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The problem is

Prove that the series $$\sum_{n=1}^\infty (-1)^n\frac{x^2+n}{n^2}$$ converges uniformly in every bounded interval, but does not converge absolutely for any value of $x$.

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My attempt is:

(a) Let $[a,b]\subset\mathbb{R}$. Let $\epsilon>0$. Choose $N\in\mathbb{N}\ni \forall n\geq N$, $$\sum_{n=k}^\infty\frac{1}{k^2}<\frac{\epsilon}{2b^2} \qquad\text{and}\qquad \sum_{k=n}^\infty\frac{(-1)^k}{k}<\frac{\epsilon}{2}.$$ Let $n\geq N$. Then $$\sum_{k=n}^\infty (-1)^k\frac{x^2+k}{k^2}=\sum_{k=n}^\infty\left((-1)^k\frac{x^2}{k^2}+(-1)^k\frac{1}{k^2}\right)\leq\sum_{k=n}^\infty(-1)^k\frac{x^2}{k^2}+\sum_{k=n}^\infty(-1)^k\frac{1}{k^2}$$ $$\leq\sum_{k=n}^\infty \frac{b^2}{k^2}+\sum_{k=n}^\infty(-1)^k\frac{1}{k^2}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$ $$\therefore \sum_{n=1}^\infty(-1)^n\frac{x^2+n}{n^2} \quad\text{converges uniformly on }[a,b]$$

(b) $$\sum_{n=1}^\infty\left|(-1)^n\frac{x^2+n}{n^2}\right|=\sum_{n=1}^\infty \left|\frac{x^2+n}{n^2}\right|\leq\sum_{n=1}^\infty\frac{1}{n}$$ hence the series is not absolutely convergent.

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Is my procedure correct?

Thanks for your help

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In your proof of part b, the last inequality should be reversed. Which is OK since $|x^2+n| \ge n$. The part a proof looks right. – coffeemath Dec 24 '12 at 5:36
A similar question:… – Jonas Meyer Dec 24 '12 at 5:47

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