1}}$$

Answer 2

If we take the logarithm of $$\left(1+\frac{1}{n}\right)^2 \left(1-\frac{1}{n^2}\right)^{n-1} $$ we get: $$ \log\left(\frac{1+\frac{1}{n}}{1-\frac{1}{n}}\right)+n \log\left(1-\frac{1}{n^2}\right)= 2\,\text{arctanh}\left(\frac{1}{n}\right)+n\log\left(1-\frac{1}{n^2}\right) $$ and since in a neighbourhood of the origin: $$ \text{arctanh}(x) = \sum_{n\geq 0}\frac{x^{2n+1}}{2n+1},\qquad \log\left(1-x^2\right)=-\sum_{n\geq 1}\frac{x^{2n}}{n} $$ it follows that:

$$ \log\left[\left(1+\frac{1}{n}\right)^2 \left(1-\frac{1}{n^2}\right)^{n-1}\right]=\color{red}{\sum_{k\geq 0}\frac{1}{(k+1)(2k+1)\,n^{2k+1}}} $$

as claimed by Claude Leibovici.