# pointwise and uniform convergence of $\sum_{n=1}^{+\infty} \frac{(3^x-2)^n}{n+n^x}$

I study the absolute convergence: $$\sum_{n=1}^{+\infty} \frac{|3^x-2|^n}{n+n^x}$$:

If $$x=1$$ $$\sum_{n=1}^{+\infty} \frac{1}{2n}$$ diverges; if x>1 $$\frac{(3^x-2)^n}{n+n^x} \sim_{+\infty} \frac{(3^x-2)^n}{n^x}$$ and for the ratio rule not converges;

If $$x<1$$ $$\frac{|3^x-2|^n}{n+n^x}\sim_{+\infty} \frac{|3^x-2|^n}{n}$$ that converges for $$0.

If $$x <0$$ the general term of the series is non infinitesimal so there isn't convergence.

If $$x=0$$ $$\sum_{n=1}^{+\infty} \frac{(-1)^n}{n+1}$$ converges for Leibniz test.

The series pointwise converges in $$[0,1)$$.But for the uniform convergence?

• You went from "uniform" to "absolute"--which is it? – zhw. Feb 10 at 22:58
• I study the absolute convergence to find pointwise convergence. – GiulyB Feb 11 at 15:18

If the series converges uniformly on $$(0,1)$$ then there exists $$n_0$$ such that $$\sum\limits_{k=N_1}^{N_2} \frac {(3^{x}-2)^{n}} {n+n^{x}} <1$$ for all $$x \in (0,1)$$ whenever $$N_2 >N_1 >n_0$$. Let $$x \to 1$$ in this to get $$\sum\limits_{k=N_1}^{N_2} \frac 1 {2n} \leq 1$$ whenever $$N_2 >N_1 >n_0$$. This is a contradiction.