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.
 A: 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.
A: Fun with Stirling.
I get
$\ln(A_n)
=\frac12+\frac1{3n}+O(\frac1{n^2})
$.
Numerically this checks.
Here's how.
$\begin{array}\\
A_n
&=\prod_{k=1}^n\left(1+\dfrac{k}{n^2}\right)\\
B_n
&=\ln(A_n)\\
&=\sum_{k=1}^n\ln\left(1+\dfrac{k}{n^2}\right)\\
&=\sum_{k=1}^n\ln\left(\dfrac{n^2+k}{n^2}\right)\\
&=\sum_{k=1}^n\ln(n^2+k)-\sum_{k=1}^n\ln(n^2)\\
&=\sum_{k=n^2+1}^{n^2+n}\ln(k)-n\ln(n^2)\\
&=\sum_{k=1}^{n^2+n}\ln(k)-\sum_{k=1}^{n^2}\ln(k)-2n\ln(n)\\
&=\ln((n^2+n)!)-\ln((n^2)!)-2n\ln(n)\\
&\approx\frac12\ln(n^2+n)+(n^2+n)(\ln(n^2+n)-1)+O(1/n^2))-(\frac12\ln(n^2)+n^2(\ln(n^2)-1)+O(1/n^2))-2n\ln(n)\\
&=\frac12\ln(1+1/n)+(n^2+n)(\ln(n^2+n)-1))-(n^2(\ln(n^2)-1)+O(1/n^2))-2n\ln(n)\\
&=\frac12\ln(1+1/n)+n^2(\ln(n^2+n)-1))-(n^2(\ln(n^2)-1)+n(\ln(n^2+n)-1))+O(1/n^2))-2n\ln(n)\\
&=\frac12\ln(1+1/n)+n^2\ln(1+1/n)+n(\ln(n^2+n)-1))+O(1/n^2))-2n\ln(n)\\
&=\frac12\ln(1+1/n)+n^2\ln(1+1/n)+n(2\ln(n)+\ln(1+1/n)-1))+O(1/n^2))-2n\ln(n)\\
&=\frac12\ln(1+1/n)+n^2\ln(1+1/n)+n(\ln(1+1/n)-1))+O(1/n^2))\\
&=(n^2+n+\frac12)\ln(1+1/n)-n+O(1/n^2))\\
&=(n^2+n+\frac12)(\frac1{n}-\frac1{2n^2}+\frac1{3n^3}+O(\frac1{n^4}))-n+O(1/n^2))\\
&=(n+\frac12+\frac1{3n}+O(\frac1{n^2}))-n+O(1/n^2))\\
&=\frac12+\frac1{3n}+O(\frac1{n^2})\\
\end{array}
$
