# Identity

• Feb 24th 2008, 10:54 AM
red_dog
Identity
There is another proof, without induction, for this identity?

$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\f rac{1}{2n}$
• Feb 24th 2008, 11:10 AM
Krizalid
This is Catalan's identity.

Consider $k_n = 1 + \frac{1}
{2} + \cdots + \frac{1}
{n}.$
Note that $\frac{1}
{2}k_n = \frac{1}
{2} + \frac{1}
{4} + \cdots + \frac{1}
{{2n}}.$
The LHS is $k_{2n} - 2\left( {\frac{1}
{2}k_n } \right)$
& the RHS is $k_{2n}-k_n,$ as required $\blacksquare$