## anonymous one year ago Does the Series ke^-k^2 converge or diverge?

1. anonymous

$\sum_{k=1}^{infinity}k e^{-k ^{2}}$

2. anonymous

$$\large \sum_{k=1}^{\infty}k e^{-k ^{2}}=$$ notice that is close to the derivative of $$\large e^{-x^2}$$

3. anonymous

Oh, right, there's a thing in the book about an integral test. We haven't discussed it in class yet, but, I'm guessing that's what this is?

4. anonymous

sort of, I was thinking of this http://prntscr.com/8j57eg

5. anonymous

That looks significantly more complicated than what we've been doing thus far.

6. anonymous

but now that you mention it, you can use integral test

7. anonymous

Okay then, that's good. I'm reading up on it now, might ask some questions

8. IrishBoy123

applied the ratio test yet?

9. anonymous

Ratio test didn't seem to work, unless I made some algebraic errors

10. IrishBoy123

$\large \frac{(k+1) e^{-(k+1) ^{2} }}{k e^{-k ^{2}}}=$

11. anonymous

Right, then I ended up with some mess with the exponents that didn't seem to work out nicely

12. IrishBoy123

$\large \frac{k+1}{k} e^{-(2k+1)}$

13. anonymous

To save you the work of typing out a whole bunch of equations, if there's some way to do it with the ratios, I should be able to handle it if you can help me through the algebra

14. IrishBoy123

right so far?

15. anonymous

Umm one moment. That's not what I did so that's probably where I went wrong

16. IrishBoy123

expand the top exponent and subtract the bottom one

17. anonymous

Right that makes sense. I dunno why I didn't see that.

18. anonymous

because the e^(-2k+1) is in the denominator. nice work irishboy