Does every bounded sequence converge? Prove. A sequence is bounded if there exists an M>=0 such that |Sn|<= M

Question

Does every bounded sequence converge? Prove. A sequence is bounded if there exists an M>=0 such that |Sn|<= M for all n belonging to N.

Answer 1

No, just take a_n = (-1)^n. It's bounderd because for every n |a_n| <=1. (actually, =1)., But it's terms are -1, 1, -1, 1.....so that it oscillates and therefore diverges.

What is true is that every bounded sequence of real numbers (or even of vectors in R^n, or of complex numbers) contains a convergent subsequence. This is known as Bolzano Weierstrass Theorem. In the case of (-1)^n, there's a subsequence that converges to 1 and another that converges to -1.

Answer 2

No. S_n=1 if n is prime, otherwise =0 does not converge.

To see if a sequence converges, form two auxiliary sequences:

A_1=the largest number that is less than all of the S_1,S_2,S_3,...
A_2=the largest number that is less than all of the S_2,S_3,S_4,...
A_3=the largest number that is less than all of the S_3,S_4,S_5,...
and so on.

B_1=the smallest number that is greater than all of the S_1,S_2,S_3,...
B_2=the smallest number that is greater than all of the S_2,S_3,S_4,...
B_3=the smallest number that is greater than all of the S_3,S_4,S_5,...
and so on.

The A series will be an nondecreasing series and the B series will be nonincreasing. If they converge to the same value, the original S series converges, otherwise it doesn't.

For example, suppose the series S={-1,1,-1,1,-1,...}

then A={-1,-1,-1,...}-->-1
and B={1,1,1,...}--> 1

So S doesn't converge.

On the other hand, suppose the series
S={1,-1/2,1/3,-1/4,...}

then
A
={-1/2,-1/2,-1/4,-1/4,...}-->0
and
B
={1,1/3,1/3,1/5,1/5...}-->0

So S converges to 0.

Answer 3

No, a bounded sequence may oscillate, such as

{1, -1, 1, -1, ...}

Answer 4

No

Here's your counterexample:

-1,1,-1,1,...