<s>[INST] 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. 
 [/INST] 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.
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