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What is a convergent sequence and a divergent sequence?

Is the sequence {((-1)^n)/2n} convergent? If so, what is the limit? (I'm thinking that it is convergent)?

  • I believe that it would be convergent by the alternating series test, but I am not certain. Can you please tell me if I am on the right track, I would greatly appreciate it! Also, I am somewhat confused on the limit part. Any hints or suggestions for that would also be very appreciated. Thank you in advanced!

  • Answer:

    The sequence is bounded and it is a monotone sequence ("monotone" means either "only increasing" or "only decreasing", not both. This one is decreasing.) By the Monotone Convergence Theorem, the sequence converges. The numerator is always +/- 1, so the sequence is bounded by -1/(2n)<[(-1)^n]/(2n)<1/(2n) These other two sequences converge to 0, by the squeeze theorem, our sequence does also converge to 0.

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Sequence or series? Is it just the sequence of terms, or are we talking about the infinite sum of all the terms? Well, in both cases, it converges. Let's consider it first as a sequence. Since (-1)^n is either -1 or 1, we therefore have: -1 <= (-1)^n <= 1 ===> -1 / (2n) <= (-1)^n / (2n) <= 1 / (2n) Both -1 / (2n) and 1 / (2n) converge to 0 (they're just multiples of 1/n), so by squeeze theorem, (-1)^n / (2n) converges to 0. Consider it now as a series. You are right that the alternating series test will make this series converge, since the absolute value of the terms (i.e. |(-1)^n / (2n)| = 1 / (2n)) converges monotonically to 0, and the series' terms alternates positive and negative. Finding the limit is trickier. and requires a bit of analysis. Essentially, it will come to (1/2)ln(2).

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