Exercise 7.6

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Exercise 6: 
    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 $x$.

analambanomenos · Accepted Answer

\def\eps{\varepsilon} Since $$\lim_{n\rightarrow\infty}\frac{x^2+n}{n^2}=0,$$ the alternating series converges for all $x$ by Theorem 3.43. It doesn't converge absolutely for any $x$ since $$\sum_{n=1}^\infty \frac{x^2+n}{n^2}\ge\sum_{n=1}^\infty\frac{1}{n}$$ which diverges. The partial sums are $$f_m(x)=\sum_{n=1}^m(-1)^n\frac{x^2+n}{n^2}=x^2\sum_{n=1}^m\frac{(-1)^n}{n^2}+ \sum_{n=1}^m\frac{(-1)^n}{n}=A_mx^2+B_m\rightarrow Ax^2+B.$$ Let $\varepsilon>0$ and let $N$ be large enough so that $|A-A_n|<\varepsilon$ and $|B-B_n|<\varepsilon$ for all $n>N$. Let $[a,b]$ be a bounded interval, and let $c=\max\bigl(|a|,|b|\bigr)$. Then for $n>N$ and $x\in[a,b]$ we have $\big|f(x)-f_n(x)\big|

Exercise 7.6

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Exercise 7.6

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