Question

Determine series convergence or divergence:

a) Sum from 0 to inifinity (ne^(-n^2))

Answer 1

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

Answer 2

\[limt \frac{n}{e ^{n2}}\]
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.

Answer 3

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