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By the Alternating Test: The series $\sum_{n=1}^{\infty}(-1)^{n+1}\cdot b_n$ converges if all three of the following conditions are satisfied:

  1. The $b_n$'s are all positive.
  2. The positive $b_n$'s are (eventually) decreasing: $b_n\ge b_{n+1}$ $\forall n\ge N$.
  3. $b_n \to 0$

But in the Divergence Test: Given $\sum_{n=1}^{\infty} a_n$, iff $\lim_{n \to \infty} a_n \neq 0 \implies$ $\sum_{n=1}^{\infty} a_n$ diverges


Now given an arbitrary alternating series:

$$S = \sum_{n=1}^{\infty}(-1)^{n+1}\cdot b_n$$

If we take the limit of $a_n$ in the series above

$$\lim_{n \to \infty}(-1)^{n+1}\cdot b_n = \underbrace{\left(\lim_{n \to \infty }(-1)^{n+1}\right)}_\text{This limit doesn't exist}\left(\lim_{n \to \infty} b_n\right)$$

Therefore by the Divergence test, $S$ should be a divergent series, regardless of the conditions needed for the alternating test. But by the Alternating Series Test, $S$ is a convergent series provided the three conditions stipulated initially are met.

So how is this seeming contradiction resolved, by the alternating series test?

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    $\begingroup$ You can't distribute the limits like that when they don't both exist. $\endgroup$ – Alexis Olson Aug 22 '16 at 15:48
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if $a_n b_n$ converges, then this does not imply that $a_n$ converges. In case $(-1)^n b_n$ the first factor is bounded. In this case, if the second factor converges to $0$, the product also converges to $0$.

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The algebra of limits says that if $\lim a_n$ and $\lim b_n$ exists then you can write:

$$\lim a_n b_n=(\lim a_n)(\lim b_n)$$

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Note: If $b_n\to 0,$ then $(-1)^nb_n \to 0.$ Proof: $|(-1)^nb_n - 0| = |(-1)^nb_n| = |b_n| = |b_n-0|.$

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$$\lim_{n\rightarrow\infty} 1 = \lim_{n\rightarrow\infty} (-1)^n(-1)^n =\left(\lim_{n\rightarrow\infty}(-1)^n\right)\left(\lim_{n\rightarrow\infty} (-1)^n\right).$$

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    $\begingroup$ An example of illegal manipulation? $\endgroup$ – Did Aug 23 '16 at 15:27
  • $\begingroup$ Yes. <insert Jackie Chan meme> $\endgroup$ – B. Goddard Aug 23 '16 at 15:46

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