How do you know if sumn/(e(n^2))ne(n2) converges from 1 to infinity?

2 Answers
May 25, 2015

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/nn=0ne(n2)=1en=onn2=1en=o1n

Thus we can see that sum_(n=0)^oo 1/nn=01n 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/n1en=01n is divergent

Proof of the divergence of the Harmonic Series.

May 26, 2015

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

sum_1^oo n/e^(n^2)1nen2

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

For f(x) = x/e^(x^2)f(x)=xex2, 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.