# Uniform and pointwise convergence of $\sum_{n=1}^\infty \frac{x^n+n}{1+n^2}$

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?

• The first series is a power series and the radius of convergence is $+\infty$ ? – GiulyB Feb 12 at 11:09
• 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. – emacs drives me nuts Feb 12 at 11:13

## 2 Answers

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.

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.

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