Examine convergence $\sum_{n=1}^{\infty} \left( \frac{1}{\sqrt{n}} - \sqrt{\ln{\frac{n+1}{n}}} \right) $

Question

I need some help with following series:

$$\sum_{n=1}^{\infty} \left( \frac{1}{\sqrt{n}} - \sqrt{\ln{\frac{n+1}{n}}} \right)$$

I have no idea how I should modify or compare it with something else.

Answer 1

$$\ln\frac{n+1}{n}=\frac1n\left(1-\frac1{2n}+O(n^{-2})\right)$$ and so $$\sqrt{\ln\frac{n+1}{n}}=\frac1{\sqrt{n}}\left(1-\frac1{4n}+O(n^{-2})\right).$$ Then $$\frac1{\sqrt n}-\sqrt{\ln\frac{n+1}{n}}=\frac1{4n^{3/2}}+O(n^{-5/2}).$$

Answer 2

Also, from $$\frac{x}{1+x}\leq \ln{(1+x)}\leq x, \forall x>-1$$ we have $$\sqrt{\frac{1}{n+1}}\leq \sqrt{\ln{\left(1+\frac{1}{n}\right)}}\leq \frac{1}{\sqrt{n}}$$ or $$0\leq \frac{1}{\sqrt{n}}-\sqrt{\ln{\left(1+\frac{1}{n}\right)}} \leq \frac{1}{\sqrt{n}}-\sqrt{\frac{1}{n+1}}$$ where $$0\leq \frac{1}{\sqrt{n}}-\sqrt{\frac{1}{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}\sqrt{n}}< \frac{\sqrt{n+1}-\sqrt{n}}{n}=\frac{1}{n(\sqrt{n+1}+\sqrt{n})}< \frac{1}{2\sqrt{n^3}}$$ As a result $$0\leq\sum\limits_{n=1}^{\infty} \left( \frac{1}{\sqrt{n}} - \sqrt{\ln{\frac{n+1}{n}}} \right)< \frac{1}{2}\sum\limits_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}$$ which converges, since $\frac{3}{2}>1$.