Proving Convergence of {S_n/n} for Bounded Sequence {S_n}

[Shayes]

Homework Statement

If {S_n} is a sequence whose values lie inside an interval [a,b], prove {S_n/n} is convergent.

We don't know Cauchy sequence yet. All we know is the definition of a bounded sequence, and convergence and divergence of sequences. Along with comparison tests and Squeeze theorem.

Homework Equations

Limit of a sequence: abs(S_n - L) < Epsilon whenever n>=N, provided for Epsilon>0.

The Attempt at a Solution

I see that every convergent sequence is bounded, but the opposite isn't always true, so how do I show that the smaller sequence is convergent given that {S_n} is bounded?

 

[susskind_leon]

I think you can use the squeeze theorem, since a/n <= S_n/n <= b/n and they both converge to 0. Do you need to proove that a/n converges or can you use that?

 

[Shayes]

I think you can use the squeeze theorem, since a/n <= S_n/n <= b/n and they both converge to 0. Do you need to proove that a/n converges or can you use that?

i thought of that, but is that the same as saying the entire sequence converges?

i feel like this statement says that each term is squeezed between the interval, but it doesn't say anything about if the limit of the sequence's terms is taken to infinity.

 

[susskind_leon]

Yes it does! Check the squeeze theorem carefully!

 

[HallsofIvy]

i thought of that, but is that the same as saying the entire sequence converges?

i feel like this statement says that each term is squeezed between the interval

I have no idea what this means, but the crucial point is the "n" in the denominator. What are the linits of a/n and b/n?

, but it doesn't say anything about if the limit of the sequence's terms is taken to infinity.