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I'm new to poof based math and I just wanted to make sure that the style I use is rigorous enough or whether it has to be more so. E.g. when rewriting inequalities below I don't mention that denominators are non-zero and therefore such manipulation is allowed. Is that considered rigorous enough? Furthermore, do we really have to show the base case calculation when using mathematical induction or it is sufficient to perform the calculation yourself and mention that the proof is obvious?


Prove the following (B. Demidovich, Problems in Mathematical Analysis, §1, ex. 10):

$$\prod_{k=1}^{n} \frac{2k-1}{2k} < \frac{1}{\sqrt {2n + 1}}, n\in N\ \ \ \ \ \ \ (1)$$

Proof will be done using mathematical induction for $$n \ge 1$$ The case of n = 1 is trivial and holds if we replace n with 1 in (1), since $$ \frac {1}{2} = 0.5 < \frac {1}{\sqrt{3}} \approx 0.57 \ \ \ \ \ (2)$$

Now, let's prove that if (1) holds for an arbitrary n then it will hold for n+1. Assuming that (1) holds for some n we then have to prove:

$$\prod_{k=1}^{n+1} \frac{2k-1}{2k} < \frac{1}{\sqrt {2n + 3}}\ \ \ \ \ \ \ (3)$$

Using right side of (1), we can conclude that it is sufficient to prove:

$$ \frac{2n+1}{2n + 2} \cdot\frac{1}{\sqrt {2n + 1}} < \frac{1}{\sqrt {2n + 3}} $$

The inequality can be rewritten as:

$$ \frac{\sqrt{2n+1}}{2n + 2} < \frac{1}{\sqrt {2n + 3}} $$

Or further like this:

$$ \frac{1}{2n + 2} < \frac{1}{\sqrt {2n + 3}\cdot\sqrt {2n + 1} } $$

It is left to show that:

$$ {2n + 2} > \sqrt {2n + 3}\cdot\sqrt {2n + 1} \ \ \ \ \ \ (a) $$

By squaring both sides we can clearly see that the inequality holds for any n.

$$4n^2 + 8n + 4 > 4n^2 +8n + 3\ \ \ \ \ \ (b) $$

Using (2) and (3) we prove that (1) holds for any $n \in N$.

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    $\begingroup$ The proof is fine but to say that $\frac 1 2 <\frac 1 {\sqrt3}$ you should say $3<4$ so $\sqrt 3 <2$ and $\frac 1 2 <\frac 1 {\sqrt3}$. Using approximation to two decimal places makes it less rigorous. $\endgroup$ Apr 9, 2019 at 9:54
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    $\begingroup$ @KaviRamaMurthy If "approximation" means "rounding" then this is no less rigorous. Number rounded to $0.57$ has to be greater then $0.5$, there is no other choice. But proving the rounding, i.e. $|\frac{1}{\sqrt{3}}-0.57|<0.01$ may be too hard. $\endgroup$
    – freakish
    Apr 9, 2019 at 11:13
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    $\begingroup$ @freakish $\frac 1 {\sqrt 3} \approx 0.57$ requires a proof. $\endgroup$ Apr 9, 2019 at 11:42
  • $\begingroup$ Ok, I see the mistake now. $\endgroup$
    – user464118
    Apr 9, 2019 at 12:04
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    $\begingroup$ @Eval I repeat: the standard convention is that $y=\sqrt{x}$ if and only if $y\ge0$ and $y^2=x$. $\endgroup$
    – egreg
    Apr 9, 2019 at 22:06

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The proof is correct.

For $n=1$, the left-hand side is $1/2$ and the right-hand side is $1/\sqrt{3}$. Since $1/4<1/3$, you also have $1/2<1/\sqrt{3}$.

Suppose the inequality holds for $n$; then, by the induction hypothesis, $$ \prod_{k=1}^{n+1}\frac{(2k-1)}{2k}<\frac{1}{\sqrt{2n+1}}\frac{2n+1}{2n+2} $$ and we just need to check that $$ \frac{1}{\sqrt{2n+1}}\frac{2n+1}{2n+2}\le\frac{1}{\sqrt{2n+3}} $$ that is, $$ \sqrt{(2n+1)(2n+3)}\le2n+2=\frac{(2n+1)+(2n+3)}{2} $$ which is a case of AM-GM.

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