Decreasing sequence proof

Question

I would like to know if this proof is valid or whether there is a more elegant solution. $$a_n=(1+\frac{1}{n})^{n+1}$$
I want to prove that it is decreasing.
$$\frac{a_{n+1}}{a_n}=\frac{n+2}{n}(1-\frac{1}{(n+1)^2})^{n+2}$$ Using bernoulli inequality $$\frac{a_{n+1}}{a_n}\leq e^{-\frac{n+2}{(n+1)^2}}$$ $e$ exponent is $<0 $ and we know that $e^x<1$ with $x<0$ so $$\frac{a_{n+1}}{a_n}<1$$

Answer 1

Here is my working out of that proof.

I am not sure what you have as Bernoulli inequality, but I take it as $$1+na \leq (1+a)^n$$

I show $1 \leq \frac{a_n}{a_{n+1}}$. Now, $$ \frac{a_n}{a_{n+1}}=\frac{ \left(1+\frac{1}{n}\right)^{n+1}}{\left(1+\frac{1}{n+1}\right)^{n+2}} =\left( \frac{(n+1)^2}{n(n+2)} \right)^{n+2} \left(\frac{n}{n+1}\right). $$ Now by the Bernoulli inequality, (with $a=\frac{1}{n^2+2n}$) $$\frac{n+1}{n} \leq \left( \frac{(n+1)^2}{n(n+2)} \right)^{n+2}$$