How to prove that $(1 + \frac{1}{n})^{\sqrt{n(n+1)}} < e$?

Question

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 $e$.

Answer 1

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*}

Answer 2

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): $$\log(x) = \int_1^x\frac{\mathrm{d}t}t < \left(\int_1^x\frac{\mathrm{d}t}{t^2}\right)^{\frac12} \left(\int_1^x\mathrm{d}t\right)^{\frac12}=\sqrt{x}-\frac1{\sqrt{x}}.$$