Here's the question you clicked on:
tux
Decide convergence / divergence: sum n=1 to infinity 1/(n*sqrt(n)+2)
since exponent in denominator is greater than 1
How to solve using comparison test or integral test?
integral test is just integrate from 1 to infinity
comparison is a limit of the ratio of .... \[\frac{a_{n+1}}{a_n}\]
actually that is "ratio test" comparison means compare to something else you know converges (or diverges)
of is the comparison a b_n of another ..... yeah lol
\[\sum_1^{\infty}\frac{1}{n\sqrt{n}+2}\]compare \[\frac{1}{n\sqrt{n}+2}<\frac{1}{n^{\frac{3}{2}}}\] second sum converges