Showing Whether a Sequence is Bounded Above or Not
I am trying to solve the following problem about a sequence: Consider the sequence ${a_n}$ where $a_n = 1 + \frac{1}{1 \cdot 3} + \frac {1}{1 \cdot 3 \cdot 5} + \frac {1}{1 \cdot 3 \cdot 5 \cdot 7} + ... + \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)}.$ Decide whether ${a_n}$ is bounded above or not, and prove your answer is correct.
I started to solve it in the following manner: Let $a_n \geq a_{n+1} \Rightarrow \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)} \geq \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1) \cdot 2(n+1)-1} \Rightarrow \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)} \geq \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1) \cdot 2n+1}.$ Multiplying by the reciprocal of $a_n$, we have $1 \geq \frac{1}{2n+1}$ which shows that $a_n$ is increasing. This is where I am stuck since I do not know how to proceed in showing it is bounded from what I have so far. Any assistance or criticism is welcome.
I am using the following textbook: Introduction to Analysis by Arthur Mattuck.
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$\begingroup$Hint: Note that $$a_n\le 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{2^{n-1}}\lt 2.$$
$\endgroup$ 5 $\begingroup$HINT: Let $b_n=\frac1{1\cdot3\cdot\ldots\cdot(2n-1)}$; for all $n\ge 1$ we have $b_{n+1}\le\frac13b_n$, so $$a_n=\sum_{k=1}^nb_k\le\sum_{k=1}^n\left(\frac13\right)^k\;.$$
Now use what you know about geometric series.
$\endgroup$ $\begingroup$If you want a direct proof that doesn't use geometric series outright, note that if you consider the terms in $a_n$ shifted forward one position and then compare to the terms in $a_{n+1}$, you get $a_{n+1} \leq 1 + a_n/3$. Also note $a_1 = 1 < 3/2$. If $a_n < 3/2$ then $a_{n+1} \leq 1 + a_n/3 < 1 + 1/2 = 3/2$ as well. So the entire sequence is bounded above by $3/2$ by induction.
$\endgroup$ 1 $\begingroup$This may be informal, but just showing that the sequence is monotone decreasing would mean the first term is the upper bound for the sequence.
We may rewrite the sequence in the following way $a_{n+1}=\frac{a_n}{2n+1}$. We can then see whether the sequence is infact monotone decreasing
$a_n-a_{n+1}=a_n-\frac{a_n}{2n+1}=\frac{2na_n}{2n+1}$
We know $ n>0, ~a_n>0,~ 2>0,~2n+1>0$ which means
$\frac{2na_n}{2n+1}>0 \implies a_n>a_{n+1}$
This sequence is monotone decreasing and therefore is bounded inclusively by the first term of the sequence itself, which is 1.
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