sum...

Dec 2009
1,506
434
Russia
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
yes, but conditionally, it's the alternating harmonic series.
 

chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
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

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 

simplependulum

MHF Hall of Honor
Jan 2009
715
427
\(\displaystyle 1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \frac{1}{8} - \frac{1}{9} + ... \)

\(\displaystyle = 1 + \left( \frac{1}{2} - \frac{1}{3}\right) + \frac{1}{4} + \left( \frac{1}{5} - \frac{1}{6} \right) + \frac{1}{7} + \left( \frac{1}{8} - \frac{1}{9} \right) + ... \)

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 .
 
Nov 2009
485
184
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
oh, i misread the problem, i thought the signs were alternated, my bad.