<s>[INST] Prove that $\displaystyle \lim_{N\to \infty}(1+\frac{0}{N^2})(1+\frac{1}{N^2})..(1+\frac{N-1}{N^2}) = \sqrt e$ Given that $\dot x(t) = tx$, $x(0) = 1$. Then we find the solution $x(t)=e^{t^2/2}$. This implies $x(1)=\sqrt e$.
Then I use Euler's method to find $x(1)$. Dividing the unit time interval into $N$
pieces, we get the step size $h = \frac{1}{N}$.
$x_{n+1}=x_n+f(t_n,x_n)h=x_n+\frac{n}{N}\cdot x_n\cdot \frac{1}{N}=(1+\frac{n}{N^2})\cdot x_n$.
Since $x(0)=1$, then I get $x_N=(1+\frac{0}{N^2})(1+\frac{1}{N^2})..(1+\frac{(N-1)}{N^2})$.
So, $\displaystyle \lim_{N\to\infty}x_N=\sqrt e$.
Is this limit correct? Are there other ways to show  $(1+\frac{0}{N^2})(1+\frac{1}{N^2})..(1+\frac{(N-1)}{N^2}) \to \sqrt e$ as $N\to \infty.$
 [/INST] Noting for small $x>0$,
$$ \frac{x}{1+x}<\ln(1+x)<x $$
then one has
\begin{eqnarray}
\sum_{i=1}^{N-1}\frac{i}{N^2+i}\le\sum_{i=1}^{N-1}\ln\bigg(1+\frac{i}{N^2}\bigg)\le\sum_{i=1}^{N-1}\frac{i}{N^2}.
\end{eqnarray}
But
$$ \sum_{i=1}^{N-1}\frac{i}{N^2+i}\ge\sum_{i=1}^{N-1}\frac{i}{N^2+N}=\frac{N(N-1)}{2(N^2+N)}, \sum_{i=1}^{N-1}\frac{i}{N^2}=\frac{N(N-1)}{2N^2}$$
and letting $N\to\infty$, one can obtain
$$ \lim_{N\to\infty}\sum_{i=1}^{N-1}\ln\bigg(1+\frac{i}{N^2}\bigg)=\frac12.$$
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