How Can I Prove That the Sequence {sin nπ/3} is Not Convergent?

To prove that the sequence {sin nπ/3} is not convergent, we need to examine the values of the sequence as n takes on different integer values. By analyzing these values, we can determine whether the sequence approaches a single limit or oscillates.

Calculating Sequence Values

Let's calculate the first few terms of the sequence {sin nπ/3}:

n 0: sin(0) 0 n 1: sin(π/3) √3/2 n 2: sin(2π/3) √3/2 n 3: sin(π) 0 n 4: sin(4π/3) -√3/2 n 5: sin(5π/3) -√3/2 n 6: sin(2π) 0

By examining these values, we observe a repeating pattern every 6 terms.

Identifying the Pattern

The sequence {sin nπ/3} shows the following:

For n ≡ 0 (mod 6): sin(nπ/3) 0 For n ≡ 1 (mod 6) and n ≡ 2 (mod 6): sin(nπ/3) √3/2 For n ≡ 3 (mod 6): sin(nπ/3) 0 For n ≡ 4 (mod 6) and n ≡ 5 (mod 6): sin(nπ/3) -√3/2

The sequence takes on the values 0, √3/2, and -√3/2 repeatedly in a cycle of 6 terms.

List of Distinct Values

The sequence has distinct values that repeat as:

0 √3/2 -√3/2

Show Non-Convergence

A sequence is convergent if and only if it approaches a single limit as n goes to infinity. In this case, the sequence does not approach a single value but oscillates between 0, √3/2, and -√3/2. This indicates that the sequence is not convergent.

A More Elegant Proof

Another elegant way to prove the sequence {sin nπ/3} is not convergent is to identify two subsequences that converge to different limits. Consider the following two subsequences:

Subsequence 1: {sin(3mπ/3)} {sin(mπ)}, where m is a positive integer. Subsequence 2: {sin[(6m-1)π/3]} {sin(π/3 - 2mπ)}, where m is a positive integer.

Subsequence 1: This subsequence simplifies to:

sin(mπ) 0

Therefore, the limit of this subsequence is 0.

Subsequence 2: This subsequence simplifies to:

sin(π/3 - 2mπ) sin(π/3) √3/2

Therefore, the limit of this subsequence is √3/2.

Since the original sequence has two subsequences with different limits (0 and √3/2), the sequence {sin nπ/3} cannot be convergent.

Conclusion

Based on the above analysis and the fact that the sequence does not settle on a single value and oscillates indefinitely, we conclude that the sequence {sin nπ/3} is not convergent.