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(-1)^n/sqrt(n) Converge Or Diverge

3 min read Jun 16, 2024
(-1)^n/sqrt(n) Converge Or Diverge

Does the sequence (-1)^n/sqrt(n) converge or diverge?

The sequence (-1)^n/sqrt(n) is a fascinating example of a sequence that converges to 0. Here's why:

Understanding the Sequence

The sequence oscillates between positive and negative values due to the (-1)^n term.

  • When n is even, (-1)^n = 1, so the term is positive.
  • When n is odd, (-1)^n = -1, so the term is negative.

The 1/sqrt(n) part ensures that the absolute value of the terms decreases as n gets larger.

Using the Squeeze Theorem

To prove convergence, we can use the Squeeze Theorem.

  1. Find two sequences that "squeeze" our sequence.

    • Consider the sequence -1/sqrt(n). Since -1/sqrt(n) <= (-1)^n/sqrt(n) for all n, this sequence is a lower bound.
    • Similarly, consider the sequence 1/sqrt(n). Since 1/sqrt(n) >= (-1)^n/sqrt(n) for all n, this sequence is an upper bound.

  2. Show that the bounding sequences converge to the same limit.

    • Both -1/sqrt(n) and 1/sqrt(n) converge to 0 as n approaches infinity. This is because as n gets larger, 1/sqrt(n) gets smaller and approaches 0.

  3. Apply the Squeeze Theorem.

    • Since the sequence (-1)^n/sqrt(n) is squeezed between two sequences that converge to 0, the sequence (-1)^n/sqrt(n) also converges to 0.

Conclusion

The sequence (-1)^n/sqrt(n) converges to 0. While it oscillates between positive and negative values, the decreasing magnitude of the terms due to 1/sqrt(n) ensures that the sequence ultimately approaches 0.

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