I study first the pointwise convergence :
1) if $x=0$ $\sum_{n=1}^{+\infty} \log(2)$ diverges
2) if $x>0$ using ratio test the series diverges
3) if $x<0$ I study absolute convergence and I find $|x+1|^n \log(1+n^x)\sim_{+\infty} |x+1| n^x $ and using ratio test I have absolute convergence and so pointwise convergence in $(-2,0)$
4) if $x=-2$ I have convergence for Leibnitz test
But for $x<-2$?
If $x < -2$, then $|x+1|^n = |(-1)(-x-1)|^n =(-x-1)^n = \alpha ^n$ where $\alpha = -x - 1 > 1$. We also have
$$\log(1+n^x) = \int_1^{1 + n^x} \frac{dt}{t}> \frac{n^x}{1+n^x} = \frac{1}{1+n^{-x}}= \frac{1}{1 + n^{1+\alpha}}> \frac{1}{2n^{1+\alpha}}$$
Thus,
$$|x+1|^n\log(1+n^x) > \frac{\alpha^n}{2n^{1+\alpha}} = \frac{e^{n \log \alpha}}{2n^{1+\alpha}}$$
where $\alpha > 1$ and $1 + \alpha > 2$. To what does the RHS converge as $n \to \infty$? What does that tell you about convergence of the series when $x < -2$?
