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.
 A: The derivative of the function being
$$f(n)=\log \left(1+\frac{1}{n}\right)-\frac{n+\frac{1}{2}}{\left(1+\frac{1}{n}\right)
   n^2}$$
$f(1)=\log (2)-\frac{3}{4} <0$ and when $n$ is large
$$f(n)=-\frac{1}{6 n^3}+O\left(\frac{1}{n^4}\right)$$
Morover
$$f'(n)=\frac{1}{2 n^2 (n+1)^2}$$ make that the function never changes concavity.
Then, the function is always decreasing.
A: When in doubt, plot the function. As you can see, the function is strictly decreasing.
A: As $n\to\infty$, $\log\left(1 + \dfrac1n\right)\to0$. Similarly, as $n\to\infty$, $\dfrac1n\to0$ and $\dfrac1{n+1}\to0$ and, therefore, $\dfrac12\left(\dfrac1n + \dfrac1{n + 1}\right)\to0$. So, $p_n^{'}\to0$ as $n\to\infty$ and, therefore, $p_n$ is not increasing in $n$.
A: 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.
