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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 $0<x<1$.

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?

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

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