Is the sequence $a_{n}=\prod\limits_{i=1}^{n}\left(1+\frac{i}{n^2}\right)$ decreasing? 
let $$a_{n}=\left(1+\dfrac{1}{n^2}\right)\left(1+\dfrac{2}{n^2}\right)\cdots \left(1+\dfrac{n}{n^2}\right)$$
since $$a_{1}=2,a_{2}=\dfrac{15}{8},a_{3}=\dfrac{1320}{729}\cdots $$

I found
$$a_{1}>a_{2}>a_{3}>\cdots$$

I conjecture
$\{a_{n}\}$ be monotone decreasing?

remark: It is well known this following limts
$$\lim_{n\to+\infty}a_{n}=e^{\lim_{n\to\infty}\sum_{i=1}^{n}\ln{\left(1+\frac{i}{n}\right)}}=e^{1/2}$$But the result seems to speculation regarding the inequality of useless,so how to prove inequality
$$a_{n}>a_{n+1},\forall n\in N^{+}$$
 A: Taking logarithms is often helpful when dealing with products, so let's try that here too. Grouping terms in the difference of the logarithms, we obtain
$$\log a_n - \log a_{n+1}
= \sum_{k = 1}^n \Biggl(\log\biggl(1 + \frac{k}{n^2}\biggr) - \log \biggl(1 + \frac{k}{(n+1)^2}\biggr)\Biggr) - \log \biggl(1 + \frac{1}{n+1}\biggr).$$
Now we want to bound the differences $\log \bigl(1 + \frac{k}{n^2}\bigr) - \log \bigl(1 + \frac{k}{(n+1)^2}\bigr)$ below, and $\log \bigl(1 + \frac{1}{n+1}\bigr)$ above in such a way that we obtain the desired inequality $\log a_n - \log a_{n+1} > 0$. We note that for $0 \leqslant x < y$ we have
$$\log (1 + y) - \log (1 + x) > \frac{y-x}{1+y}$$
by the mean value theorem. Using this for $x_k = \frac{k}{(n+1)^2},\, y_k = \frac{k}{n^2}$, we get
\begin{align}
\sum_{k = 1}^n \bigl(\log (1 + y_k) - \log (1 + x_k)\bigr)
&> \sum_{k = 1}^n \frac{y_k - x_k}{1+y_k} \\
&= \biggl(\frac{1}{n^2} - \frac{1}{(n+1)^2}\biggr) \sum_{k = 1}^n \frac{k}{1+ y_k} \\
&= \frac{2n+1}{(n+1)^2}\sum_{k = 1}^n \frac{y_k}{1+y_k} \\
&> \frac{2n+1}{(n+1)^2}\sum_{k = 1}^n \bigl(y_k - y_k^2\bigr) \\
&= \frac{2n+1}{(n+1)^2}\biggl(\frac{n+1}{2n} - \frac{(n+1)(2n+1)}{6n^3}\biggr) \\
&= \frac{2n+1}{(n+1)^2}\biggl(\frac{1}{2} + \frac{1}{6n} - \frac{1}{2n^2} - \frac{1}{6n^3}\biggr) \\
&= \frac{1}{(n+1)^2}\biggl(n + \frac{1}{2} + \frac{1}{3} -\frac{5}{6n} - \frac{1}{3n^2} - \frac{1}{6n^3}\biggr) \\
&= \frac{1}{(n+1)^2}\biggl((n+1) - \frac{1}{2} + \frac{1}{3} - \frac{5}{6n} - \frac{1}{3n^2} - \frac{1}{6n^3}\biggr),
\end{align}
and an easy verification shows that the final expression is greater than
$$\frac{1}{n+1} - \frac{1}{2(n+1)^2} + \frac{1}{3(n+1)^3} > \log \biggl(1 + \frac{1}{n+1}\biggr)$$
for $n \geqslant 5$.
The remaining cases are verified by hand:
$$a_1 = 2 > a_2 = \frac{15}{8} > a_3 = \frac{440}{243} > a_4 = \frac{14535}{8192} > a_5 = \frac{3420144}{1953125}.$$
So indeed the sequence $(a_n)$ is strictly decreasing.
