Solving Infinite Series: a_n = \frac{2n}{3n+1}

[ziddy83]

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

Determinte whether a_n is convergent for...

a_n = \frac{2n}{3n+1}

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? I am asking this because in this section we're studying geometric series, and that doesn't look like a geo series to me since n is not in the exponent. Any help would be appreciated, thanks.

 

[steven187]

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 I am sure you should know what happens to 1/n as n goes to infinty

steven

 

[steven187]

one more thing

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

steven

 

[HallsofIvy]

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

Determinte whether a_n is convergent for...

a_n = \frac{2n}{3n+1}

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? I am asking this because in this section we're studying geometric series, and that doesn't 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 \frac{2n}{3n+1}, just divide both numerator and denominator by n to get \frac{2}{3+\frac{1}{n}}. What happens to that as n goes to infinity?

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