- #1

- 426

- 25

^{n}

I have proven that the sequence converges numerically, but I can't do it analytically, and can't do anything for the series (maybe it the series doesn't converge?)

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- Thread starter swampwiz
- Start date

- #1

- 426

- 25

I have proven that the sequence converges numerically, but I can't do it analytically, and can't do anything for the series (maybe it the series doesn't converge?)

- #2

- 27

- 0

Or if you want the asymptotic behavior you can use the stirling's approximation

- #3

- 371

- 0

Use the ratio test

- #4

- 129

- 0

If the series is going to converge, the sequence needs to go to 0, which it does as n!/n^n <= 1/n^2 (so use the squeze theorem).

By the p test(you could also use ratio test), we know the series where x_n = 1/n^2 converges.

Thus by the comparision test we see that the series with x_n = n!/n^n converges.

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