# uniform and pointwise convergence of $\sum_{n=1}^{+\infty} (x+1)^n \log(1+n^x)$

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 $$x<-2$$?

## 1 Answer

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 $$x < -2$$?

• Exponential $e^{n \log \alpha}$ grows faster than polynomial $n^{1+\alpha}$ as $n \to \infty$. – RRL Feb 12 '20 at 17:21