Series and Sequences?

Question

One of tests commonly used to detmine if an infinite series is divergent or not, is the n-th term test, which is basically determining if the SEQUENCE converges or not.

We know that if the SEQUENCE a_n is divergent as n goes to infinity, then the series a_n also has to diverge. But the converse is not true. Just because a sequence a_n converges does not imply that the series a_n will converge also.

One such example is a_n=1/n.
My question is, does anybody else know of any other examples. Because this is only one I know and this is the only one my professors know of.

Did everybody even read the question before answering it?

Sal, I wasn't asking different tests. I know which ones there are. But thanks for answering anyway.

oktobejustme and site23, if you have no idea what am I talking about, then why the hell did you answer. You only made yourself look stupid.

Pascal, good try, but the sequence 1/p where p is a prime, is just a special case of 1/n where n is an integer. So it doesn't count but thanks for trying.

Anymouse, you are a GENIUS. That is exactly what I was looking for. I can't believe that it has been right there in my face for all these years and never bothered to think about it.

Anybody else there?????

Answer 1

According to what is commonly referred to as the "p-test", we know that the sum of (1/n)^p is convergent for p > 1, and divergent for p < 1.

This gives a whole family divergent series, such that the terms themselves converge to zero. ( by changing the value of 0 < p < 1 )

Answer 2

Some remarks:
1) you say series which has the general term a_n. It's not corectly series a_n.
2) Pascal's example is pretty nice . It's not really a special case. Since it contains only prime numbers the series could be "smaller" but it is not.
3) Yes you are right if a_n is divergent the series is divergent. Even more: the sequence a_n should be convergent to zero
4) sal gave you other tests that allow you to construct other examples
for example: sum 1/(ln n) since 1/n < 1/(ln n) and sum 1/n diverges.
Of course you can take any power of ln n:
1/(ln n)^p

Answer 3

First of all, the divergence test says that if the sequence a_n does not converge to 0,then the series diverges. So a_n=(n+1)/n will make the sequence converge but not the series.

Other examples where the sequence converges to 0 but the series diverges:

1/sqrt(n)
1/[nln(n)]
1/n^(1+1/n)
1/[nsqrt(ln(n)]

There are many, many more.

Answer 4

Here's an interesting one: consider the series 1/2+1/3+1/5+1/7+1/11+1/13... where the nth term in the sequence is the reciprocal of the nth prime number. It can be proven that this series also diverges: see http://en.wikipedia.org/wiki/Proof_that_the_sum_of_the_reciprocals_of_the_primes_diverges

Answer 5

If I understand your question correctly, the simplest set of series that answer your question are p-series.

Generally, a p-series is a series of the form \sum_{n=1}^{\infty} 1/n^p, where p is a positive number.

In calculus, we can prove that the p-series diverges if p <=1 and converges if p>1. Regardless, \lim_{n \to \infty} 1/n^p=0.

Answer 6

there are a few others like

integral test

comprassion test

absolute test

so other test

this website contains all test for determinig the divergence or convergence and examples

http://tutorial.math.lamar.edu/AllBrowsers/2414/ConvergenceOfSeries.asp

Answer 7

oh sure
a plane can travel east to west in a continuous circle
a plance can travel north to south in a continuouse circle
some times they may cross paths some time not...

Answer 8

Here is another:
a_n = coth n
^_^

Answer 9

elvis has left the classroom