Given a series $$ \sum^\infty_{n=0} (-1)^n b_n$$ the Alternating Series test states that if $b_n$ is convergent if

1) $b_n$ is positive

2) $b_n$ is monotonically decreasing

3) $\lim_{n\to\infty} b_n = 0$

Every proof of this theorem requires these to all be true, but what I wonder is whether this theorem would still hold if (2) was omitted.

In essence, can you think of an example where the alternating series diverges if (1) and (3) are satisfied but not (2)?

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    $\begingroup$ The convergence holds without condition $1)$ and under the relaxed condition $2)$ where $$\sum_{n=1}^\infty |b_{n+1}-b_n| \le L<\infty$$ $\endgroup$ – Mark Viola Apr 6 '18 at 3:18

$$b_n=\left\{ \begin{array}{lc} \frac{1}{n} &\mbox{ if } n \mbox{ is even } \\ \frac{1}{n^2} & \mbox{ if } n \mbox{ is odd } \end{array} \right.$$

One another hand, if $b_n$ is decreasing and $\lim_n b_n=0$ then $b_n \geq 0$, so 1) is actually redundant.

Added :the partial sum $S_{2n}$ is

$$S_{2n}=\sum_{k=0}^{2n} (-1)^kb_k=\sum_{k=1}^n \frac{1}{2n}-\sum_{k=0}^{n-1} \frac{1}{(2n+1)^2}$$

Now, since $\sum_{k=1}^n \frac{1}{2n} \to \infty$ and $\sum_{k=0}^{n-1} \frac{1}{(2n+1)^2}$ is convergent, you get that $S_{2n} \to \infty$, which shows that the series is divergent.

  • $\begingroup$ Could you expand on why that choice of $b_n$ is divergent? I don't seem to see it. $\endgroup$ – Аlon Djurinsky Apr 6 '18 at 3:15
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    $\begingroup$ @АlonDjurinsky Added. In short, the positive terms add to $\infty$ and the negative part is convergent. $\endgroup$ – N. S. Apr 6 '18 at 3:21
  • $\begingroup$ The sum should begin at $k=1$, not $k=0$. $\endgroup$ – Mark Viola Apr 6 '18 at 3:26

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