Determine series convergence or divergence: a) Sum from 0 to inifinity (ne^(-n^2))

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Determine series convergence or divergence: a) Sum from 0 to inifinity (ne^(-n^2))

Mathematics
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[f(x)=n \times e ^{-n ^{2}}\] is this the function ?
\[limt \frac{n}{e ^{n2}}\] if we sub value of n then we got \[\frac{\infty}{\infty}\] drive The numerator, and denominator \[\lim \frac{1}{2n \times e^{n^{2}}}\] if we sub infinity \[\frac{1}{\infty} = zero\] also its looks like true by the plot http://www.wolframalpha.com/input/?i=n%2F%28e%5En%5E2%29 Wish I'm right ... you are welcome.
Use a comparison test; for example compare the series with 1/n^2 (which is a convergent series) since (((1/(n²)))/((n/(e^{n²}))))=((e^{n²})/(n³))→∞ you get that eventually (ie beginning with a certain index) 1/n^2 dominates your general term; since both are positive and the bigger one converges, it follows that your series is also convergent The above argument (waleed kha) doesn't work: if the general term goes to 0, then then anything may happen.... for example the general term 1/n goes to 0 but the series diverges

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Not the answer you are looking for?

Search for more explanations.

Ask your own question