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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

Struggling_Student
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    $10<2$ isn't true and $\frac{1}{100}$ is less than $1$, but still $\frac{1}{100}\cdot10<2$ is true. Similarly, $10<2$ isn't true, and $\frac{1}{5}$ is less than $1$, but $\frac{1}{5}\cdot 10<2$ isn't true. The fact that we can have these two examples with different outcomes shows that the two observations that you made of $\frac{1}{2}\cdot\frac{3}{4}...$ being less than $1$ and of $\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}$ not being true, have no deductive power over whether $\frac{1}{2}\cdot\frac{3}{4}...\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}$ is true or not. –  Mar 08 '20 at 10:33

2 Answers2

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I realized that, I can't just ignore the terms: $$ \frac{1}{2} \cdot \frac{3}{4} \cdot\cdot\cdot\cdot $$ since these do contribute to the result

Struggling_Student
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The problem lies in the first step; $$kA \lt B$$ does not imply $A \lt B$ for $0 < k < 1$. Instead, it implies $A \lt \frac B k$.

Trelzevir
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