Alternating series convergence test

Question

I have a series $\sum_{n=1}^\infty c_n x^n$ where $c \le c_n \le C$.

I can determine radius of convergence easily by the root test, but how does one determine convergence for $x = -1$?

It is not a positive series, so divergence test does not work, and it is not absolutely convergent either, so that does not work.

I don't see how we have enough information to solve this problem?

I know nothing but that which is posted :) I am beginning to think the problem was posed incorrectly, and that perhaps they merely wanted radius of convergence? — Peter Felafe, May 18 '15 at 20:13
Hint: If the terms do not have limit $0$, the series does not converge. The simplest, but often useful, divergence test. — André Nicolas, May 18 '15 at 20:20

score 0 · Answer 1 · answered May 18 '15 at 20:24

If $c > 0$ or $C < 0$ then the series cannot converge for $x = -1$ because the terms do not go to zero in absolute value. If $c \leq 0 \leq C$ then it is more subtle. If the $c_n$ alternate between $-1/n$ and $1/n$ for example then you will get divergence at $x = -1$ even though the terms go to zero. However if the $c_n$ are all $1/n$ you will get convergence.

Alternating series convergence test

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