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.