<s>[INST] How to prove $(n + \frac{1}{2})\log(1+\frac{1}{n})$ is increasing in $n$? Is $(n + \frac{1}{2})\log(1+\frac{1}{n})$ increasing in $n$?
I attempted to differentiate $p_n=(n + \frac{1}{2})\log(1+\frac{1}{n})$ with respect to $n$.
$p_n^{'} = \log(1+\frac{1}{n}) + \frac{(n+\frac{1}{2})}{(1+\frac{1}{n})}(-\frac{1}{n^2}) = \log(1+\frac{1}{n}) - \frac{1}{2}(\frac{1}{n}+\frac{1}{n+1})$.
I am stuck at this point and I cannot prove that $\log(1+\frac{1}{n}) - \frac{1}{2}(\frac{1}{n}+\frac{1}{n+1}) > 0$.
I know by the definition $\log (1+\frac{1}{n}) =\int_1^{1+\frac{1}{n}}\frac{1}{x}dx$.
But I can only reach to the point where $ \frac{1}{n+1} < \log (1+\frac{1}{n}) < \frac{1}{n}$.
Thanks for your help in advance.
 [/INST] Follow your argument. If you can show that $f(x)=(x+\frac{1}{2}) ln (1+\frac{1}{x})$ is decreasing when $x\ge 1$, then your problem is solved. Define $g(x)=f^\prime (x)=ln (1+\frac{1}{x})-\frac{1}{2}(\frac{1}{x}+\frac{1}{x+1})=ln(x+1)-ln(x)-\frac{1}{2}(\frac{1}{x}+\frac{1}{x+1})$. Then $g^\prime (x)=\frac{1}{x+1}-\frac{1}{x}+\frac{1}{2x^2}+\frac{1}{2(x+1)^2}=\frac{1}{2x^2 (x+1)^2}>0$. Since $g(1)=ln(2)-0.75<0$ and $g(+\infty )=0$, $g(x)<0$ when $x\ge 1$, i.e., $f(x)$ is decreasing.
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