# Sequences, convergence

• Aug 31st 2011, 10:03 AM
Arron
Sequences, convergence
I need to show if the following sequence converges and its limit.

3n^2+4^n+2^n/n^23^n+4n^7

The dominant term is 4^n.

Dividing both numerator and denominator by 4^n, we get

3n^2/4^n + 1 + (2/4)^n / n^2(3/4)^n + 4n^7/4^n

Since n^2/4^n, (2/4)^n, n^2(3/4)^n and n^7/4^n are basic null sequences, we find, by the combination rules, that

lim = 0+1+0/0+0 = ?

not sure what to do. Can anyone help?
• Aug 31st 2011, 10:29 AM
Siron
Re: Sequences, convergence
Is the limit:
$\lim_{x\to \infty} \frac{3n^2+4^n+2^n}{(n^{23})^n+4n^7}$
?
• Aug 31st 2011, 11:13 AM
Arron
Re: Sequences, convergence
Sorry the it should read n^2*3^n not (n^23)^n.
• Aug 31st 2011, 11:58 AM
Siron
Re: Sequences, convergence
modified.
• Aug 31st 2011, 12:24 PM
Ted
Re: Sequences, convergence
$\displaystyle \lim_{n\to\infty} \frac{3n^2+4^n+2^n}{n^2 \, 3^n +4n^7}$

Start by dividing top & bottom by $4^n$
• Sep 2nd 2011, 02:07 AM
Arron
Re: Sequences, convergence
I did this and got 3n^2/4^n + 1 + (2/4)^n. I am not sure what to do next.
• Sep 2nd 2011, 03:37 AM
Ted
Re: Sequences, convergence
You should get : $\displaystyle \lim_{n\to\infty} \frac{ \frac{3n^2}{4^n} + 1 + \left( \frac{1}{2} \right)^n } { n^2 \left( \frac{3}{4} \right)^n + \frac{4n^7}{4^n} }$

You need two things here :

(1) L'Hospital's Rule.

(2) $\lim_{n\to\infty} r^n = 0$ if $|r| < 1$.