Just started going over sequences, and was wondering why how these 2 problems are convergent. Thanks.
1.)
solution posted:
2.)
solution posted:
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
$\displaystyle \displaystyle\lim_{n\rightarrow\infty}\left(\frac{ 2}{3}\right)^n=\lim_{n\rightarrow\infty}\left(\fra c{\left[\frac{2}{2}\right]}{\left[\frac{3}{2}\right]}\right)^n=\lim_{n\rightarrow\infty}\frac{1}{\left (\frac{3}{2}\right)^n}$
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.
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.