anonymous
  • anonymous
Do the following series converge?
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\sum_{n=1}^{\infty} \frac{ e^{-\sqrt{n}} }{ \sqrt{n} }\] \[\sum_{n=2}^{\infty} \frac{ 1 }{ n(\ln n)^3 }\]
geerky42
  • geerky42
I don't do this thing yet so I may be wrong... Ask yourself a question. Which increase faster as n approaches to infinity? \(\large e^{-\sqrt{n}}\) or \(\sqrt{n}\). If \(\large e^{-\sqrt{n}}\) increase faster or approach to something greater than \(\sqrt{n}\), then it diverges. Otherwise it converges. What does \(n(\ln n )^3\) approach to when n approach to infinity? If it approaches to something under one, then it diverges, otherwise it converges.
amoodarya
  • amoodarya
for the first use ratio test that is converegent and for the second use integral test see what i draw|dw:1360527817600:dw| that is also conv

Looking for something else?

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

More answers

anonymous
  • anonymous
Thank you!

Looking for something else?

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