# sum...

#### Also sprach Zarathustra

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

#### chisigma

MHF Hall of Honor
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
$$\displaystyle 1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \frac{1}{8} - \frac{1}{9} + ...$$

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

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

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

#### Krizalid

MHF Hall of Honor
oh, i misread the problem, i thought the signs were alternated, my bad.

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