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I'm working on a problem where I need to show that the series of functions, $$ f(x) = \sum_{n\geq 1} \frac{x^n}{n^2}, $$ converges to some $f(x)$, and that $f(x)$ is continuous, differentiable, and integrable on $[-1,1]$.

I know how to show that $f(x)$ is continuous, since each $f_n(x)$ is continuous, and $f_n(x)$ converges uniformly. Because each $f_n(x)$ is also integrable, I can also show $f(x)$ is integrable.

The trouble I'm having is proving that $f(x)$ is differentiable. I need to show that the series of derivatives converges uniformly. However, I don't think I can use the Weierstrass M-Test in this scenario. Any ideas?

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    Can you show that the derived series $\sum_{n=0}^{\infty}\frac{x^{n-1}}{n}$ has the same radius of convergence of the original series?2012-11-18
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    @Nameless I think you mean to index the series at 1 - there is no $x^{-1}$ term when you differentiate, right?2012-11-18
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    @JohnMartin Of course...2012-11-18
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    @Nameless cool. My apologies if that was overly pedantic.2012-11-18
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    @JohnMartin No worries. Now the OP said that (s)he can show $f$ is integrable. The same can be done for $g(x)=\sum_{n=1}^{\infty}\frac{x^{n-1}}{n}$. This is continuous and integrable and $f(x)=\int g(x)\, dx$...2012-11-18
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    I'm trying to show that f(x) is differentiable. I know that a x=1, the series of derivatives does not converge. I believe that I can show the series does converge on [-1,1). However, does that mean that f(x) is not differentiable at 1?2012-11-18
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    You will have to search that separately. First check the differentiability on $(-1,1)$ and then on the end points.2012-11-18

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