tux
  • tux
Decide convergence / divergence: sum n=1 to infinity 1/(n*sqrt(n)+2)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
converges for sure
anonymous
  • anonymous
since exponent in denominator is greater than 1
tux
  • tux
How to solve using comparison test or integral test?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
integral test is just integrate from 1 to infinity
amistre64
  • amistre64
comparison is a limit of the ratio of .... \[\frac{a_{n+1}}{a_n}\]
anonymous
  • anonymous
actually that is "ratio test" comparison means compare to something else you know converges (or diverges)
amistre64
  • amistre64
of is the comparison a b_n of another ..... yeah lol
anonymous
  • anonymous
\[\sum_1^{\infty}\frac{1}{n\sqrt{n}+2}\]compare \[\frac{1}{n\sqrt{n}+2}<\frac{1}{n^{\frac{3}{2}}}\] second sum converges

Looking for something else?

Not the answer you are looking for? Search for more explanations.