What is the rational series question for (n + 1) / (3n - 1)?

[rcmango]

Homework Statement

(n + 1) / (3n - 1)

Homework Equations

A_n = L

The Attempt at a Solution

lim n-> infinity
(n/n + 1/n) / (3/n - 1/n)

= (1 + 0) / (3 - 0)

= 1/3

Thats the solution, however i have questions..

1.) If a series is in rational form like this, is it typical to always divide by the largest n in the denominator?

2.) What is 1/3? Is that the limit of the series? In other words, 1/3 is what it converges to?

3.) I'm confused as to why the series approaches infinity, so why is 1/3 the limit?

any help to explain what is going on, would help greatly. Thankyou.

 

[ircdan]

1) yes
2) 1/3 is the limit of the sequence a_n = (n + 1)/(3n - 1) as n approaches infinity, and yes, we say a_n converges to 1/3
3) a_n is sequence not a series, and it converges to 1/3 as n->infinity

 

[HallsofIvy]

To clarify the last part, (3), {a_n} is not a series (or sum) but a sequence. Further, the sequence does not "approach infinity", it is only the index, n, that "goes to infinity". In fact, {a_n} decreases steadily from a₁= 1 down to 1/3.

 

[rcmango]

thankyou.

i have two other problems that i believe both converge if I'm doing them correctly below:

both n_infinity: 1/(n^2 + 4) and 1/(n^2 + 1)

so it seems that for both of these problems that the numerator will be 1/n^2
which is 0? so both converge to 0? just want to make sure they do not diverge.
thanks

also......

Okay, as for a problem like this: 1/(3n +1)

where I'm asked to solve it using the intergral test. Do i first want to reduce it like we've done above?

thanks so far.