Show that the sequence converges

  • Thread starter Iyan Makusa
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In summary: Since the next factors are always ##<\frac 1 2## of the previous term, then the values are decreasing, to I assume is ##\frac 1 2##. So the limit of the sequence is ##\frac 1 2##.
  • #1
Iyan Makusa
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Homework Statement
Use the Monotone Convergence Theorem to show that the sequence converges.
Relevant Equations
$$\left\{\frac{n!}{1\cdot3\cdot5\cdot\dots\cdot(2n+1)}\right\}^{+\infty}_{n=0}$$
So what I know about the Monotone Convergence Theorem is that it states that: if a sequence is bounded and monotone, then it is convergent. So all I have to show is that the sequence is bounded and monotone.

My attempt at showing that it is bounded:

The sequence can be expanded as:
$$={\frac 1 1}\cdot{\frac 2 3}\cdot{\frac 3 5}\cdot\dots\cdot{\frac {(n-1)} {(2n-1)}}\cdot{\frac n {(2n+1)}}$$
I'm not really sure how to show this mathematically, but I saw that on all the factors, the highest value it has obtained is ##1##, and that is from the first factor, ##\frac 1 1##. Every next factor has been ##<1##. So I think its upper bound is 1:
$${\frac 1 1}\cdot{\frac 2 3}\cdot{\frac 3 5}\cdot\dots\cdot{\frac {(n-1)} {(2n-1)}}\cdot{\frac n {(2n+1)}}\leq1$$
Next, we saw that ##n## starts at ##0##. And having ##a_0=1##. Since ##n\rightarrow+\infty## then the value of each factor can't become negative. So I think it is always postive e.g. ##>0##:
$$0\lt{\frac 1 1}\cdot{\frac 2 3}\cdot{\frac 3 5}\cdot\dots\cdot{\frac {(n-1)} {(2n-1)}}\cdot{\frac n {(2n+1)}}\leq1$$
So I guess this shows that it is bounded...?

Now I have to show that it is monotone:

A monotone sequence to my understanding is a sequence that is either increasing or decreasing. I think I can show this by equating:
$$\frac{a_{n+1}}{a_n}$$
to find out if the sequence is increasing or decreasing. So I do:
$$\begin{align}
a_n&=\frac{n!}{1\cdot3\cdot5\cdot\dots\cdot(2n+1)} \\
a_{n+1}&=\frac{(n+1)(n!)}{1\cdot3\cdot5\cdot\dots\cdot(2n+1)\cdot(2n+3)} \\
\frac{a_{n+1}}{a_n}&=\frac {n+1}{2n+3}
\end{align}$$

We can see here that for all ##n\geq0, \frac {n+1}{2n+3}\lt1##, so I think that proves that the sequence is decreasing.

Since it seems to satisfy the two conditions of the Monotone Convergence Theorem, I guess it proves that it is convergent?
I think I managed to solve it, but I wanted to post it here because I am really unsure of how I proved it, and if there's a better or simpler way of solving this.
 
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  • #2
The main idea certainly is correct. I didn't check for details though. Note that once you have shown that the sequence is decreasing, it is automatically bounded above (by the first sequence term), so you could make your answer shorter.

As an interesting side question: do you have any idea what the limit could be?
 
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  • #3
Math_QED said:
Note that once you have shown that the sequence is decreasing, it is automatically bounded above (by the first sequence term), so you could make your answer shorter.
Oh well that's definitely shorter! I'll keep that in mind.

Math_QED said:
As an interesting side question: do you have any idea what the limit could be?
Hmm, this is an interesting side question. I'll give it a try:

So I think I can isolate the entire sequence to just the last factor ##\frac{n}{2n+1}## and evaluate its limit:
$$\begin{align}
\lim_{n\rightarrow\infty}\frac{n}{2n+1}
\end{align}$$
Applying L'Hopital's rule because both the numerator and denominator evaluate to infinity:
$$\begin{align}
\lim_{n\rightarrow\infty}\frac{n}{2n+1} = \lim_{n\rightarrow\infty}\frac1 2 = \frac 1 2
\end{align}$$
Did I do this correctly?
 
  • #4
Iyan Makusa said:
Oh well that's definitely shorter! I'll keep that in mind.Hmm, this is an interesting side question. I'll give it a try:

So I think I can isolate the entire sequence to just the last factor ##\frac{n}{2n+1}## and evaluate its limit:
$$\begin{align}
\lim_{n\rightarrow\infty}\frac{n}{2n+1}
\end{align}$$
Applying L'Hopital's rule because both the numerator and denominator evaluate to infinity:
$$\begin{align}
\lim_{n\rightarrow\infty}\frac{n}{2n+1} = \lim_{n\rightarrow\infty}\frac1 2 = \frac 1 2
\end{align}$$
Did I do this correctly?

You are ignoring all of the other factors.

You found that to obtain [itex]a_{n+1}[/itex] from [itex]a_n[/itex], you multiply by a factor [itex](n+1)/(2n+3)[/itex] which is always strictly less than 1, and the sequence was therefore decreasing. But in fact [itex](n+1)/(2n+3) < \frac12[/itex], so the value of each term in the sequence is slightly less than half the value of the previous term. What does that tell you about the limit?
 
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  • #5
pasmith said:
You are ignoring all of the other factors.

You found that to obtain [itex]a_{n+1}[/itex] from [itex]a_n[/itex], you multiply by a factor [itex](n+1)/(2n+3)[/itex] which is always strictly less than 1, and the sequence was therefore decreasing. But in fact [itex](n+1)/(2n+3) < \frac12[/itex], so the value of each term in the sequence is slightly less than half the value of the previous term. What does that tell you about the limit?
My bad, I was under the impression that I can ignore the other factors because only the "last" factor matters because...well it's the last, so that's got to be where the limit will be. Guess that was wrong!

Since the next factors are always ##<\frac 1 2## of the previous term, then the values are decreasing, to I assume is 0?
 
  • #6
Iyan Makusa said:
My bad, I was under the impression that I can ignore the other factors because only the "last" factor matters because...well it's the last, so that's got to be where the limit will be. Guess that was wrong!

Since the next factors are always ##<\frac 1 2## of the previous term, then the values are decreasing, to I assume is 0?

Good intuition! Now, make it formal!

Hint:

##a_1 < a_0/2, a_2 < a_1/2 < a_0/2^2, \dots##

Continue this process. What estimate can you make for ##a_n##?
 
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  • #7
Iyan Makusa said:
Oh well that's definitely shorter! I'll keep that in mind.Hmm, this is an interesting side question. I'll give it a try:

So I think I can isolate the entire sequence to just the last factor ##\frac{n}{2n+1}## and evaluate its limit:
$$\begin{align}
\lim_{n\rightarrow\infty}\frac{n}{2n+1}
\end{align}$$
Applying L'Hopital's rule because both the numerator and denominator evaluate to infinity:
$$\begin{align}
\lim_{n\rightarrow\infty}\frac{n}{2n+1} = \lim_{n\rightarrow\infty}\frac1 2 = \frac 1 2
\end{align}$$
Did I do this correctly?
I get something different. First, write
$$1 \cdot 3 \cdots (2n+1) =\frac{ 1 \cdot 2 \cdot 3 \cdot 4 \cdots (2n) \cdot (2n+1)}{2 \cdot 4 \cdots (2n)}\\
\hspace{3em} = \frac{(2n+1)!}{2^n n!}$$ so your ratio is
$$\text{ratio} = \frac{2^n (n!)^2}{(2n+1)!}$$ Now one can apply Stirling's formula ##k! \sim \sqrt{2 \pi k}\, k^k e^{-k}## for large ##k##.
 
  • #8
Another approach (assuming you have studied infinite series, which maybe you haven't). You have shown$$

\frac{a_{n+1}}{a_n}=\frac {n+1}{2n+3} \rightarrow \frac 1 2$$which shows the series ##\sum a_n## converges by the ratio test, which implies ##a_n \to 0##.
 
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  • #9
A slightly different finish (that is close to MathQEDs) is to note that we have a positive sequence that is monotone decreasing and bounded below by zero -- using this proves the limit exists. i.e. we know ##a_n \to c ## for some ##c \in [0,1]## but this also implies ##a_{n+1} \to c##

so the sequence is
##0 \leq a_{n+1} \leq \frac{1}{2}a_n ##
which implies

##0 \leq c\leq \frac{1}{2}c ##

which simplifies to get the limitting value.
 
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  • #10
Alternatively, you can write [tex]
0 < \frac{n!}{1 \cdot 3 \cdot \, \dots \, \cdot (2n+1)} = \prod_{k=1}^n \frac{k}{2k+1} < \frac1{2^n}.
[/tex]
 

What does it mean for a sequence to converge?

A sequence converges when its terms get closer and closer to a single value as the sequence progresses, and this single value is called the limit of the sequence.

How can you show that a sequence converges?

To show that a sequence converges, you need to prove that the terms of the sequence get arbitrarily close to the limit value as the sequence progresses. This can be done through various methods such as the epsilon-delta definition, the squeeze theorem, or the monotone convergence theorem.

What is the epsilon-delta definition of convergence?

The epsilon-delta definition of convergence states that for any given small value epsilon, there exists a corresponding term in the sequence after which all subsequent terms will be within epsilon distance from the limit value. In other words, the terms of the sequence get arbitrarily close to the limit value as the sequence progresses.

What is the squeeze theorem and how is it used to show convergence?

The squeeze theorem states that if two sequences have the same limit and a third sequence is always between them, then the third sequence also converges to the same limit. This theorem is used to show convergence by finding two sequences that have the same limit as the given sequence and then showing that the given sequence is always between them.

What is the monotone convergence theorem and how is it used to show convergence?

The monotone convergence theorem states that if a sequence is either increasing and bounded above or decreasing and bounded below, then it must converge to a limit. This theorem is used to show convergence by proving that the given sequence is either increasing and bounded above or decreasing and bounded below, and then finding the limit of the sequence.

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