<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] We first (directly) prove:
$$\frac{1}{\sqrt{k+1}}\cdot\frac{2k+1}{2k+2} < \frac{1}{\sqrt{k+2}}$$
for $k>1$. Starting with:
$$9k+2<12k+4$$
which is obviously true for $k>1$. Then, we add $4k^3+12k^2$ to both sides:
$$4k^3+12k^2+9k+2<4k^3+12k^2+12k+4$$
Splitting and factoring:
$$4k^3+4k^2+k+8k^2+8k+2<4k^3+8k^2+4k+4k^2+8k+4$$
$$(4k^2+4k+1)(k+2)<(4k^2+8k+4)(k+1)$$
$$(2k+1)^2(k+2)<(2k+2)^2(k+1)$$
Taking square roots:
$$(2k+1)\sqrt{k+2}<(2k+2)\sqrt{k+1}$$
Dividing by $(2k+2)\sqrt{k+1}\sqrt{k+2}$ on both sides:
$$\frac{1}{\sqrt{k+1}}\cdot\frac{2k+1}{2k+2} < \frac{1}{\sqrt{k+2}}$$
Now we start the induction step:
$$\frac{1 \cdot 3 \cdot 5 \cdots[2k-1] [2(k + 1) - 1]}{2 \cdot 4 \cdot 6 \cdots [2k][2(k + 1)]} \leq \frac{1}{\sqrt{k + 1}}\cdot \frac{2(k+1)-1}{2(k+1)}=\frac{1}{\sqrt{k + 1}}\cdot \frac{2k+1}{2k+2}<\frac{1}{\sqrt{k + 2}}$$
There is our desired result! Q.E.D.
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