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Determine the convergence of the series (-1)n / (n + (-1)n)s

by RoRi

Consider the series

    \[ \sum_{n=2}^{\infty} \frac{(-1)^n}{(n+(-1)^n)^s}. \]

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

If s \leq 0 then the series diverges, if 0 < s \leq 1 the series converges conditionally, and if s \geq 1, then the series converges absolutely.

Proof. Case 1. Assume s \leq 0. Then

    \[ \lim_{n \to \infty} \frac{(-1)^n}{(n+(-1)^n)^s} \neq 0 \]

so the series must diverge.

Cases 2 & 3. Assume s > 0. The series converges since

    \begin{align*} \sum_{n=2}^{\infty} \frac{(-1)^n}{(n+(-1)^n)^s} &= \sum_{n=2}^{\infty} \left( \frac{1}{(2n-1)^s} - \frac{1}{(2n-2)^s} \right) \\[9pt] &= - \sum_{n=2}^{\infty} \left( \frac{1}{(2n-2)^s} - \frac{1}{(2n-1)^s} \right). \] \end{align*}

Now, we can apply the Leibniz rule since (for s>0),

    \[ \lim_{n \to \infty} \left( \frac{1}{(2n-2)^s} - \frac{1}{(2n-1)^s} \right) = 0. \]

Further, the sequence is monotonically decreasing since

    \begin{align*} \left( \frac{1}{(2x-2)^s} - \frac{1}{(2x-1)^s} \right)' &= \frac{-s}{(2x-2)^{s-1}} + \frac{s}{(2x-1)^{s-1}} \\[9pt] &= \frac{s(2x-2)^{s-1} - s (2x-1)^{s-1}}{(4x^2-6x+2)^{s-1}}. \end{align*}

But,

    \[ s(2x-2)^{s-1} - s (2x-1)^{s-1} < 0 \]

since 2x-2 < 2x-1, and for x \geq 2

    \[ (4x^2-6x+2)^{s-1} > 0. \]

Hence, the derivative is always negative, so the function is decreasing. Therefore, the series of terms taking this function value on the integers must be decreasing as well. By the Leibniz rule, this means the series converges for all s > 0.

Finally, we must show this convergence is conditional if 0 < s \leq 1 and absolute if s > 1. So, we consider,

    \[ \sum_{n=2}^{\infty} \left| \frac{(-1)^n}{(n+(-1)^n)^s} \right| = \sum_{n=2}^{\infty} \frac{1}{(n+(-1)^n)^s} < \sum_{n=2}^{\infty} \frac{1}{(n-1)^s}. \]

Considering the series \sum \frac{1}{n^s} we have

    \[ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{n^s}{(n-1)^s} = 1. \]

This means the two series will converge or diverge together. But, we know \sum \frac{1}{n^s} converges for s > 1 and diverges for 0 < s \leq 1. Hence, the series in the exercise converges conditionally if 0 < s \leq 1 and absolutely if s > 1. \qquad \blacksquare

4 comments

  1. Artem says:

    Ok, for everyone doing this exercise: REARRANGMENT only allowed when the series is ABSOLUTELY CONVERGENT. It means, for any s 1, is solved by doing the limit-ratio test with 1/n^s. Case s = 0 is separate. s< 0: separate case too, where terms do not tend to 0. Case s 0: for this use the dirichlet’s test – very simple if you manage to notice it.

  2. tom says:

    Looks like some mistakes on this one. Firstly, the derivatives are wrong, but more over, Leibniz’ rule can’t be used as stated here because the (-1)^n term was incorporated into the series (no longer alternating). Fix is easy- odd and even terms get ADDED, the series is still monotonic decreasing so the same result.

    • tom says:

      Okay correction: No need to differentiate at all because the series is equivalent to -Σ(-1)^n/n^s , which as we know converges for s>0. The series as stated is not alternating at all- the partial sums are all positive- until the odd-even terms are grouped slightly different. This rearrangement apparently is allowed because the terms alternate, so I’m curious why the rearrangement is allowed yet the series is not absolutely convergent (I know rearrangements are covered in next chapter).

      So summarizing, I see from where you separated into odd-even on the first line how things would reassemble into -Σ(-1)^n/n^s but am having trouble seeing the intermediary steps using sigma notation, or do we just make the logical jump at this point?

      • tom says:

        READ THIS ONE: Okay, so the sums do alternate, but Σ(-1)^n/(n+(-1)^n)^s is the same as -Σ(-1)^n/n^s, which is the correct form. (your method was incorrect as the alternating nature of the series was ‘baked into’ the two series for odd and even terms.) I see how to transform now the sums, by changing the index from n=2 to n=1. My question, which may be obvious, is isn’t thus a rearrangement technically? And is this rearrangement allowed when it involves alternating odd-even?

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