## anonymous 4 years ago Determine series convergence or divergence: a) Sum from 0 to inifinity (ne^(-n^2))

1. anonymous

$f(x)=n \times e ^{-n ^{2}}$ is this the function ?

2. anonymous

$limt \frac{n}{e ^{n2}}$ 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.

3. cristiann

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