A combinatorial identity (teaser?)

$\sum_{k=1}^{n-1} H_{k}H_{n-k} \equiv 2\sum_{k=1}^{n-1} H_{k}\frac{n-k}{k+1}$

Try to prove it! If you want to know how I proved it, get this thread to hit 20 replies saying "Asuka best grill :^)" and the question mark will go away ;)

( $H_{n} = \frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{n}$ )

#Combinatorics

Easy Math Editor

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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`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

The only question mark in your note is in the title.

Lee Gao - 5 years, 7 months ago

Rei best grill

Goran Nand - 5 years, 10 months ago

What shit taste ;c

Jake Lai - 5 years, 10 months ago

Likewise my friend

Goran Nand - 5 years, 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}`	$\boxed{123}$