# Convergence

• Feb 22nd 2010, 01:24 AM
harish21
Convergence
The sequence { 1/n } converges to 0 and that it does not converge to any other number. Using this fact prove that NONE of the following assertions is equivalent to the definition of convergence of a sequence {an} to the number a.

a. for some e>0, there is an index N such that |an -a| < e for all indices n>=N

b. for each e>0 and each index N, |an - a| < e for all indices n>=n

• Feb 22nd 2010, 01:34 AM
HallsofIvy
Quote:

Originally Posted by harish21
The sequence { 1/n } converges to 0 and that it does not converge to any other number. Using this fact prove that NONE of the following assertions is equivalent to the definition of convergence of a sequence {an} to the number a.

a. for some e>0, there is an index N such that |an -a| < e for all indices n>=N

Show that this is true for $\displaystyle a_n= \frac{1}{n}$, a= 1, and e= 2. (The point is that this says "for some e> 0" rather than "for all e> 0".)

Quote:

b. for each e>0 and each index N, |an - a| < e for all indices n>=n
Did you mean "n> N"? Show that this is NOT true if you take $\displaystyle a_n= \frac{1}{n}$, a= 0, e= .001, and N= 1. (The point is that this says "for each index N" rather than "for some index N".)

Quote:

• Feb 22nd 2010, 01:44 AM
harish21
Quote:

Originally Posted by HallsofIvy
Show that this is true for $\displaystyle a_n= \frac{1}{n}$, a= 1, and e= 2. (The point is that this says "for some e> 0" rather than "for all e> 0".)

Did you mean "n> N"? Show that this is NOT true if you take $\displaystyle a_n= \frac{1}{n}$, a= 0, e= .001, and N= 1. (The point is that this says "for each index N" rather than "for some index N".)

Yes that was supposed to be n>=N in b. Thank you.

Likewise there is one more question that states there is an index N such that for every number e>o |an - a| < e for all indices n>=N. What should I show here?