Determinăm termenul general:
[tex] \bigg| \dfrac{k}{k+1} -\dfrac{k+1}{k+2} \bigg| \\ = \bigg| \dfrac{k(k+2)-(k+1)(k+1)}{(k+1)(k+2)} \bigg| \\ = \bigg| \dfrac{k^2+2k -k^2-2k-1}{(k+1)(k+2)} \bigg| \\ =\bigg| -\dfrac{1}{(k+1)(k+2)} \bigg| \\ = \dfrac{1}{(k+1)(k+2)} \\ = \dfrac{k+2-(k+1)}{(k+1)(k+2)} \\ =\dfrac{k+2}{(k+1)(k+2)}-\dfrac{k+1}{(k+1)(k+2)} \\ = \dfrac{1}{k+1} -\dfrac{1}{k+2} [/tex]
Bun.Termenul meu general începe cu k(primul termen) iar în suma ta , k =1, deoarece începe cu 1.
[tex] |\tfrac{1}{2} -\tfrac{2}{3}| +|\tfrac{2}{3} -\tfrac{3}{4}| +\ldots+ |\tfrac{2007}{2008} -\tfrac{2008}{2009}| \\ = \dfrac{1}{2} -\dfrac{1}{3} +\dfrac{1}{3} -\dfrac{1}{4} +\ldots +\dfrac{1}{2008} -\dfrac{1}{2009} \\ = \dfrac{1}{2} -\dfrac{1}{2009} \\ =\dfrac{2009-2}{2\cdot 2009} \\ =\tt \dfrac{2007}{4018} [/tex]
