Let P(n) be 1/2 * 3/4 ... (2n-1)/2n < 1/sqrt(3n) (note that parentheses around 2n-1 are mandatory). You verified that P(1) is true. The induction step consists of proving that P(k) implies P(k + 1) for all k >= 1. Strictly speaking, this is a true statement just because the conclusion P(k + 1) is true. After all, the overall problem is to prove P(n) for all n. What fails is a particular, most natural way of proving the step. Indeed, the natural way is to say

\(\displaystyle \frac{1}{2} \cdot \frac{3}{4} \cdot \dots\cdot\frac{2n-1}{2n}\cdot \frac{2n+1}{2n+2} < \frac{1}{\sqrt{3n}}\cdot \frac{2n+1}{2n+2}\)

by the induction hypothesis and then to try proving that the right-hand side is < \(\displaystyle \frac{1}{\sqrt{3(n+1)}}\). However,

\(\displaystyle \frac{1}{\sqrt{3n}}\cdot\frac{2n+1}{2n+2} < \frac{1}{\sqrt{3(n+1)}}\)

has no solutions.

When we strengthen the induction hypothesis, the method above works. Note that the new strict inequality fails for n = 1, so you should either prove a non-strict inequality

\(\displaystyle \frac{1}{2}\cdot\frac{3}{4}\cdot\dots\cdot\frac{2n-1}{2n}\le\frac{1}{\sqrt{3n+1}}\)

for all positive integers n and then say that \(\displaystyle \frac{1}{\sqrt{3n+1}}<\frac{1}{\sqrt{3n}}\), or consider n = 2 in the base step.