<s>[INST] Part of an induction question I might have done or not realize something stupid, but I can't seem to prove the following...
Inductive hypothesis
Assume $\exists$k$\in$N such that P(k) is true.
P(k): $\frac{1 \cdot 3 \cdot 5 \cdot\cdot\cdot (2k - 1)}{2 \cdot 4 \cdot 6 \cdot\cdot\cdot (2k)}$ $\leq$ $\frac{1}{\sqrt{k + 1}}$
Inductive step
Prove P(k+1) is true.
P(k+1): $\frac{1 \cdot 3 \cdot 5 \cdot\cdot\cdot [2(k + 1) - 1]}{2 \cdot 4 \cdot 6 \cdot\cdot\cdot [2(k + 1)]}$ $\leq$ $\frac{1}{\sqrt{(k + 1) + 1}}$
$\frac{1 \cdot 3 \cdot 5 \cdot\cdot\cdot [2(k + 1) - 1]}{2 \cdot 4 \cdot 6 \cdot\cdot\cdot [2(k + 1)]}$
= $\frac{1 \cdot 3 \cdot 5 \cdot\cdot\cdot (2k - 1) \cdot [2(k + 1) - 1]}{2 \cdot 4 \cdot 6 \cdot\cdot\cdot (2k) \cdot [2(k + 1)]}$
$\leq$ $\frac{1}{\sqrt{k + 1}}$ $\cdot$ $\frac{2(k + 1) - 1}{2(k + 1)}$ #by IH
= $\frac{2(k + 1) - 1}{2(k + 1) \cdot \sqrt{k + 1}}$
$\leq$ $\frac{2(k + 1)}{2(k + 1) \cdot \sqrt{k + 1}}$
= $\frac{1}{\sqrt{k + 1}}$
So what I seem to have done is proven that $\frac{1 \cdot 3 \cdot 5 \cdot\cdot\cdot [2(k + 1) - 1]}{2 \cdot 4 \cdot 6 \cdot\cdot\cdot [2(k + 1)]}$ $\leq$ $\frac{1}{\sqrt{k + 1}}$ and not $\frac{1}{\sqrt{(k + 1) + 1}}$. I seem to need a minor adjustment somewhere.
 [/INST] You have given away too much. You want to prove that
$$\frac{2k+1}{2(k+1)\sqrt{k+1}}\lt \frac{1}{\sqrt{k+2}}.$$
Equivalently, you want to prove that $(2k+1)^2(k+2)\lt 4(k+1)^3$. Expand, and the result will drop out.
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