My professor showed following problem and he solved through induction.
However, I'm not sure why this problem can't be solved the way I did it, since every step seems to be true.
For all natural numbers $n$ following holds true: $$ \frac{1}{2} \cdot \frac{3}{4} \cdot \cdot \cdot \frac{2n-1}{2n} \lt \frac{1}{\sqrt{2n +1}} $$
my idea was that since,
$$
\frac{1}{2} \cdot \frac{3}{4} \cdot \cdot \cdot \cdot
$$
has to be smaller then $1$, then following has to be true:
$$
\frac{2n-1}{2n} \lt \frac{1}{\sqrt{2n +1}}
\\
\Leftrightarrow \sqrt{2n +1}{} \cdot(2n-1) \lt 2n
\\
\Leftrightarrow (2n+1) \cdot(2n-1)^2 \lt 4n^2
\\
\Leftrightarrow 8n^3 - 2n +1 \lt 4n^2
$$
which isn't the case.
any help would be appreciated
The problem lies in the first step; $$kA \lt B$$ does not imply $A \lt B$ for $0 < k < 1$. Instead, it implies $A \lt \frac B k$.