Question

does the sum of e^(-n^2) from 0 to infinity converge or diverge? how do you use the direct comparison test?

Answer 1

\[\huge \sum_{n=0}^{\infty}e^{-n^2}=\sum_{n=0}^{\infty}\frac{1}{e^{n^2}}\]

Answer 2

First of all, what do you suspect? Having a vague idea can be useful...
Do you suspect that this converges or diverges?

Answer 3

And while you're thinking about that, allow me to refresh you on direct comparison test... suppose you have two non-negative series
\[\huge \sum_{n=k}^{\infty}a_n \ \ \ \ \ \text{and} \ \ \ \ \ \ \sum_{n=k}^{\infty}b_n\]Such that there is an N>0 where for all n>N
\[\huge a_n \le b_n\]
Then if
\[\huge \sum_{n=k}^{\infty}a_n\]is divergent, then so is \[\huge \sum_{n=k}^{\infty}b_n\]







And if
\[\huge \sum_{n=k}^{\infty}b_n\]is convergent, then so is \[\huge \sum_{n=k}^{\infty}a_n\]

Answer 4

it seems like it should be convergent because n is to 2 which is greater than 1

Answer 5

Stop at "It should be convergent" LOL
You're asked to do a direct comparison test, right? What are the convergent series that you know of that have its terms greater than or equal to the terms of this series?

Answer 6

im not sure that is the problem

Answer 7

To use the Direct Comparison test, if you suspect a series is convergent, try looking for a known convergent series having terms greater than or equal to your original series.

If, on the other hand, you suspect a series is divergent, try looking for a known divergent series having terms less than or equal to your original series.

Answer 8

would the series 1/n^2 work?

Answer 9

Let's see :)
Is it true that
\[\huge \frac1{e^{n^2}}\le\frac1{n^2}\]?

Answer 10

yes that is true

Answer 11

You know, of course, that \[\Large e^{n^2}\]increases exponentially, and \[\Large n^2\]is just a polynomial, which doesn't increase nearly as quickly.

So... in fact, for n = 1, onwards
\[\huge n^2<e^{n^2}\]

Answer 12

So, dividing both sides by n squared... and since n squared is positive
\[\huge 1 < \frac{e^{n^2}}{n^2}\]and now dividing both sides by e to the n squared
\[\huge \frac1{e^{n^2}}<\frac1{n^2}\]

Answer 13

but since it is 1/e^n^2 and 1/n^2 that would make 1/(e^(n^2)) less than 1/(n^2) right?

Answer 14

That's right :)
And so... by the comparison test...? ^.^

Answer 15

e^(-n^2) is convergent?

Answer 16

Yeap.
That was fun :D

Answer 17

thank you:)