Question

Calculus: Convergence test (ratio test): How to simplify this further?

Answer 1

\[\sum_{k=1}^{\infty} \frac{k^{60}}{e^k} = \lim_{k\to\infty} \frac{(k+1)^{60}}{e^{k+1}}*\frac{e^k}{k^{60}} = \lim_{k\to\infty} \frac{(k+1)^{60}}{e^{k}e}*\frac{e^k}{k^{60}} = \lim_{k\to\infty} \frac{(k+1)^{60}}{e}*\frac{1}{k^{60}} \]

Answer 2

I can't see the code

Answer 3

It is very very wrong to say that the series equals that limit

Answer 4

ah, I was just checking for convergence.

Answer 5

well 1/e,

because limit n->0 of { (k+1)/k }^60 is 1

Answer 6

I mean k -> ∞. ops

Answer 7

(if you had k instead of 60 in the exponent, then it would be 1/e^2)

Answer 8

but wit 60, it is still conv based ratio test, since |r|<1

Answer 9

ahh, that never occurred to me.

Answer 10

what do you mean?

Answer 11

the ^k case?

Answer 12

that I could just combine k+1 and 1/k to form ((k+1)/k)^(60)

Answer 13

Thanks! I kept trying to find a factor to cancel out k^60, forgot about that property. I will close this now.

Answer 14

Well, if we do algebra with limit properties:

\[\large \lim_{k \rightarrow \infty} \frac{(k+1)^{60}e^{k}}{k^{60}e^{k+1}}\]
\[\large \lim_{k \rightarrow \infty} \frac{(k+1)^{60}}{k^{60}e^{1}}\]
\[(1/e) \times \left(\large \lim_{k \rightarrow \infty} \frac{(k+1)^{60}}{k^{60}} \right)\]
\[(1/e) \times \left(\large \lim_{k \rightarrow \infty} (\frac{k+1}{k})^{60} \right)\]
\[(1/e) \times \left(\large \lim_{k \rightarrow \infty} \frac{k+1}{k} \right)^{60}\]

Answer 15

(1/e) times 1^(60) = 1/e

Answer 16

yep yep. well thanks a lot, I can go to sleep now.

Answer 17

lol, good night