## Calculus (3rd Edition)

The series $\Sigma_{n=1}^{\infty} \frac{1}{\sqrt n+n}$ diverges.
Take $a_n= \frac{1}{\sqrt n+n}$ and $b_n= \frac{1}{n}$; now we have the limit comparison test $$L=\lim_{n \to \infty}\frac{a_n}{ b_n}=\lim_{n \to \infty}\frac{1/(\sqrt n+n)}{1/n}\\ =\lim_{n \to \infty}\frac{n}{\sqrt n+n}=\lim_{n \to \infty}\frac{1}{(1/\sqrt n)+1}=1$$ Since $L\gt 0$ and the series $\Sigma_{n=1}^{\infty} \frac{1}{n}$ diverges then the series $\Sigma_{n=1}^{\infty} \frac{1}{\sqrt n+n}$ diverges.