Paul's Online Notes vs. James Stewart Essential Calculus Early Transcendentals

In the case of these two sources I am deeply lost whether the textbook is even right about this particular matter.

Alternating Series Test

The text book by James Stewart states on page 455, that:

If the alternating series $$\sum_\limits{n=1}^\infty(-1)^{n-1} b_n = b_1-b_2+b_3-b_4+b_5-b_6+... b_n>0$$ satisfies

(i) $b_{n+1} \le b_n$ for all n

(ii) $\lim_\limits{n \to \infty}b_n=0$

Then the series is convergent.

However, I started to do this problem:

\begin{align} \sum_\limits{n=1}^\infty (-1)^n \frac{n}{\sqrt{n^3+2}} &=-\frac{\sqrt{3}}3+\frac{\sqrt{10}}{5}-\frac{3\sqrt{29}}{29} && \mathbf{Given} \\ \mathbf{Condition \ 1 \ False \ (to \ my \ belief)} \end{align}

At right then, and there I thought game over the series is divergent; however, I ended up going to Symbolab, and Wolfram Alpha, and found out that by the Alternating Series Test this series is convergent. I did some digging, and found Paul's Online Notes and it looked at the series expanded more terms and found that the first condition was valid in the long run. Is my textbook wrong or is my understanding of it wrong?

What is Wrong?

I believe the first condition is wrong in this case since it is not valid for all n.

  • 1
    $\begingroup$ I have no idea what you are asking. Is what wrong? $\endgroup$
    – copper.hat
    Dec 25, 2019 at 1:24
  • $\begingroup$ The correct definition would certainly be $\forall n \ge N$ for some $N$. $\endgroup$
    – gt6989b
    Dec 25, 2019 at 1:25
  • $\begingroup$ @copper.hat OP is asking if the textbook definition of alternating series is correct $\endgroup$
    – gt6989b
    Dec 25, 2019 at 1:25
  • 1
    $\begingroup$ The alternating series test provides a sufficient test for convergence, but not a necessary one. There are series which converge which don't satisfy the test - I'm sure you can think of some (eg geometric series with positive ratio) where all the terms have the same sign. That would make them invalid on the first condition (the $b_n$ would alternate in sign rather than being monotonic). $\endgroup$ Dec 25, 2019 at 1:37
  • 2
    $\begingroup$ One thing to keep in mind: there are many, many tests like this that only work in one direction. When you find a "test" or other theorem that says "if $P$ then $Q$," the only way this can be wrong is if there is some way for $P$ to be true and $Q$ to be false. Moreover, when you find a theorem stated that way in a textbook, very often there is some way for $Q$ to be true though $P$ is false; otherwise the theorem likely would have have been given as "$P$ if and only if $Q$". $\endgroup$
    – David K
    Dec 25, 2019 at 2:05

2 Answers 2


Your understanding is wrong. First, and most importantly, the alternating series test (AST) says that if these conditions hold, then we can guarantee convergence. But that does not mean that in order to guarantee convergence, we need the conditions to hold or that if the conditions don't hold then the series is divergent.

Second, with any series, the behaviour does not depend on the first $N$ terms for any finite $N$. That is, if you were to change $a_0, \dots, a_N$ to whatever you want, it doesn't affect the convergence behaviour. This means that if the AST conditions hold for $n \ge $ some finite $N$ then you can guarantee convergence. This also works for any of the tests.

For example, the integral test normally says that if $a_n = f(n)$ and $f$ is decreasing then $\sum a_n$ converges if and only if $\int f$ converges. But, like with the AST, we just need $f$ to be decreasing on $(N,\infty)$.


Your textbook is correct when it says that if $b_1\geqslant b_2\geqslant b_3\geqslant\cdots$ and if $\lim_{n\to\infty}b_n=0$, then the series $\sum_{n=1}^\infty(-1)^{n-1}b_n$ converges. But this doesn't mean that otherwise it diverges.

Note that if the first condition gets replaced by$$\text{for some $N\in\mathbb N$, }b_N\geqslant b_{N+1}\geqslant b_{N+2}\geqslant\cdots$$it is still true that the series converges. And it is natural to call alternate series test to this statement.


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