- 3

- 2

- 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:

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.

*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.