Okay whats going on here....

• May 29th 2008, 05:30 PM
evilpostingmongy
Okay whats going on here....
I have no idea how to solve sequence problems. What is the trick? I'm
at explaining them-it has all this jarjon like "well if X>a and n>epsilon then
for all numbers when plotted to infinity" yadda (lol I know I posted jarjon
on purpose) sorry if I sound like a jerk, I'm just frusturated at the moment.(Headbang)
Here's where it didn't make sense:
an=2^n/(3^n+1)

and (2n-1)!/2n+1! The author did some weird manipulation to get the solutions (trying to find out divergence or convergence) and I have no idea
what or even why he did what he did lol. Any help would be greatly appreciated!
• May 29th 2008, 07:03 PM
angel.white
Quote:

Originally Posted by evilpostingmongy
I have no idea how to solve sequence problems. What is the trick? I'm
at explaining them-it has all this jarjon like "well if X>a and n>epsilon then
for all numbers when plotted to infinity" yadda (lol I know I posted jarjon
on purpose) sorry if I sound like a jerk, I'm just frusturated at the moment.(Headbang)
Here's where it didn't make sense:
an=2^n/(3^n+1)

and (2n-1)!/2n+1! The author did some weird manipulation to get the solutions (trying to find out divergence or convergence) and I have no idea
what or even why he did what he did lol. Any help would be greatly appreciated!

you are trying to find out whether the sum converges or diverges for
$\sum_{n=1}^\infty \frac {2^n}{1+3^n}$

Is that the correct question?
• May 29th 2008, 07:07 PM
evilpostingmongy
I'm trying to figure out what I need to find when trying to determine whether or not a sequence converges or diverges and those two problems posted
got me confused.
• May 29th 2008, 07:29 PM
ThePerfectHacker
Quote:

Originally Posted by evilpostingmongy
Here's where it didn't make sense:
an=2^n/(3^n+1)

The first step is to guess what the limit is, it seems to be $\tfrac{2}{3}$.
Next we need to show $|a_n - \tfrac{2}{3}|$ can be made small.
Write, $\left| \tfrac{2^n}{3^n+1} - \tfrac{2}{3} \right| = \left| \tfrac{-2}{3(3^n+1)} \right| = \tfrac{2}{3(3^n+1)}$.
Now it remains to show this quantity gets small, it is easier to bound it.
Note, $\tfrac{2}{3(3^n+1)} \leq \tfrac{2}{3(3^n+0)} \leq \tfrac{3}{3(3^n)} = \tfrac{1}{3^n}$.
Finally note that $\tfrac{1}{3^n}$ can be made sufficiently small.

Here is how the formal proof looks like:
Given $\epsilon > 0$ choose $N\in \mathbb{N}$ so that $\tfrac{1}{3^N} < \epsilon$.
Then if $n\geq N$ it would imply $|a_n - \tfrac{2}{3}| < \epsilon$.
• May 29th 2008, 07:39 PM
Mathstud28
Quote:

Originally Posted by evilpostingmongy
I have no idea how to solve sequence problems. What is the trick? I'm
at explaining them-it has all this jarjon like "well if X>a and n>epsilon then
for all numbers when plotted to infinity" yadda (lol I know I posted jarjon
on purpose) sorry if I sound like a jerk, I'm just frusturated at the moment.(Headbang)
Here's where it didn't make sense:
an=2^n/(3^n+1)

and (2n-1)!/2n+1! The author did some weird manipulation to get the solutions (trying to find out divergence or convergence) and I have no idea
what or even why he did what he did lol. Any help would be greatly appreciated!

As for the second one if you mean

$\lim_{n\to\infty}\frac{(2n-1)!}{(2n+1)!}$

all you need to remember is that

$(2n+1)!=(2n+1)(2n)(2n-1)!$

$\therefore\frac{(2n-1)!}{(2n+1)!}=\frac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\frac{1}{(2n+1)(2n)}$

$\therefore\lim_{n\to\infty}\frac{(2n-1)!}{(2n+1)!}=\lim_{n\to\infty}\frac{1}{(2n+1)(2n) }=0$

EDIT: I forgot to note that a sequence is convergent if $\lim_{n\to\infty}a_n=c$ where c is an arbitrary finite constant
• May 29th 2008, 07:57 PM
evilpostingmongy
Quote:

Originally Posted by ThePerfectHacker
The first step is to guess what the limit is, it seems to be $\tfrac{2}{3}$.
Next we need to show $|a_n - \tfrac{2}{3}|$ can be made small.
Write, $\left| \tfrac{2^n}{3^n+1} - \tfrac{2}{3} \right| = \left| \tfrac{-2}{3(3^n+1)} \right| = \tfrac{2}{3(3^n+1)}$.
Now it remains to show this quantity gets small, it is easier to bound it.
Note, $\tfrac{2}{3(3^n+1)} \leq \tfrac{2}{3(3^n+0)} \leq \tfrac{3}{3(3^n)} = \tfrac{1}{3^n}$.
Finally note that $\tfrac{1}{3^n}$ can be made sufficiently small.

Here is how the formal proof looks like:
Given $\epsilon > 0$ choose $N\in \mathbb{N}$ so that $\tfrac{1}{3^N} < \epsilon$.
Then if $n\geq N$ it would imply $|a_n - \tfrac{2}{3}| < \epsilon$.

Oh okay but how did you get to this tfrac{3}{3(3^n)} and how do you determine whether or not the sequence diverges or converges?
• May 29th 2008, 08:02 PM
ThePerfectHacker
Quote:

Originally Posted by evilpostingmongy
Oh okay but how did you get to this tfrac{3}{3(3^n)} and how do you determine whether or not the sequence diverges or converges?

Once you pick $N$ so that $\tfrac{1}{3^N} < \epsilon$. Then (by induction) if $n\geq N$ it would mean $\tfrac{1}{3^n} < \epsilon$.
And so if $n\geq N$ we have $|a_n - \tfrac{2}{3}| = \left| \tfrac{2^n}{3^n+1} - \tfrac{2}{3} \right| = \tfrac{2}{3(3^n+1)} \leq \tfrac{1}{3^n} < \epsilon$.

Whenever you do a delta/epsilon proof there are two parts. The mental part and the stuff you write down on paper. The mental part is what you work out on paper, that is what I did previously. What I wrote now is how it is presented on paper. The only reason why I knew how to pick such an N is because I worked it out in the other post.