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 $e$.
 A: 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*}
A: Take logarithms on both sides. Use $$\log(x) < \sqrt{x} - \frac1{\sqrt{x}}$$ for $x>1$ and simplify.
The logarithmic inequality can be proved in the following two ways. First one follows from overestimating the function $1/t$ on the interval $[1, x]$ and the trapezoidal rule:
$$\log(x)=\int_1^x\frac{\mathrm{d}t}t< \frac12\left(1+\frac1x\right)(x-1)= \frac12 \left(x - \frac1x\right).$$
Then substitute $x \leftarrow \sqrt{x}$ to get the desired inequality. Second one uses the Cauchy-Schwarz inequality (using strict inequality since the functions are not linearly dependent):
$$\log(x) = \int_1^x\frac{\mathrm{d}t}t < \left(\int_1^x\frac{\mathrm{d}t}{t^2}\right)^{\frac12} \left(\int_1^x\mathrm{d}t\right)^{\frac12}=\sqrt{x}-\frac1{\sqrt{x}}.$$
