Infinite series please help.

  • Thread starter ziddy83
  • Start date
I'm having a little trouble with the following problem. here it is...

Determinte whether a_n is convergent for...

[tex] a_n = \frac{2n}{3n+1}[/tex]

How would i go about solving this? Can i just simply take the limit and use L'Hospital's rule to see if it diverges or converges? Or is it a little more complicated than that? Im asking this because in this section we're studying geometric series, and that doesnt look like a geo series to me since n is not in the exponent. Any help would be appreciated, thanks.
 
hello there

that aint a series, its a sequence

well to find if a_n is convergent, you would want to see what happens as n goes to infinity, and to do that divide the numerator and the denominator by the highest power in a_n, and im sure you should know what happens to 1/n as n goes to infinty

steven
 
one more thing

if it is a series i would use either the
comparision test
or the
limit comparision test

steven
 

HallsofIvy

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ziddy83 said:
I'm having a little trouble with the following problem. here it is...

Determinte whether a_n is convergent for...

[tex] a_n = \frac{2n}{3n+1}[/tex]

How would i go about solving this? Can i just simply take the limit and use L'Hospital's rule to see if it diverges or converges? Or is it a little more complicated than that? Im asking this because in this section we're studying geometric series, and that doesnt look like a geo series to me since n is not in the exponent. Any help would be appreciated, thanks.
Series or sequence?

You don't need anything as complicated as L'Hospital's rule to determine this. If you mean the sequence [tex]\frac{2n}{3n+1}[/tex], just divide both numerator and denominator by n to get [tex]\frac{2}{3+\frac{1}{n}}[/tex]. What happens to that as n goes to infinity?

If you mean the series [tex]\Sigma_{n=1}^{\infty}\frac{3n}{3n+1}[/tex], that's also easy after you know the limit of the sequence- in order that the series [tex]\Sigma a_n[/tex] converge, the sequence {an} must converge to 0.
 

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