<s>[INST] Show that $f(n) = -\sqrt{n} + n\log\left(1 + \frac{1}{\sqrt{n}}\right), f:\mathbb{N}\rightarrow \mathbb{R}$ is a decreasing function. The way I tried to approach this question was to consider it as a continuous $f: \mathbb{R} \rightarrow \mathbb{R}$. I then tried to show that $\frac{d}{dx}f < 0$. So
\begin{align*}
\frac{d}{dx}f(x) &= -\frac{1}{2\sqrt{x}} + \log\left(1 + \frac{1}{\sqrt{x}}\right) -\frac{1}{2} \frac{1}{1+ \sqrt{x}} \left(x^{-\frac{3}{2}}\right) \\
&= \log\left(1 + \frac{1}{\sqrt{x}}\right) - \frac{1}{2}\left(\frac{2 + \sqrt{x}}{x + \sqrt{x}}\right).
\end{align*}
Using the substitution $x = \frac{1}{u^2}$ gives
$$\log(1+u) - \frac{1}{2}\left(\frac{u^2}{u+1} + u\right).$$
At this point I'm stuck. How do I show this thing is negative for $u>0$?
 [/INST] Hint: To complete what you had begun, compute the second derivative and study the function $f'$. Deduce its sign from its table of variations.
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