My professor was discussing the Alternating Series Test today in class and it goes as follows:

Given a sequence of positive numbers $\{a_n\}$:

$\sum_{n=1}^{\infty}(-1)^{n+1}a_n\quad \text {converges if:}$

1) $\lim_{n\to \infty}a_n=0$


2) $a_n≥a_{n+1}$

My question is, why is the second condition greater than or equal to instead of just greater than? How can it be continually decreasing if it is equal to the next term?

  • $\begingroup$ Note: I reformatted your question and added the condition that $a_n>0$ which I think you forgot. $\endgroup$ – lulu Oct 19 '16 at 18:05
  • $\begingroup$ The Leibniz test as this is sometimes known, works with $≥$. The proof (available on that link) is fairly straight forward. $\endgroup$ – lulu Oct 19 '16 at 18:07
  • $\begingroup$ @lulu thank you, I would've done that but I'm currently on mobile. $\endgroup$ – user10984587 Oct 19 '16 at 18:08
  • $\begingroup$ No problem. I hope the link is clear. The way I view this, the inequality means that the partial sums are confined to the interval $[a_0-a_1,a_0]$ and the limit tells us that the distance between the partial sums goes to $0$. Easy to see that these two observations imply convergence. $\endgroup$ – lulu Oct 19 '16 at 18:10
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    $\begingroup$ @user10984587 by the way, I upvoted your question. It's nice that you're asking about the assumptions for random theorems you're given in class :) $\endgroup$ – Andres Mejia Oct 19 '16 at 19:11

Consider the sequence $(1,1,0,0,0,0....)$. The alternating sum over this sequence will converge, but it is the case that $a_n=a_{n+1}$ for infinitely many $n \in \mathbb N$.

A short remark: if you consider a sequence so that $a_n=a_{n+1} \neq 0$ for all $n>N$ for some $N$, we will have a problem, since the sum will be:

$$\sum_{n=1}^{N}(-1)^{n} a_n+\sum_{n=N+1}^{\infty}(-1^n)a_n=a_n\left(\sum_{n=1}^{\infty}(-1)^n\right),$$ which diverges. So, the first situation described only occurs if $a_n=a_{n+1}=0.$ But this is okay since in the case where equality occurs for some nonzero $a_n$, the sequence does not converge to zero, and the theorem does not apply.

However, suffice it to say, that the inequality $\geq$ is slightly more general, and if we can prove the theorem (for example, one of the comments) for this setting, why only prove it for sequences that satisfy $a_n>a_{n+1}$?


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