1. ## sum...

Is the next sum convergence?

1 + 1/2 - 1/3 + 1/4 + 1/5 - 1/6 + 1/7+ 1/8 - 1/9+...

thanks!

2. yes, but conditionally, it's the alternating harmonic series.

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

$\displaystyle \chi$ $\displaystyle \sigma$

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 .

5. Wait a second - if a series does not converge absolutely, is it acceptable to group terms like that?

6. oh, i misread the problem, i thought the signs were alternated, my bad.