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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 $x<-2$?

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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 $x < -2$?

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  • $\begingroup$ Exponential $e^{n \log \alpha}$ grows faster than polynomial $n^{1+\alpha}$ as $n \to \infty$. $\endgroup$ – RRL Feb 12 at 17:21

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