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Let $A_n = \sum a_n$ be a sum that converges absolutely.

This sum can be separated into its even terms and odd terms, $\sum a_n=\sum a_{2n}+\sum a_{2n+1}$, but, must each individual sub-series converge on its own? In other words, if $\sum(|a_n|)$ converges, does it imply that individually, $\sum a_{2n+1}$ converges or that $\sum a_{2n }$ converges?

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  • $\begingroup$ You could use the monotone convergence theorem for this. $\endgroup$ – Cameron Williams Jul 22 at 3:35
  • $\begingroup$ But there's no certainty or criteria that any of the sequences must be monotone, so how could that be useful? You could have a series that acts sinusoidal and decays with its sums converging absolutely. $\endgroup$ – cheesemonkey102 Jul 22 at 3:38
  • $\begingroup$ Absolute convergence means you can assume each $a_n \geq 0$ $\endgroup$ – D_S Jul 22 at 3:42
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    $\begingroup$ You can still just displace a sinusoidal sequence above 0 and then divide it by something that allows the sums to converge. $\cos(n)+2$ is such a sequence. Also, that isn't what absolutely convergence means, it just means that the sum of the absolute value of the terms converges, it doesn't mean you get to assume every term of the sequence is always non-negative. $\endgroup$ – cheesemonkey102 Jul 22 at 3:44
  • $\begingroup$ It doesn't matter tho $\endgroup$ – D_S Jul 22 at 3:49
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The monotone convergence theorem says if $0 \leq c_n \leq d_n$, and $\sum d_n$ converges, then so does $\sum c_n$.

Assume $\sum a_n$ converges absolutely. This means that $\sum\limits_n |a_n|$ converges. To show that, say, $\sum\limits a_{2n+1}$ converges, define $b_n$ by

$$b_n = \begin{cases} a_n & \textrm{ if $n$ is odd} \\ 0 & \textrm{ if $n$ is even} \end{cases}$$ Then $\sum\limits_n |a_{2n+1}| = \sum\limits_n |b_n|$ converges by the monotone convergence theorem, because $|b_n| \leq |a_n|$ for all $n$.

This shows that $\sum\limits_n a_{2n+1}$ converges absolutely, and in particular, $\sum\limits_n a_{2n+1}$ converges.

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  • $\begingroup$ How did you determine $|b_n| \leq |a_n|$? If you have a sequence with a sinusoidal quality, you could pick out terms that exceed their same-indexed counterpart. $\endgroup$ – cheesemonkey102 Jul 22 at 3:54
  • $\begingroup$ Look carefully... $\endgroup$ – D_S Jul 22 at 3:55
  • $\begingroup$ $b_n$ is just a sub-sequence of odd terms, that condition itself says nothing about the order of their sum in relation to the sum of all even and all odd terms. $\endgroup$ – cheesemonkey102 Jul 22 at 3:56
  • $\begingroup$ So either $|b_n| = |a_n|$ or $|b_n| = 0$. Which implies that $|b_n| \leq |a_n|$ for all $n$ as I claimed. $\endgroup$ – D_S Jul 22 at 3:58
  • $\begingroup$ Okay, so you are slapping a summation operation on both sides of that inequality to show the sum of odd terms is less than the sum of total terms? $\endgroup$ – cheesemonkey102 Jul 22 at 3:59
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Yes. Each sum is an increasing function, bounded above by the overall sum, so it converges to its supremum. The series may not be monotone, but the sum is.

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