Say I have this series:

$$\sum_{n=1}^{\infty} (-1)^{n} \frac{3n-1}{2n+1}$$

so $b = \frac{3n-1}{2n+1}$

I can't tell if this is decreasing by looking at it since the numerator is smaller but the denominator is smaller too. So the first derivative is:

$$f'(x) = \frac{3(2n+1) - 2(3n-1)}{2n+1}$$

$$= \frac{6n+3-6n+2}{2n+1}$$

$$= \frac{5}{2n+1}$$

which is positive for $n \geq 1$

So is is true that if it fails the AST, then we only know it doesn't converge according to the AST? It still may converge or diverge so we have to use alternate tests right? It's a sufficient condition for convergence only but not a necessary one for convergence and failure doesn't sufficiently mean divergence either?

Let's use the Divergence test:

$$\lim_{n \to \infty} -1^n \frac{3n-1}{2n+1}$$

$$\lim_{n \to \infty} -1^n \frac{n-\frac{1}{2n}}{1 + \frac{1}{2n}}$$

$$\lim_{n \to \infty} -1^n n \neq 0$$

which is not equal to 0 so it diverges by divergence test right?

  • $\begingroup$ Indeed. What you've done here looks all right to me. $\endgroup$ – Don Thousand Apr 3 '19 at 18:26
  • $\begingroup$ Is the theory right? The failure of the AST shows really nothing? $\endgroup$ – Jwan622 Apr 3 '19 at 18:34
  • 1
    $\begingroup$ @Jwan622 Yes, the AST is only a sufficient condition. $\endgroup$ – Jan Apr 3 '19 at 18:37

The Alternating Series Test says that an alternating series $\sum_{n = 0}^\infty a_n$ converges if

  1. $|a_n|$ is a monotonically decreasing function
  2. $\lim_{n \to \infty} a_n = 0$.

Your work in fact shows that both of these conditions fail for the given series, $$\sum_{n = 0}^{\infty} \frac{3 n - 1}{2 n + 1} :$$

The computation of the derivative of $x \mapsto \frac{3 x - 1}{2 n + 1}$ shows that (1) fails, and a corrected version of your latter computation shows that (2) fails. (In the first step, multiplying both the numerator and denominator by $\frac{1}{n}$ gives $\lim_{n \to \infty} (-1)^n \frac{3 - \frac{1}{n}}{2 + \frac{1}{n}}$, and this limit does not exist because $\frac{3 - \frac{1}{n}}{2 + \frac{1}{n}} \to \frac{3}{2} \neq 0$.)

Condition (2) is actually a necessary condition for the convergence of any series, not just an alternating series. So, because the series in this example fails condition (2), we conclude that the series does not converge.

But even when (2) is satisfied, (1) is not a necessary condition for convergence of an alternating series, and hence the Alternating Series Test is only a sufficient condition for an alternating series to converge, not a necessary one.

For example, consider the alternating series $$ \sum_{n = 1}^{\infty} a_n, \qquad a_n = [1 + 2 (-1)^n] 2^{-n} . $$ The Comparison Test shows that (a) the series converges (in fact, absolutely), but (b) for any even integer $2 m$, $$|a_{2m}| = 3 \cdot 4^{-m} > 2 \cdot 4^{-m} = |a_{2 m - 1}|,$$ so condition (1) fails.


But by the Leibnitz test for alternating series $$a_n$$ must have the limit zero. But $$\frac{3n-1}{2n+1}=\frac{3-\frac{1}{n}}{2+\frac{1}{n}}$$ has the limit $$\frac{3}{2}$$ so your series is not convergent.


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