# convergent sequences proof

• Mar 11th 2007, 05:45 PM
luckyc1423
convergent sequences proof
Prove the following by using this definition:
A sequence (Sn) is said to converge to the real number S provided that for each e > 0 there exists a real number N such that for all n element of N, n > N implies that |Sn - S| < e
If (Sn) converges to S, then S is called the limit of sequence (Sn), and we write lim n -> infinity Sn = S, lim Sn = S, or Sn -> S. If a sequence does not converge to a real number, it is said to diverge

Prove this from the definition above:
a) lim (3n + 1)/(n + 2) = 3
b) lim (sin n) / n = 0
c) lim (n+2)/(n^2 - 3) = 0
• Mar 11th 2007, 06:28 PM
ThePerfectHacker
Quote:

Originally Posted by luckyc1423
a) lim (3n + 1)/(n + 2) = 3

For any e>0 choose N=5/e

Then if n>N we have n>5/e thus, 5/n<e.

But then,

|(3n+1)/(n+2)-3|=|-5/(n+2)|=5/(n+2)<=5/n<e
Quote:

b) lim (sin n) / n = 0
For any e>0 choose N=1/e then if n>N we have 1/n<e.

Thus,
|sin(n)/n-0|=|sin(n)/n|<=1/n<e
• Mar 11th 2007, 06:33 PM
ThePerfectHacker
Quote:

Originally Posted by luckyc1423
c) lim (n+2)/(n^2 - 3) = 0

For any e>0 choose N=max{2,4/n}

Then if n>N* we have 4/n<e.

But then!

|(n+2)/(n^2-3)|=(n+2)/(n^2-3)<=2n/(n^2-3)<=2n/(.5n^2)=4/n<e

Q.E.D.

*)You also need to consider the case max{2,4/n}=2.
But that case is easy.