I am trying to find under which conditions does the sum below converge to 0,
$\sum_{n=1}^{\infty}{(-1)}^{n+1}\cos(b\log(n))({n}^{a-1}-{n}^{-a})=0,$
with b and a real numbers, with 0 < a < 1.
I know that this sum is conditionally convergent as its absolute value diverges. Furthermore I cannot use the alternating series test as the terms of the sum are not all positive and not all negative. Any help would be appreciated. Thank you.
I'd rather write this as an answer. Now we have an alternating series and add together every 2 term, that is $$\cos(b\log (2n-1))((2n-1)^{a-1}-(2n-1)^{-a})-\cos(b\log 2n)(2n^{a-1}-2n^{-a})\\ =\cos(b\log (2n-1))[((2n-1)^{a-1}-2n^{a-1})-((2n-1)^{-a}-2n^{-a})]\\ -[\cos(b\log 2n)-\cos(b\log (2n-1))](2n^{a-1}-2n^{-a})$$
For the first term $\cos(b\log (2n-1))$ is bounded by $1$, $((2n-1)^{a-1}-2n^{a-1})$ grows at the same rate as $a(2n)^{a-2}$ and similarly $((2n-1)^{-a}-2n^{-a})$ at $a(2n)^{-a-1}$, so the first part absolutely converge.
For the second part, use the inequality $$|\cos(b\log(2n))-\cos(b\log (2n-1)|=|\int_{2n-1}^{2n} -\sin(\cos(b\log x))\frac{b}{x} dx|<\int_{2n-1}^{2n}|\frac{b}{2n}| dx<\frac{b}{2n}$$ So the second term grows at the rate of $O(n^{a-2})+O(n^{-a-1})$, also absolutely convergent.
I don't know if the original series absolutely converges, which needs far more effort to verify.