# Thread: Sequences: why does this converge?

1. ## Sequences: why does this converge?

Just started going over sequences, and was wondering why how these 2 problems are convergent. Thanks.

1.)

solution posted:

2.)

solution posted:

2. 1)

$\displaystyle a_n=\frac{2^n}{3^{n+1}}$

$\displaystyle a_{n+1}=\frac{2^{n+1}}{3^{n+2}}$

so u have

$\displaystyle \displaystyle \lim_{n\to\infty} \frac {a_{n+1}}{a_n}= \lim_{n\to\infty} \frac {\frac{2^{n+1}}{3^{n+2}}}{\frac{2^n}{3^{n+1}}}= \lim_{n\to\infty} \frac {\frac{2^n\cdot2}{3^{n+1}\cdot3}}{\frac{2^n}{3^{n+ 1}}}=\lim_{n\to\infty}\frac {2}{3}= \frac {2}{3}<1$

so it converge

for the second I don't see where u have any problems
same thing do as in first (maybe u have problems with "$\displaystyle !$") but i think it's not that, because you're doing sequences, and factorials must be done way more before sequences

Edit : this is wrong sorry

4. Originally Posted by Plato
i'm so sorry i just got a little a bit confused , while translating it's just the same sorry and i forgot what is which and didn't quite look at it sorry again

5. Originally Posted by Evan.Kimia
Just started going over sequences, and was wondering why how these 2 problems are convergent. Thanks.

1.)

solution posted:

2.)

solution posted:
$\displaystyle \displaystyle\lim_{n\rightarrow\infty}\left(\frac{ 2}{3}\right)^n=\lim_{n\rightarrow\infty}\left(\fra c{\left[\frac{2}{2}\right]}{\left[\frac{3}{2}\right]}\right)^n=\lim_{n\rightarrow\infty}\frac{1}{\left (\frac{3}{2}\right)^n}$

As $\displaystyle n\rightarrow\infty$ the denominator keeps increasing.

$\displaystyle \frac{1}{infinitely\ large}=infinitely\ small$

The second one has an infinitely large value in the denominator position as n increases without limit.

6. I pasted this same reply at the wrong place and got bit by the lion in the den. Here I am in the right place and doing it once again, just for you..

I can’t see what you posted, but if you want to know why the sequence $\displaystyle a_n=\frac{2^n}{3^{n+1}}$ converges, you can imagine 1 remains stationary, while the $\displaystyle 3^n$ in the denominator increases at a much faster pace than the numerator $\displaystyle 2^n$, and eventually it converges to 0.
Once we know that it converges to 0, we can prove that with a given $\displaystyle \epsilon>0$, there is a positive integer $\displaystyle N$ such that if $\displaystyle |\frac{2^n}{3^{n+1}}-0|=|\frac{2^n}{3^{n+1}}|<\epsilon$, but we know that $\displaystyle |\frac{2^n}{3^{n+1}}<|\frac{2^n}{3^n}|<\epsilon$, so we can use $\displaystyle |\frac{2^n}{3^n}|=|(\frac{2}{3})^n|=|(\frac{1}{(3/2)^n})|<\epsilon$ to find n.

Since $\displaystyle (3/2)^n> \frac{1}{\epsilon}$, which is equivalent to $\displaystyle n \log 1.5>\log \frac{1}{\epsilon}$, which also is equivalent to $\displaystyle n>\log_{1.5} (\frac{1}{\epsilon})$, we are now ready to prove.

Proof:
Let $\displaystyle \epsilon >0$. Choose $\displaystyle N=max(1,\lceil \log_{1.5} \frac{1}{\epsilon}\rceil )$ and let $\displaystyle n$ be any integer such that $\displaystyle n>N$ . Thus $\displaystyle n>\log_{1.5} (\frac{1}{\epsilon})$, and so $\displaystyle |\frac{2^n}{3^{n+1}}-0|=|\frac{2^n}{3^{n}}|<|\frac{1}{(3/2)^n}<\epsilon$ .
Q.E.D.