<s>[INST] Proving by induction: $ \frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{1\cdot2\cdot3\cdot\ldots\cdot n} \leq 2^n $ WTS $ \frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{1\cdot2\cdot3\cdot\ldots\cdot n} \leq 2^n  $ for all natural $n$.
Have checked $P_1$, and assumed $P_k$.
Trying the following argument: 
$P_{k+1} = \frac{1\cdot3\cdot5\cdot \ldots \cdot (2k-1)\cdot(2k+1)}{1\cdot2\cdot3\cdot\ldots\cdot k \cdot (k+1)} = \frac{1\cdot3\cdot5\cdot \ldots \cdot (2k-1)}{1\cdot2\cdot3\cdot\ldots\cdot k}\cdot \frac{2k+1}{k+1} \leq 2^k \cdot\frac{2k+1}{k+1} \leq 2^{k+1}\cdot \frac{2k+1}{k+1}$ using $P_k$ and that these are all positive numbers.
Now dividing through by $\frac{2k+1}{k+1}$ would give the desired result.
I just feel like I've done something wrong here. Can anyone tell me if I've overlooked anything?
 [/INST] All you need is to prove that: $\dfrac{2k+1}{k+1} \leq 2$
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