# Compute the limit (3^n)/(2^n)+1

• January 22nd 2009, 08:46 AM
sebjory
Compute the limit (3^n)/(2^n)+1
Compute the limit of the sequence

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

I have simplified a much harder problem to this limit and i am having a mind blank... helpplease? cheers
• January 22nd 2009, 08:58 AM
h2osprey
The $n$th term tends to $(3/2)^n$.
• January 22nd 2009, 09:43 AM
ThePerfectHacker
Quote:

Originally Posted by sebjory
Compute the limit of the sequence

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

I have simplified a much harder problem to this limit and i am having a mind blank... helpplease? cheers

$\frac{3^n}{2^n + 1} \geq \frac{3^n}{2^n + 2^n} = \tfrac{1}{2}\left( \tfrac{3}{2} \right)^n \to \infty$
• January 22nd 2009, 09:51 AM
Jester
Quote:

Originally Posted by sebjory
Compute the limit of the sequence

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

I have simplified a much harder problem to this limit and i am having a mind blank... helpplease? cheers

Could also use L'Hoptial's rule on

$\lim_{x \to \infty} \frac{3^x}{2^x+1} = \lim_{x \to \infty} \frac{3^x \ln 3}{2^x \ln 2} = \frac{\ln 3}{\ln 2} \lim_{x \to \infty} \left( \frac{3}{2} \right)^x \to \infty$ as seen in the ThePerfectHacker post.
• January 22nd 2009, 10:23 AM
Soroban
Hello, sebjory!

Yet another approach . . .

Quote:

$\lim_{n\to\infty} \frac{3^n}{2^n + 1}$

Divide top and bottom by $3^n\!:\;\;\frac{\frac{1}{3^n}\cdot3^n}{\frac{1}{3^ n}(2^n+1)} \;=\;\frac{1}{\frac{2^n}{3^n} + \frac{1}{3^n}}$

Therefore: . $\lim_{n\to\infty}\left[\frac{1}{\left(\frac{2}{3}\right)^n + \frac{1}{3^n}}\right] \;=\; \frac{1}{0+0} \;=\;\infty$

• January 22nd 2009, 03:23 PM
sebjory
thanks a hell of a lot guys that was exactly what i was looking for. Think i might use Sorobans solution nice and elegant. cant believe i didnt see it (Headbang)(Giggle)
• January 22nd 2009, 04:56 PM
Mathstud28
Quote:

Originally Posted by Soroban
Hello, sebjory!

Yet another approach . . .

$\frac{1}{0+0} \;=\;\infty$

Soroban! Nooo! (Rofl)