Some arithmetic Identities - Printable Version +- OMath! (http://math.elinkage.net) +-- Forum: Math Forums (/forumdisplay.php?fid=4) +--- Forum: Algebra (/forumdisplay.php?fid=5) +--- Thread: Some arithmetic Identities (/showthread.php?tid=104) Some arithmetic Identities - elim - 10-09-2010 06:30 PM (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) $\displaystyle{\sum_{k=2}^{2n}\frac{(-1)^k (2n+1-k)}{k}=\sum_{k=1}^n\frac{2k-1}{n+k}}$ $\quad\quad\displaystyle{\frac{2n+1-2}{2}-\frac{2n+1-3}{3}+\cdots-\frac{2}{2n-1}+\frac{1}{2n}=\frac{1}{n+1}+\cdots+\frac{2k-1}{n+k}+\cdots+\frac{2n-1}{2n}}$ RE: Identities (1) and (2) - elim - 10-13-2010 06:22 PM (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 $\displaystyle{\frac{2009}{2}-\frac{2008}{3}+\frac{2007}{4}-\cdots+\frac{1}{2010} =\frac{1}{1006}+\frac{3}{1007}+\frac{5}{1008}+\cdots+\frac{2009}{2010}}$