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Telling if a series is convergent or divergent?

m3dicat3d

New member
Jan 11, 2013
19
Just had a question from a coworker regarding how to tell if a series is convergent or divergent.

Been a while since I've dealt with this so I thought I'd ask here.

I *think* I remember that arithmatic series were convergent by nature, but a geometric series could be either convergent or divergent.

And that's about all I know (assuming that is even correct).

Could anyone clear this up for me?

Thanks again :)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Telling if a series in convergent or divergent?

In general, an arithmetic series will be divergent (even the $n$th term is divergent as $n\to\infty$), while a geometric series will be convergent iff $|r|<1$.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,503
Re: Telling if a series in convergent or divergent?

Infinite arithmetic series $\sum_{n=0}^\infty(an+b)$ diverges unless $a=b=0$: this follows from the $n$th term test for divergence. Infinite geometric series $\sum_{n=0}^\infty ar^n$ converges and its sum equals $a/(1-r)$ iff $|r|<1$.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: Telling if a series in convergent or divergent?

Just had a question from a coworker regarding how to tell if a series is convergent or divergent.

Been a while since I've dealt with this so I thought I'd ask here.

I *think* I remember that arithmatic series were convergent by nature, but a geometric series could be either convergent or divergent.

And that's about all I know (assuming that is even correct).

Could anyone clear this up for me?

Thanks again :)
It is essential to specify if we intend a series as a finite or infinite sum. If we intend it as infinite sum then, according to...

Arithmetic Series -- from Wolfram MathWorld

... an arithmetic series is the solution of the difference equation...

$\displaystyle a_{n+1}= a_{n}+ d,\ a_{0}= \alpha$ (1)

... which converges only if is d=0. In other word, an arithmetic series seems to be 'divergent by nature'...

Kind regards

$\chi$ $\sigma$