Thread Rating:
• 0 Votes - 0 Average
• 1
• 2
• 3
• 4
• 5
 Some arithmetic Identities
10-09-2010, 06:30 PM
Post: #1
 elim Moderator Posts: 581 Joined: Feb 2010 Reputation: 0
Some arithmetic Identities
(1) $\displaystyle{\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}=\sum_{k=1}^n \frac{1}{n+k}}$
$\quad\quad\displaystyle{1-\frac{1}{2}+\frac{1}{3}-\cdots+\frac{1}{2n-1}-\frac{1}{2n} = \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+k}+\cdots+\frac{1}{2n}}$

(2) $\displaystyle{\sum_{k=2}^{2n}\frac{(-1)^k (2n+1-k)}{k}=\sum_{k=1}^n\frac{2k-1}{n+k}}$
$\quad\quad\displaystyle{\frac{2n+1-2}{2}-\frac{2n+1-3}{3}+\cdots-\frac{2}{2n-1}+\frac{1}{2n}=\frac{1}{n+1}+\cdots+\frac{2k-1}{n+k}+\cdots+\frac{2n-1}{2n}}$
10-13-2010, 06:22 PM
Post: #2
 elim Moderator Posts: 581 Joined: Feb 2010 Reputation: 0
RE: Identities (1) and (2)
(1) $\displaystyle{\sum_{k=1}^n \frac{1}{n+k} = \sum_{k=1}^{2n}\frac{1}{k} -\sum_{k=1}^n \frac{1}{k}= \sum_{k=1}^{2n}\frac{1}{k} - 2\sum_{k=1}^n \frac{1}{2k}= \sum_{k=1}^{2n}\frac{1}{k} - \sum_{k=1}^{2n} \frac{1+(-1)^n}{k}=\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}}$

(2) $\displaystyle{\sum_{k=2}^{2n} \frac{(-1)^k(2n+1-k)}{k} = 2n+\sum_{k=1}^{2n}\frac{(-1)^k(2n+1-k)}{k}=2n-(2n+1)\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}-\sum_{k=1}^{2n}(-1)^k}$
$\quad\quad \displaystyle{= 2n-(2n+1)\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}=2n-(2n+1)\sum_{k=1}^n\frac{1}{n+k}=\sum_{k=1}^n (2-\frac{2n+1}{k+n})=\sum_{k=1}^n\frac{2k-1}{n+k}}$

$\quad\quad$ Let $n=995$ we get $\displaystyle{\frac{1989}{2}-\frac{1988}{3}+\frac{1987}{4}-\cdots+\frac{1}{1990} =\frac{1}{996}+\frac{3}{997}+\frac{5}{998}+\cdots+\frac{1989}{1990}}$

$\quad\quad$ Let $n=1005$ we get $\displaystyle{\frac{2009}{2}-\frac{2008}{3}+\frac{2007}{4}-\cdots+\frac{1}{2010} =\frac{1}{1006}+\frac{3}{1007}+\frac{5}{1008}+\cdots+\frac{2009}{2010}}$

 « Next Oldest | Next Newest »

Forum Jump: