Is the next sum convergence? 1 + 1/2 - 1/3 + 1/4 + 1/5 - 1/6 + 1/7+ 1/8 - 1/9+... thanks!
Also sprach Zarathustra Dec 2009 1,506 434 Russia May 21, 2010 #1 Is the next sum convergence? 1 + 1/2 - 1/3 + 1/4 + 1/5 - 1/6 + 1/7+ 1/8 - 1/9+... thanks!
Krizalid MHF Hall of Honor Mar 2007 3,656 1,699 Santiago, Chile May 21, 2010 #2 yes, but conditionally, it's the alternating harmonic series.
chisigma MHF Hall of Honor Mar 2009 2,162 994 near Piacenza (Italy) May 21, 2010 #3 The series can be written as... \(\displaystyle \sum_{n=0}^{\infty} (\frac{1}{3n+1}+ \frac{1}{3n+2} - \frac{1}{3n+3})\) (1) ... where the general term for 'large n' can be approximated as... \(\displaystyle a_{n} \approx \frac{1}{3n}\) (2) ... so that the series (1) diverges... Kind regards \(\displaystyle \chi\) \(\displaystyle \sigma\) Reactions: Also sprach Zarathustra
The series can be written as... \(\displaystyle \sum_{n=0}^{\infty} (\frac{1}{3n+1}+ \frac{1}{3n+2} - \frac{1}{3n+3})\) (1) ... where the general term for 'large n' can be approximated as... \(\displaystyle a_{n} \approx \frac{1}{3n}\) (2) ... so that the series (1) diverges... Kind regards \(\displaystyle \chi\) \(\displaystyle \sigma\)
S simplependulum MHF Hall of Honor Jan 2009 715 427 May 21, 2010 #4 \(\displaystyle 1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \frac{1}{8} - \frac{1}{9} + ... \) \(\displaystyle = 1 + \left( \frac{1}{2} - \frac{1}{3}\right) + \frac{1}{4} + \left( \frac{1}{5} - \frac{1}{6} \right) + \frac{1}{7} + \left( \frac{1}{8} - \frac{1}{9} \right) + ... \) We have \(\displaystyle \frac{1}{2} - \frac{1}{3} > 0 \) \(\displaystyle \frac{1}{5} - \frac{1}{6} > 0 \) ... The series is thus greater than \(\displaystyle 1 + \frac{1}{4} + \frac{1}{7} + ... = \sum_{k=0}^{\infty} \frac{1}{ 3k+1 } \) Divergent .
\(\displaystyle 1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \frac{1}{8} - \frac{1}{9} + ... \) \(\displaystyle = 1 + \left( \frac{1}{2} - \frac{1}{3}\right) + \frac{1}{4} + \left( \frac{1}{5} - \frac{1}{6} \right) + \frac{1}{7} + \left( \frac{1}{8} - \frac{1}{9} \right) + ... \) We have \(\displaystyle \frac{1}{2} - \frac{1}{3} > 0 \) \(\displaystyle \frac{1}{5} - \frac{1}{6} > 0 \) ... The series is thus greater than \(\displaystyle 1 + \frac{1}{4} + \frac{1}{7} + ... = \sum_{k=0}^{\infty} \frac{1}{ 3k+1 } \) Divergent .
roninpro Nov 2009 485 184 May 21, 2010 #5 Wait a second - if a series does not converge absolutely, is it acceptable to group terms like that?
Krizalid MHF Hall of Honor Mar 2007 3,656 1,699 Santiago, Chile May 22, 2010 #6 oh, i misread the problem, i thought the signs were alternated, my bad.