Sequences

• Nov 3rd 2008, 08:03 PM
veronicak5678
Sequences
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, 08:52 PM
Jameson
This uses a couple of neat tricks.

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

$\displaystyle \ln(A)=4n\ln(1+\frac{3}{n})$

$\displaystyle \frac{\ln(1+\frac{3}{n})}{(4n)^{-1}}=\ln(A)$

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

$\displaystyle \frac{(\frac{n}{n+3}) (\frac{-3}{n^2})}{\frac{-4}{n^2}}=\ln(A)$

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:

$\displaystyle \frac{3n}{4n+12}=\ln(A)$

Raise both sides to the power of e and you get $\displaystyle A=e^{\frac{3}{4}}$
• Nov 4th 2008, 08:37 AM
veronicak5678
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, 08:58 AM
veronicak5678
I never learned the part you're asking me to remember. Can you show me why that is true?
• Nov 4th 2008, 11:26 AM
Soroban
Hello, veronicak5678!

We're expected to know this fact:

. . $\displaystyle \lim_{z\to\infty}\left(1 + \tfrac{1}{z}\right)^z \;=\;e$

Quote:

$\displaystyle A_n \:= \:\left(1 + \tfrac{3}{n}\right)^{4n}$
We want: .$\displaystyle \lim_{n\to\infty}\left(1 + \tfrac{3}{n}\right)^{4n}$

Multiply the exponent by $\displaystyle \tfrac{3}{3}$

. . $\displaystyle \left(1 + \tfrac{3}{n}\right)^{4n\cdot\frac{3}{3}} \;=\;\left(1 + \tfrac{3}{n}\right)^{\frac{n}{3}\cdot12} \;=\; \bigg[(1 + \tfrac{3}{n})^{\frac{n}{3}}\bigg]^{12} \;= \bigg[\left(1 + \frac{1}{\frac{n}{3}}\right)^{\frac{n}{3}}\bigg]^{12}$

$\displaystyle \text{Let }\:z \,=\,\tfrac{n}{3}$

. . and we have: .$\displaystyle \lim_{z\to\infty}\bigg[\left(1 + \frac{1}{z}\right)\bigg]^{12} \;=\;\bigg[\lim_{z\to\infty}\left(1 + \tfrac{1}{z}\right)^z\bigg]^{12} \;=\;e^{12}$

• Nov 4th 2008, 11:34 AM
veronicak5678
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, 12:42 PM
Jameson
I made a mistake obviously somewhere. My apologies. Soroban was there to save the day as usual.