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Rossinole
Jul9-08, 11:24 AM
1. The problem statement, all variables and given/known data

Is the series from n=1 to infinity of 3/n converging or diverging?

2. Relevant equations


3. The attempt at a solution

Since 3/n is not a geometric series, my guess is that we can just use the Test for Divergence and take it's limit to see if it's converging or diverging. As n->infinity, 3/n -> 0 and lim = 0, so it's converging.

However, I am not sure if this is right way to go about it.

Tom Mattson
Jul9-08, 11:30 AM
Since 3/n is not a geometric series,


Correct.


my guess is that we can just use the Test for Divergence and take it's limit to see if it's converging or diverging.


Not a bad guess, but beware that the Test for Divergence cannot tell you if a series converges (hence, its name).


As n->infinity, 3/n -> 0 and lim = 0, so it's converging.


Wrong. The Test for Divergence says that:

\lim_{n\rightarrow\infty}a_n \neq 0 \Rightarrow \sum_{n=1}^\infty a_n diverges.

Equivalently, it says that:

\sum_{n=1}^\infty a_n converges \Rightarrow \lim_{n\rightarrow\infty}a_n = 0

If the limit is zero, then the test yields no information and you have to use another test.

Rossinole
Jul9-08, 11:42 AM
So I would have to treat it as a Harmonic Series?

Tom Mattson
Jul9-08, 11:43 AM
It is a harmonic series.

Rossinole
Jul9-08, 11:45 AM
Alright, thank you for your help.