# Show that the sequence converges

#### Iyan Makusa

Problem 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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#### Math_QED

Science Advisor
Homework Helper
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?

• Iyan Makusa

#### Iyan Makusa

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.

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?

#### pasmith

Homework Helper
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 $a_{n+1}$ from $a_n$, you multiply by a factor $(n+1)/(2n+3)$ which is always strictly less than 1, and the sequence was therefore decreasing. But in fact $(n+1)/(2n+3) < \frac12$, 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?

• Math_QED

#### Iyan Makusa

You are ignoring all of the other factors.

You found that to obtain $a_{n+1}$ from $a_n$, you multiply by a factor $(n+1)/(2n+3)$ which is always strictly less than 1, and the sequence was therefore decreasing. But in fact $(n+1)/(2n+3) < \frac12$, 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 gotta 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?

#### Math_QED

Science Advisor
Homework Helper
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 gotta 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 $a_n$?

• SammyS

#### Ray Vickson

Science Advisor
Homework Helper
Dearly Missed
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 $k$.

#### LCKurtz

Science Advisor
Homework Helper
Gold Member
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 $a_n \to 0$.

• Math_QED

#### StoneTemplePython

Science Advisor
Gold Member
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.

• Math_QED

#### pasmith

Homework Helper
Alternatively, you can write $$0 < \frac{n!}{1 \cdot 3 \cdot \, \dots \, \cdot (2n+1)} = \prod_{k=1}^n \frac{k}{2k+1} < \frac1{2^n}.$$

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