A_n = (1 + (3/n) )^4n

I don't know hot to find if this converges. I feel like I should break up the exponent, but I don't know how that would help.

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- Nov 3rd 2008, 09:03 PMveronicak5678Sequences
A_n = (1 + (3/n) )^4n

I don't know hot to find if this converges. I feel like I should break up the exponent, but I don't know how that would help. - Nov 3rd 2008, 09:52 PMJameson
This uses a couple of neat tricks.

Let this limit equal A. Then take the natural log of both sides and you get:

Now if you try to evaluate this limit for n -> infinity, you get an indeterminate form, so you can use L'Hopitals rule.

The right hand side stays the same because after we apply the rule, nothing changes about the limit.

Now it's just simplifying and you should get it down to:

Raise both sides to the power of e and you get - Nov 4th 2008, 09:37 AMveronicak5678
That is pretty neat, but is there any other way to do that? Since we haven't done anything like that yet in class, I feel like there should be a more obvious way...

- Nov 4th 2008, 09:58 AMveronicak5678
I never learned the part you're asking me to remember. Can you show me why that is true?

- Nov 4th 2008, 12:26 PMSoroban
Hello, veronicak5678!

We're expected to know this fact:

. .

Quote:

Multiply the exponent by

. .

. . and we have: .

- Nov 4th 2008, 12:34 PMveronicak5678
I see! I'm beginning to think this was in the part of the lecture my teacher never got to before time ran out. Thanks so much for your help!

- Nov 4th 2008, 01:42 PMJameson
I made a mistake obviously somewhere. My apologies. Soroban was there to save the day as usual.