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. 

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. 
