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anonymous

  • 4 years ago

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

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  1. anonymous
    • 4 years ago
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    \[f(x)=n \times e ^{-n ^{2}}\] is this the function ?

  2. anonymous
    • 4 years ago
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    \[limt \frac{n}{e ^{n2}}\] 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
    • 4 years ago
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    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

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