Is the next sum convergence?

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

thanks!

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- May 21st 2010, 04:17 PM #1

- May 21st 2010, 05:01 PM #2

- May 21st 2010, 06:54 PM #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

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

- May 21st 2010, 07:50 PM #4

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$\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 .

- May 21st 2010, 08:52 PM #5

- May 22nd 2010, 12:09 PM #6