sequences.covergence.and limitss

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maeveoneill

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Use a graph of the sequence to decide whether the sequence is convergent or divergent. If the sequence is convergent, guess the value of the limit from the graph and then prove your guess.

an = [1.3.5......(2n-1)] / [(2n)^n] *all the . are multiplication

I didn't graph the function but found the values for a1, a2 .... and am making the guess that the function is convergent and has a limit of 1.

just wondering how i prove the limit of the function..

lim = [(2n-1)! ] / [(2n)^n] ???
n-->infiinity

Thanks
 
I assume that there are two different problems here.

The first is a sequence.
\(\displaystyle a_n = \frac{{\prod\limits_{k = 1}^n {\left( {2k - 1} \right)} }}{{\left( {2n} \right)^n }} = \frac{1}{{2^n }}\prod\limits_{k = 1}^n {\frac{{\left( {2k - 1} \right)}}{n}} < \frac{1}{{2^n }}\)
Is that correct? The product of the first n odd positive integers divided by \(\displaystyle {\left( {2n} \right)^n }\)???

What is the second one?
 
i dont think that is right... isnt the 1, 3, 5 the values of a1, a2, a3.. when subbed into 2n-1??
 
maeveoneill said:
i dont think that is right... isnt the 1, 3, 5 the values of a1, a2, a3.. when subbed into 2n-1??
Click to expand...
That is what I asked you!
It is your problem!
What are the exact words in the statement of the problem?
We do not have your textbooks so we don't know anymore than you tell us.
 
My textbooks says exactly this:

Use a graph of the sequence to decide whethe the sequence is convergent or divergent. If the sequence is convergent, guess the value of the limit from teh graph and then prove your guess.

an= [1 . 3. 5 ... (2n-1)] / [(2n)^n]
 
maeveoneill said:
My textbooks says exactly this:
an= [1 . 3. 5 ... (2n-1)] / [(2n)^n]
Click to expand...
Well then because \(\displaystyle \prod\limits_{k = 1}^n {\left( {2k - 1} \right)} = 1 \cdot 3 \cdot 5 \cdots (2n - 1)\)
the answer I gave you is correct.
 
okay... can you explain it at all.. i kind of understand but im a little stupid when it comes to calculus
 
maeveoneill said:
okay... can you explain it at all..
Click to expand...
Here are a few terms.
\(\displaystyle \L a_1 = \frac{1}{2},\;a_2 = \frac{{1 \cdot 3}}{{4^2 }},\;a_3 = \frac{{1 \cdot 3 \cdot 5}}{{6^3 }}\; \cdots \;a_{10} = \frac{{1 \cdot 3 \cdot 5 \cdots 19}}{{20^{10} }}\)

The general term: \(\displaystyle \L a_n = \frac{{1 \cdot 3 \cdot 5 \cdots (2n - 1)}}{{\left( {2n} \right)^n }}.\)

\(\displaystyle \L a_n = \frac{{1 \cdot 3 \cdot 5 \cdots (2n - 1)}}{{\left( {2n} \right)^n }} = \frac{{1 \cdot 3 \cdot 5 \cdots (2n - 1)}}{{2^n n^n }} = \frac{1}{{2^n }}\left( {\frac{1}{n} \cdot \frac{3}{n} \cdot \frac{5}{n} \cdots \cdot \frac{{2n - 1}}{n}} \right) < \frac{1}{{2^n }}\).

The limit is zero.
 
do we then use squeeze thereom,

0 <an < 1/2^n

and because 1/2^n approaches 0 as n appraches infiintiy.. say that an approaches zerio as well..??
 
Just for kicks, I ran it through the calculator. It shows it is unbounded.

limes9.png


Unless I am doing something wrong.
 
1*3*5*6*7*...(2n-1) (problem #1)is not equal to (2n-1)! (problem#2)

for n = 3

1*3*5 = 15....................................problem #1

and

(6-1)! = 5! = 120....................................problem #2
 
This is a famous pair of problems. They seem to be so close together. But one converges to zero and the other is unbounded.
Consider this factorization:
\(\displaystyle \L \begin{array}{rcl}
\left( {2n - 1} \right)! & = & 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot \cdots \left( {2n - 2} \right) \cdot \left( {2n - 1} \right) \\
& = & 1 \cdot 2\left[ 1 \right] \cdot 3 \cdot 2\left[ 2 \right] \cdot 5 \cdot 2\left[ 3 \right] \cdot \cdots 2\left[ {n - 1} \right] \cdot \left( {2n - 1} \right) \\
& = & 2^{n - 1} \left\{ {1 \cdot \left[ 1 \right] \cdot 3 \cdot \left[ 2 \right] \cdot 5 \cdot \left[ 3 \right] \cdot \cdots \left[ {n - 1} \right] \cdot \left( {2n - 1} \right)} \right\} \\
& = & 2^{n - 1} \left( {n - 1} \right)!\left\{ {1 \cdot 3 \cdot 5 \cdot \cdots \left( {2n - 1} \right)} \right\} \\
\end{array}\)

Doing a similar rearrangement as I did above, it is easy to see why the second problem diverges.
 
I've been thinking about this for awhile and have come up with the following argument. (its a long one!)

\(\displaystyle \L \frac{\(2n-1\)!}{\(2n\)^n} = \frac{1}{2n}\cdot\frac{\(2n\)!}{\(2n\)^{n}} \\= \(\frac{1}{2n}\) \cdot \frac{\(2\)[n]}{2n} \cdot \frac{\(3\)[n+1]}{2n} \cdot \frac{\(4\)[n+2]}{2n} \cdots \frac{\(n-1\)[2n]}{2n}\)

This has come from rearranging the way the numerator's product looks.

Note that every numerator is greater than or equal to the denominator (aside from the first).

For the kth part of the chain (again excluding the first 1/(2n)) our numerator will be exactly \(\displaystyle \(k-1\)(n + k + 1)\) larger than our denominator. These are "remainders" after devision by 2n. Thus (this is not exact as (n+1)/2 is not always an integer, but it can be modified for even n also),

\(\displaystyle \L \frac{\(2n-1\)!}{\(2n\)^n} = \(\frac{1}{2n}\) \cdot \frac{\(2\)[n]}{2n} \cdot \frac{\(3\)[n+1]}{2n} \cdot \frac{\(4\)[n+2]}{2n} \cdots \frac{\(n-1\)[2n]}{2n} = \\
\(\frac{1}{2n}\) \(1+\frac{n+3}{2n}\) \cdot \(2+\frac{8}{2n}\) \cdot \(1+\frac{3n+15}{2n}\) \cdot \(3+\frac{24}{2n}\) \cdots \(\frac{n+1}{2}+\frac{n(n+1)}{2n}\)\)


Now, if I just take two terms out of this product along with \(\displaystyle \frac{1}{2n}\), say \(\displaystyle \L \(2+\frac{8}{2n}\)\cdot\(\frac{n+1}{2}+\frac{n(n+1)}{2n}\)\)... then:

\(\displaystyle \L \frac{1}{2n}\(2+\frac{8}{2n}\)\cdot\(\frac{n+1}{2}+\frac{n(n+1)}{2n}\) = \frac{4n^2+12n+8}{4n^2} \ge 1\)

Thus by removing those 'links' we have obtained a number greater than one and the \(\displaystyle \frac{1}{2n}\) is no longer present. Now it is obvious that it tends to infinity as ALL terms are bigger than one now and almost every other one gets bigger in magnitude of about one. Going a few more terms inward you'll see the pattern (5+a)(1+b)(6+c)(1+d)(7+e)... where a,b,c,.. are all positive.

\(\displaystyle \L \(\frac{1}{2n}\) \cdot \(1+\frac{n+3}{2n}\) \cdot \(2+\frac{8}{2n}\) \cdot \(1+\frac{3n+15}{2n}\) \cdot \(3+\frac{24}{2n}\) \cdots \(\frac{n+1}{2}+\frac{n(n+1)}{2n}\) \,\, \ge \\

\(1+\frac{n+3}{2n}\) \cdot \(1+\frac{3n+15}{2n}\) \cdot \(3+\frac{24}{2n}\) \cdots \(\frac{(n-1)}{2}+\frac{(n-1)n}{2n}\) \,\,\).

So this product is bigger than 1*1*3*1*4*1*5*... = 3*4*5*6*... I'm sorry if the presentation here is poor, school has taken a serious bite into my sleep.
 
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