- #1
Iyan Makusa
- 6
- 2
- 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.
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.