Question

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.

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.