Question

Does the Series ke^-k^2 converge or diverge?

Answer 1

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

Answer 2

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

Answer 3

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?

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

applied the ratio test yet?

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

right so far?

Answer 15

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

Answer 16

expand the top exponent and subtract the bottom one

Answer 17

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

Answer 18

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