The sign of each term in the logarithmic series depends on the sign of `x`. If `x` is positive, then so is `x^k` for every `k`, and the sign of the term is the sign of `text[(]-1text[)]^(k+1)`, positive for odd `k` and negative for even `k`. In particular, the signs alternate in this case. On the other hand, if `x` is negative, then `x=- text[|] x text[|]` and `text[(]-1text[)]^(k+1)x^k=text[(]-1text[)]^(k+1)text[(]-1text[)]^k|x|^k=text[(]-1text[)]^(2k+1)|x|^k`. In this case, every term is negative, so the signs do not alternate. This means that the first condition of the Alternating Series Test is satisfied only for `x>0`.
If `0<x<=1`, then
`x^(k+1)/(k+1)=(x cdot x^k)/(k+1)<x^k/k`,
so the terms do decrease in size. Furthermore, `x^k<=1`, so `x^ktext[/]k<=1text[/]k`, and `lim_(k rarr oo) x^k/k=0`.
Thus all three conditions of the Alternating Series Test are satisfied if `0<x<=1`.
If `x` is any fixed number greater than `1`, then `x^k`, as a function of `k`, is an exponential function with base `x>1`, which must grow faster than any linear function of `k`, so the terms do not approach zero — in fact, they become arbitrarily large. Conclusion: The Alternating Series Test applies to the logarithmic series only for `0<x<=1`.