<s>[INST] The sequence $A_n=\prod_{k=1}^n\left(1+\frac{k}{n^2}\right)$ is decreasing Let $A$ be the sequence of real numbers defined by :
$$\forall n\in\mathbb{N}^\star,\,A_n=\prod_{k=1}^n\left(1+\frac{k}{n^2}\right)$$
I know how to prove that this sequence converges to $\sqrt e$, using the following inequalities :
$$\forall t>0,\,t-\frac{t^2}2\leqslant\ln(1+t)\leqslant t$$
I found numerical evidence that $(A_n)$ is decreasing, but wasn't able to prove it. Any help will be appreciated.
 [/INST] It is easy to verify $A_1 > A_2 > A_3 > A_4$.
It suffices to prove that $A_n > A_{n+1}$ for all $n\ge 4$.
It suffices to prove that, for all $n\ge 4$,
$$\sum_{k=1}^n \ln (1 + k/n^2) > \sum_{k=1}^{n+1} \ln (1 + k/(n+1)^2)$$
or
$$\sum_{k=1}^n \ln \frac{1 + k/n^2}{1 + k/(n+1)^2} > \ln (1 + 1/(n+1))$$
or
$$\sum_{k=1}^n \ln\left(1 + \frac{k(2n+1)}{n^2(n^2+k+2n+1)} \right) > \ln (1 + 1/(n+1)).$$
By using $\ln (1+x) \ge \frac{x}{1+x}$ for $x > 0$, we have
\begin{align}
\ln\left(1 + \frac{k(2n+1)}{n^2(n^2+k+2n+1)} \right)
&\ge \frac{k(2n+1)}{(n^2+k)(n+1)^2}\\
&= \frac{k(2n+1)}{n^2(n+1)^2}\, \frac{1}{1 + k/n^2}\\
&\ge \frac{k(2n+1)}{n^2(n+1)^2}(1 - k/n^2).
\end{align}
Also, by using $\ln(1+x) < \frac{x^2+6x}{6+4x}$ for $x > 0$, we have
$$\ln (1 + 1/(n+1)) < \frac{7+6n}{2(3n+5)(n+1)}.$$
Thus, it suffices to prove that, for all $n\ge 4$,
$$\sum_{k=1}^n \frac{k(2n+1)}{n^2(n+1)^2}(1 - k/n^2) > \frac{7+6n}{2(3n+5)(n+1)}$$
or
$$\frac{(n-1)(2n+1)(3n+1)}{6n^3(n+1)} > \frac{7+6n}{2(3n+5)(n+1)}$$
or
$$\frac{6n^3-17n^2-23n-5}{6n^3(n+1)(3n+5)} > 0.$$
It is true. We are done.
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