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If I consider $$\sum_{n=1}^{\infty} \frac{x^n}{1+n^2}+\sum_{n=1}^{\infty} \frac{n}{1+n^2}$$ I have that the first series converges in all $\mathbb R$ and the second diverges so the given series diverges?

If $|x|\le1$ the general term $f_n(x) \sim_{+\infty} {{1}\over{n}}$ general term of a divergent series

If $|x|>1$ the general term $f_n(x) \sim_{+\infty}{{x^n}\over{1+n^2}}$ general term of a power series that diverges. So the given series diverges?

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  • $\begingroup$ The first series is a power series and the radius of convergence is $+\infty$ ? $\endgroup$ – GiulyB Feb 12 at 11:09
  • $\begingroup$ The 1st series does not converge over all of $\mathbb R$, in particular it diverges for $|x|>1$. When they both diverge you can not conclude that the original series diverges from that. $\endgroup$ – emacs drives me nuts Feb 12 at 11:13
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Yes. If $\sum_{n=1}^\infty a_n$ converges and $\sum_{n=1}^\infty b_n$ diverges, then $\sum_{n=1}^\infty(a_n+b_n)$ always diverges.

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I have that the first series converges in all $\mathbb R$

The series $$\sum_{n=1}^{\infty} \frac{x^n}{1+n^2}$$ diverges for $|x| > 1$. Simply because the addends do not converge to 0.

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  • $\begingroup$ I have edited the solution.It's correct? $\endgroup$ – GiulyB Feb 12 at 12:13

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