How do I prove that
$\prod\limits_{i=1}^{n} (\frac{2i - 1}{2i}) \le \frac{1}{\sqrt{3n + 1}} $
using induction for n > 0 ?
In general, for (n+1) case, I split the productory in:
factor(n+1) * productory [i, n] factor(i)
given productory [i, n] factor(i) is the hypothesis, we replace for
factor(n+1) * [right_side_of_hypotesis - K]
where K is the difference between the sides of the hypothesis
I couldn't get out of this stage...
I am going to show $\qquad\frac{2n-1}{2n}\le\frac{\sqrt{3n-2}}{\sqrt{3n+1}},\quad$ which can be re-written as $$1-\frac{1}{2n}\le\sqrt{1-\frac{3}{3n+1}}\tag{1}$$
By squaring, we have ($1$) is equivalent to \begin{align} &&1-\frac{1}{n}+\frac{1}{4n^2}&\le 1-\frac{3}{3n+1}\\ \iff&&\frac{3}{3n+1}+\frac{1}{4n^2}&\le \frac{1}{n}\\ \iff&&4n^2(3)+3n+1&\le4n(3n+1)\\ \iff&&12n^2+3n+1&\le12n^2 +4n\\ \iff&&1&\le n \end{align} Last inequality holds, then we have $$\frac{2n-1}{2n}\le\frac{\sqrt{3n-2}}{\sqrt{3n+1}}\;\;\implies \;\;\prod_{i=1}^n\frac{2i-1}{2i}\le\prod_{i=1}^n\frac{\sqrt{3i-2}}{\sqrt{3i+1}}\;\;\iff\;\;\prod_{i=1}^n\frac{2i-1}{2i}\le\frac{1}{\sqrt{3n+1}}$$
What about using induction?
1) You must show that the inequality holds for $n=1$, i.e. $\frac{1}{2}\le\frac{1}{\sqrt{4}}$.
2)Inductive hypothesis: let's suppose the inequality $$\prod_{i=1}^k\frac{2i-1}{2i}\le\frac{1}{\sqrt{3k+1}}$$ holds for $k\ge 1$.
3) Show, from steps 1) and 2) that
$$\prod_{i=1}^{k+1}\frac{2i-1}{2i}\le\frac{1}{\sqrt{3(k+1)+1}}\;\; \text{holds}$$
It can be done as I did above for showing that ($1$) holds.
