Re: [Math4u] Sequence : converges and diverges

Hi Brian:

I am not sure whether this note will be posted to math4u or to your personal email, as I am at a PC that does not permit direct access to the message board.  At any rate, a sequence, {an}, is said to be convergent iff limit(n-->inf)[an] = r for some real r.  For instance, {an} = 1/n converges because 1/n tends toward zero as n goes to infinity.  That said, although n - 1/n approaches n as n goes to infinity, as you correctly note, for that very same reason, n - 1/n goes to infinity.  Thus the limit does not exist (is not real), and it follows that the sequence is not convergent -- i.e., it is divergent.

Regards,

Rich B.      

-----Original Message-----
From: Brian E. Jensen <brianejensen@prodigy.net>
To: Math4u@yahoogroups.com
Sent: Mon, 8 Oct 2007 2:16 am
Subject: Re: [Math4u] Sequence : converges and diverges

If I understand the problem,

Problem 1

We have a series a(1), a(2), a(3), a(4), ...a(n) where

a(1) = 1-1/1

a(2) = 2-1/2

a(3) = 3-1/3

a(n) = n-1/n

It looks to me like the bigger n gets, the closer a(n) gets to n.

So I think we say that as n--> infinity, a(n) approaches n.

problem 2

Here we have a series where the odd terms are zero and the even terms are 2 times the series in problem 1. So this is an interesting situation. The odd terms fall on one line and the even terms approach a slanted line. So you need to look at the precise definition of converging series to see if the series must approach one line or if the series can alternate between 2 lines.

Regards,

Brian Jensen

----- Original Message -----

From: alvi_66

To: Math4u@yahoogroups.com

Sent: Sunday, October 07, 2007 2:55 AM

Subject: [Math4u] Sequence : converges and diverges

I am student of Bs-Mathematics. I need help in this question if
everybody tell me about this question and in future other questions, i
shall be very thankful for this kindness.
Q. Which of the sequence converges, and which diverges? Give reasons
for your answer.
1. an=n-1/n
2. an=((-1)^n+1)(n+1/n)

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