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

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

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\[\sum_{k=1}^{\infty} \frac{k^{60}}{e^k} = \lim_{k\to\infty} \frac{(k+1)^{60}}{e^{k+1}}*\frac{e^k}{k^{60}} = \lim_{k\to\infty} \frac{(k+1)^{60}}{e^{k}e}*\frac{e^k}{k^{60}} = \lim_{k\to\infty} \frac{(k+1)^{60}}{e}*\frac{1}{k^{60}} \]
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It is very very wrong to say that the series equals that limit
ah, I was just checking for convergence.
well 1/e, because limit n->0 of { (k+1)/k }^60 is 1
I mean k -> ∞. ops
(if you had k instead of 60 in the exponent, then it would be 1/e^2)
but wit 60, it is still conv based ratio test, since |r|<1
ahh, that never occurred to me.
what do you mean?
the ^k case?
that I could just combine k+1 and 1/k to form ((k+1)/k)^(60)
Thanks! I kept trying to find a factor to cancel out k^60, forgot about that property. I will close this now.
Well, if we do algebra with limit properties: \[\large \lim_{k \rightarrow \infty} \frac{(k+1)^{60}e^{k}}{k^{60}e^{k+1}}\] \[\large \lim_{k \rightarrow \infty} \frac{(k+1)^{60}}{k^{60}e^{1}}\] \[(1/e) \times \left(\large \lim_{k \rightarrow \infty} \frac{(k+1)^{60}}{k^{60}} \right)\] \[(1/e) \times \left(\large \lim_{k \rightarrow \infty} (\frac{k+1}{k})^{60} \right)\] \[(1/e) \times \left(\large \lim_{k \rightarrow \infty} \frac{k+1}{k} \right)^{60}\]
(1/e) times 1^(60) = 1/e
yep yep. well thanks a lot, I can go to sleep now.
lol, good night

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