Number theory

Prove that for $n\geq5$ , $(p_{n+1})^{3}<p_{1} \times p_{2}\times p_{3}\cdots p_{n}$ where $p_{i}$ is the $i^\text{th}$ prime?

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

\geq can be written inside Latex brackets to make $\geq$ @Gnanananda Shreyas

A Former Brilliant Member - 10 months, 1 week ago

Note : Bertrand's Postulate states if n> 1 there is always at least one prime p such that n < p < 2n.

In other words, $p_{k+1} < 2 p_{k}$

so for n > 5 $(p_{n+2})^{3} < 8 (p_{n+1})^3 < 8 ( p_1 \times p_2 \times p_3....\times p_n ) < p_{n+1} \times ( p_1 \times p_2 \times p_3....\times p_n )$ as $p_i > 8$ for $i >5$

Pop Wong - 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$