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I'm studying power series and their radius of convergence, and my book says

$$\sum_{k\ge0} x^k$$

has a convergence radius of 1 and doesn't converge in 1 and -1, as opposite to

$$\sum_{k\ge1} \frac {1}{k} x^k$$

which has a convergence radius of 1 too, but converges in -1 as well and diverges in 1.

To my understanding, the latter converges in -1 because we keep adding and subtracting numbers and the sum doesn't ever get much big, while if we have 1 we have the $\sum_{k\ge1} \frac {1}{k}$ number series which is a known diverging one.

Doesn't the same apply to the former series? I mean, if we have $x=1$ it is pretty obvious why it would diverge, but with -1 wouldn't we keep adding and subtracting 1, thus never diverging?

Is it that the infinite jumping between 1 and 0 that the sum would do is not considered convergence, while not being divergence, because a limit does not exist, as opposite to the second series in which the added/subtracted numbers get infinitely smaller thus getting closer and closer to a fixed value?

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  • $\begingroup$ Is $s_n=(-1)^n$ a converging sequence? $\endgroup$
    – Surb
    Apr 4, 2018 at 6:02
  • $\begingroup$ This is what I was wondering. For sure it's not a diverging one, but a limit does not exist... So perhaps it's neither converging not diverging? $\endgroup$ Apr 4, 2018 at 6:03
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    $\begingroup$ Definitely it is not converging, then what is diverging is another question. $\endgroup$
    – Surb
    Apr 4, 2018 at 6:04
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    $\begingroup$ Anyway, the radius of convergence describes convergence :). And the series, as you noticed, cleary does not converge at $-1$. $\endgroup$
    – Surb
    Apr 4, 2018 at 6:06
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    $\begingroup$ Ah Thanks, but I think the other answers answer already quite well your question. $\endgroup$
    – Surb
    Apr 4, 2018 at 6:09

2 Answers 2

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Using the $n$th term test, a requirement for a convergent series is:

$$\lim_{n\to\infty} a_n = 0$$

If not, then the series diverges. The limit, in this case, the doesn't exist as it oscillates between $1$ and $-1$ (and its partial sums $1$ and $0$), the series diverges:

$$\lim_{n\to\infty} (-1)^n = \text{DNE}$$

Divergence does not necessarily mean becoming really large, it just means it does not approach a value. Divergence essentially means the lack of convergence.

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Diverging means not converging. You're right that the sum $1-1+1-1+\dots$ doesn't explode to infinity, but it does not converge since it's not getting closer to any particular value.

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