<s>[INST] How to prove that $(1 + \frac{1}{n})^{\sqrt{n(n+1)}} < e$? I've been trying to solve the following problem:
Show that $\ln{(k+1)} - \ln{k} = \ln{(1 + \frac{1}{k})} \leq \frac{1}{\sqrt{k(k+1)}}$
EDIT: the title was inaccurate, my bad. 
So what we have to prove is that the upper limit of $(1 + \frac{1}{n})^{\sqrt{n(n+1)}}$ equals $e$.
 [/INST] More information...
As $n \to +\infty$, we have
\begin{align*}
\sqrt{n(n+1)} &= n + \frac{1}{2} - \frac{1}{8n} + O(n^{-2})
\\
\log\left(1+\frac{1}{n}\right) &= \frac{1}{n} - \frac{1}{2n^2} +
\frac{1}{3n^3} + O(n^{-4})
\\
\sqrt{n(n+1)}\log\left(1+\frac{1}{n}\right) &=
1-\frac{1}{24n^2}+O(n^{-3})
\end{align*}
Therefore, for large enough $n$, 
\begin{align*}
\sqrt{n(n+1)}\log\left(1+\frac{1}{n}\right) & <
1
\\
\left(1+\frac{1}{n}\right)^{\sqrt{n(n+1)}} &< e
\end{align*}
</s>