How do you know if #sumn/(e(n^2))# converges from 1 to infinity?

Answer 1

We can rewrite the sum as:

#sum_(n=0)^oo n/(e(n^2)) = 1/e sum_(n=o)^oo n/n^2 = 1/e sum_(n=o)^oo 1/n#
Thus we can see that #sum_(n=0)^oo 1/n# is the Divergent Harmonic Series.

Thus we have a scalar multiple of a Divergent series, thus we end up with a Divergent series.

so:

#1/e sum_(n=0)^oo 1/n# is divergent

Proof of the divergence of the Harmonic Series.

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Answer 2

It seems possible that the question was supposed to be about the series:

#sum_1^oo n/e^(n^2)#

If this is the series we're interested in, use the integral test.

For #f(x) = x/e^(x^2)#, we have #f'(x) = (1-2x^2)/e^(x^2)#, which is eventually negative (#x > 1/sqrt2#). So #f# is eventually decreasing.
#int x/e^(x^2) dx = int e^(-x^2) x dx#
can be evaluated by substitution: #u = -x^2#
#int x/e^(x^2) dx = int e^(-x^2) x dx = -1/2 e^(-x^2) + C#

So

#int_1^oo x/e^(x^2) dx = lim_(brarroo) (-1/2 e^(-b^2) -+ 1/(2e))#

which converges.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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