Differentiability of a Series of Functions

luke8ball

I'm working on a problem where I need to show that the series of functions, f(x) = Ʃ (xⁿ)/n², where n≥1, 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 I 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?

Erland

The differentiated series converges on the open interval (-1,1). Hence the original series is differentiable there and the derivative is the differentiated series. The differentated series diverges at 1 and converges conditionally at -1. Although I cannot prove it, this makes me suspect that the original series is not differentiable at 1 and -1.