The ${n}$th-term test for divergence everywhere I've seen basically states that, if the limit of the ${n}$th term as ${n}$ approaches infinity is not 0 or doesn't exist, the series diverges. However, this test does not give any precondition on the ${n}$th term (e.g. that it has to be positive), so it should be applicable to alternating series as well. So,

Shouldn't all alternating series diverge, since the limit of the nth term of the series, ${a_n =(-1)^nb_n}$ (including the ${(-1)^n}$), as n approaches infinity usually does not exist?

Won't this contradict those series that pass the Alternating Series Test (AST)? Or does the AST supersede the ${n}$th-term test in such cases (if so, why isn't this mentioned anywhere)?

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    $\begingroup$ What would make you think that the limit of $(-1)^nb_n$ does not exist? $\endgroup$ Mar 10, 2016 at 17:48
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    $\begingroup$ The limit of $b_n$ must be $0$ for the alternating series test. From this it follows that $(-1)^nb_n$ has limit $0$. $\endgroup$ Mar 10, 2016 at 17:48
  • $\begingroup$ @David I mistakenly thought that, since the limit of ${(-1)^n}$ doesn't exist, the limit of ${(-1)^nb_n}$ would also not exist. I checked Wolfram Alpha for the limits of ${a_n}$ for convergent alternating series, and they are indeed 0, so the nth-term test remains valid after all! $\endgroup$
    – Leponzo
    Mar 10, 2016 at 18:09
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    $\begingroup$ Let's stick to integers, $(-1)^x$ gets us into trouble for non-integer $x$. If $b_n$ has a non-zero limit, then sure, $(-1)^n b_n$ does not have a limit. But if $b_n$ is after a while close to $0$, then so is $(-1)^nb_n$. $\endgroup$ Mar 10, 2016 at 18:22
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    $\begingroup$ Two comments. First, of course it was clear what your error was, I was trying to get you to find it yourself. The problem was remembering things without the details: It's a Fact that if the limits of $a_n$ and $b_n$ exist then the limit of $a_nb_n$ exists, and equals the product of the limits. But that Fact says nothing about limits not existing. Second: You checked Wolfram Alpha? That's very sad. Also dangerous - WA gets things wrong sometimes. If you think about the definition of limits it's very easy to see that if $\lim b_n=0$ then $\lim (-1)^nb_n=0$. $\endgroup$ Mar 10, 2016 at 18:33

1 Answer 1


Well, obviously there are convergent alternating series, so there must be some mistake here. We know that, for $\sum a_n$ to converge, then the condition


is necessary (but not sufficient!). So if we take $a_n=(-1)^nb_n$, then if $\sum (-1)^nb_n$ converges, then


which happens if and only if


so well, we can have an alternating sequence who's series converges, but we need to have $b_n\to 0$ for $\sum (-1)^nb_n$ to converge. An example:


on the other hand, we can make a series like \begin{equation} c_n= \begin{cases} \frac{1}{n} \text{ if } n \text{ is even}\\ \frac{1}{2n} \text{ otherwise } \end{cases} \end{equation} for which actually

$$\sum_{n=1}^\infty (-1)^nc_n$$

actually diverges to $+\infty$, but $\lim c_n\to 0$. We again see, for $\sum (-1)^nb_n$ to converge, the condition $\lim b_n=0$ is necessary, but not sufficient.


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