Study the convergence of a parametrical sum. 
Determine for which $\alpha >0$ we have absolute convergence or just normal convergence of this sum: $$\sum_{n=1}^{\infty}(-1)^n(\sqrt[n]{n}-1)^{\alpha}\arctan n$$

My attempt: Let's study at first the absolute convergence of this sum: if we denote by $b_n=(\sqrt[n]{n}-1)^{\alpha}\arctan n$ we have that this is a positive succession so
$$\sum_{n=1}^{\infty}|(-1)^n(\sqrt[n]{n}-1)^{\alpha}\arctan n|=\sum_{n=1}^{\infty}(\sqrt[n]{n}-1)^{\alpha}\arctan n$$
Now we have:
$$(\sqrt[n]{n}-1)^{\alpha}\arctan n = (e^{\frac{1}{n}\log n}-1)^{\alpha}\arctan(n) \sim \frac{\log^{\alpha}n}{n^{\alpha}}\arctan n$$
and, by using Cauchy Condensation Test with $c=e$ we have:
$$\sum_{n=1}^{\infty}\frac{\log^{\alpha}n}{n^{\alpha}}\arctan n < \infty \iff \sum_{n=1}^{\infty}\frac{n^{\alpha}}{e^{n(\alpha-1)}}\arctan n < \infty$$
which is true iff $\alpha > 1$. 
For the standard convergence, using Leibnitz Criterion and reconducting us to this case we can see that the convergence is $\forall \alpha > 0$.
Am I missing something or is it a good proof? Thanks in advance.
 A: Use that
$$\arctan n=\frac{\pi}2-\arctan \frac1n$$
then
$$\sum_{n=1}^{\infty}(-1)^n(\sqrt[n]{n}-1)^{\alpha}\arctan n=\frac{\pi}2\sum_{n=1}^{\infty}(-1)^n(\sqrt[n]{n}-1)^{\alpha}+\sum_{n=1}^{\infty}(-1)^n(\sqrt[n]{n}-1)^{\alpha}\arctan \frac1n$$
and we have that
$$\sqrt[n]{n}-1=\frac{\log n}n+o\left(\frac{\log n}{n}\right)$$
then
$$(\sqrt[n]{n}-1) ^{\alpha} =\frac{\log^{\alpha} n}{n^{\alpha}}\left(1+o\left(1\right)\right)^{\alpha}=\frac{\log^{\alpha} n}{n^{\alpha}}+o\left(\frac{\log^{\alpha} n}{n^{\alpha}}\right)$$
and thus
$$\sum_{n=1}^{\infty}(-1)^n(\sqrt[n]{n}-1)^{\alpha}=\sum_{n=1}^{\infty}(-1)^n\frac{\log^{\alpha} n}{n^{\alpha}}+\sum_{n=1}^{\infty}(-1)^no\left(\frac{\log^{\alpha} n}{n^{\alpha}}\right)$$
converge by alternating series test for any $\alpha>0$ and since
$$\left|(-1)^n(\sqrt[n]{n}-1)^{\alpha}\arctan \frac1n\right|=\frac{\log^{\alpha} n}{n^{\alpha+1}}+o\left(\frac{\log^{\alpha} n}{n^{\alpha+1}}\right)$$
the series
$$\sum_{n=1}^{\infty}(-1)^n(\sqrt[n]{n}-1)^{\alpha}\arctan \frac1n$$
converges absolutely by limit comaprison test with
$$\sum_{n=1}^{\infty} \frac{\log^{\alpha} n}{n^{\alpha/2+1}}$$
