Convergence of the series $\sum_{n=1}^\infty [\frac{1}{\sqrt{n}}-\sqrt{\ln(1+\frac{1}{n})}]$. I know the series converges. I am not allowed to use any calculus to solve this problem. My professor suggested that I try and use the limit comparison test, but I do not know how to evaluate any limits involving these terms.
 A: What you could do is to use the Taylor expansion 
$$\log \left(1+\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ Now, using the binomial expansion
$$\sqrt{\log \left(1+\frac{1}{n}\right)}=\frac 1 {\sqrt n}-\frac 14 \frac 1 {n^{3/2}}+O\left(\frac{1}{n^{5/2}}\right)$$
$$\frac 1 {\sqrt n}-\sqrt{\log \left(1+\frac{1}{n}\right)}=\frac 14 \frac 1 {n^{3/2}}+O\left(\frac{1}{n^{5/2}}\right)$$ 
A: $\displaystyle 0<\sum\limits_{n=1}^\infty \left(\frac{1}{\sqrt{n}}-\sqrt{\ln\left(1+\frac{1}{n}\right)}\right)=\sum\limits_{n=1}^\infty\frac{1-\ln\left(1+\frac{1}{n}\right)^n}{\sqrt{n}\left(1+\ln\left(1+\frac{1}{n}\right)^n\right)}<$
$\displaystyle <\,\sum\limits_{n=1}^\infty\frac{\ln\left(1+\frac{1}{n}\right)^{n+1}-\ln\left(1+\frac{1}{n}\right)^n}{\sqrt{n}}=\sum\limits_{n=1}^\infty\frac{\ln\left(1+\frac{1}{n}\right)}{\sqrt{n}}\leq\sum\limits_{n=1}^\infty\frac{1}{n\sqrt{n}}=\zeta(\frac{3}{2})$
Used:
$\displaystyle \frac{1}{1+\ln\left(1+\frac{1}{n}\right)^n}<1~~$ and $~~\displaystyle 1\leq\ln\left(1+\frac{1}{n}\right)^{n+1}~~$ and $~~\displaystyle \ln\left(1+\frac{1}{n}\right)\leq\frac{1}{n}~$ for all $~n\in\mathbb{N}$
