evaluating a sum using Cauchy condensation test

Question

Let $$\sum\limits_{n\ge1}{\frac{(-1)^n}{n^\alpha \ln n}}$$

I want to check if the sum is converges absolutely. Hence, we need to check the convergence of $$\sum\limits_{n\ge1}{\frac{1}{n^\alpha \ln n}}$$

Now, I decided to use Cauchy condensation test:
$$\sum\limits_{n\ge1}{\frac{2^n}{(2)^{n\alpha} n\ln(2) }} = \sum\limits_{n\ge1}{\frac{1}{(2)^\alpha n\ln(2) }}$$

Now, I've been told the equality above is actually not true. Why?

Answer 1

You're going to kick yourself. When you divide powers you subtract the exponents:

$$ \frac{2^n}{2^{n\alpha}} = 2^{n(1-\alpha)} \neq 2^{-\alpha} $$