# Harmonic series

• Oct 11th 2009, 10:28 AM
james_bond
Harmonic series
$\displaystyle A=1-\frac 12+\frac 13-\frac 14+\ldots +\frac 1{2005}-\frac 1{2006}$
$\displaystyle B=\frac 1{1004}+\frac 1{1005}+\frac 1{1006}+\ldots +\frac 1{2005}+\frac 1{2006}$
Show that $\displaystyle \frac 12 <A,B<1$. $\displaystyle A$ or $\displaystyle B$ is larger?
• Oct 12th 2009, 02:28 AM
Opalg
Quote:

Originally Posted by james_bond
$\displaystyle A=1-\frac 12+\frac 13-\frac 14+\ldots +\frac 1{2005}-\frac 1{2006}$
$\displaystyle B=\frac 1{1004}+\frac 1{1005}+\frac 1{1006}+\ldots +\frac 1{2005}+\frac 1{2006}$
Show that $\displaystyle \frac 12 <A,B<1$. $\displaystyle A$ or $\displaystyle B$ is larger?

The part about A and B lying between 1/2 and 1 is not too hard and ought not to need a hint.

For the last part (which of A and B is larger?), let $\displaystyle A_N = 1-\tfrac 12+\tfrac 13-\tfrac 14+\ldots +\tfrac 1{2N-1}-\tfrac 1{2N}$ and let $\displaystyle B_N = \tfrac1{N+1} + \tfrac1{N+2} + \ldots + \tfrac1{2N}$. Then $\displaystyle A_{N+1} - A_N = \tfrac1{2N+1} - \tfrac1{2N+2}$ and $\displaystyle B_{N+1} - B_N = \tfrac1{2N+1} + \tfrac1{2N+2} - \tfrac1{N+1}$. Check that these two expressions are equal. So $\displaystyle A_{N+1} - B_{N+1} = A_N - B_N$. By induction, $\displaystyle A_{1003} - B_{1003} = A_1 - B_1 = 0$.