# series diverging

• July 22nd 2007, 01:52 PM
davecs77
series diverging
I am having trouble understand divergence and convergence. I am trying to solve this problem, Show that each of the following series diverges.
3 + 5/2 + 7/3 + 9/4

Do i need to find a S(n) formula then take the limit of it as n goes to infinity? I know to use limits, but I don't understand what the answer to the limit tells you. Thanks for the help!
• July 22nd 2007, 02:01 PM
Jhevon
Quote:

Originally Posted by davecs77
I am having trouble understand divergence and convergence. I am trying to solve this problem, Show that each of the following series diverges.
3 + 5/2 + 7/3 + 9/4

Do i need to find a S(n) formula then take the limit of it as n goes to infinity? I know to use limits, but I don't understand what the answer to the limit tells you. Thanks for the help!

Let $S_n = 3 + \frac {5}{2} + \frac {7}{3} + \frac {9}{4} + ...$

then $S_n = \sum_{n = 1}^{ \infty} \frac {2n + 1}{n}$

note that $\lim_{n \to \infty} S_n = 2 \neq 0$

thus the series diverges by the test for divergence

here is what the answer to the limit tells you. if the answer is zero then the series MIGHT converge, if it isn't zero, then it definitely does not converge

Theorem: Let $S_n$ be a sequence. If $\sum S_n$ converges, then $\lim_{n \to \infty} S_n = 0$

Note, this is an implication, the converse is not always true
• July 22nd 2007, 02:23 PM
davecs77
Quote:

Originally Posted by Jhevon
Let $S_n = 3 + \frac {5}{2} + \frac {7}{3} + \frac {9}{4} + ...$

then $S_n = \sum_{n = 1}^{ \infty} \frac {2n + 1}{n}$

note that $\lim_{n \to \infty} S_n = 2 \neq 0$

thus the series diverges by the test for divergence

here is what the answer to the limit tells you. if the answer is zero then the series MIGHT converge, if it isn't zero, then it definitely does not converge

Theorem: Let $S_n$ be a sequence. If $\sum S_n$ converges, then $\lim_{n \to \infty} S_n = 0$

Note, this is an implication, the converse is not always true

How would you know if the series converges if the limit is 0? Can you do a test for that?
• July 22nd 2007, 02:25 PM
Jhevon
Quote:

Originally Posted by davecs77
How would you know if the series converges if the limit is 0? Can you do a test for that?

if the limit was zero, we would need another test. there are many to choose from: the comparison test, the integral test, the limit comparison test, ...
• July 22nd 2007, 03:12 PM
ThePerfectHacker
Quote:

Originally Posted by davecs77
How would you know if the series converges if the limit is 0? Can you do a test for that?

For example,
$\sum_{n=1}^{\infty}\frac{1}{n}$ diverges even though $\lim \frac{1}{n}=0$.