Help with following equality

Question

Could you please help me understand how this is?

$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} \,.$$

Thank you.

What does this have to do with the rank-nullity theorem? – EuYu Mar 02 '13 at 07:59 — EuYu, Mar 02 '13 at 07:59

score 5 · Accepted Answer · edited Mar 02 '13 at 13:25

$\text{L.H.S.}$

$=\displaystyle \frac 1 1-\frac 1 2+\frac 1 3 - \frac 1 4 +\cdots+\frac 1 {2n-1} - \frac 1 {2n}$

$\displaystyle=\frac 1 1+\frac 1 2+\frac 1 3 + \frac 1 4 +\cdots+\frac 1 {2n-1} + \frac 1 {2n}-2\Big(\frac 1 2+\frac 1 4+\cdots+\frac 1 {2n}\Big)$

$\displaystyle=\frac 1 1+\frac 1 2+\frac 1 3 + \frac 1 4 +\cdots+\frac 1 {2n-1} + \frac 1 {2n}-\Big(\frac 1 1+\frac 1 2+\cdots+\frac 1 {n}\Big)$

$=\displaystyle\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} $

$= \text{R.H.S.}$

Q.E.D.

1 Answers1