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$\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^3+x}$

show that the series converge in $(-1,\infty)$ to a continuously differentiable function.

I know it's something with uniformly convergence but not sure how.

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  • $\begingroup$ So you didn't try anything? Weierstrass M? The theorem on uniform convergence and differentiation? Hard to believe you have no idea here ... $\endgroup$
    – zhw.
    Jul 17, 2018 at 19:46

1 Answer 1

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Hint: it is sufficient to show that the derivatives converge uniformly and that original series converges at (at least) one point. To show uniform convergence, consider using the Weierstrass M-test.


Here's a further hint for uniform convergence: the derivative of the $n$th term is $$-\frac{n\sin(nx)}{n^3 + x} + \frac{\cos(nx)}{(n^3 + x)^2}\,.$$

In order to show uniform convergence, it is sufficient to bound this term above for $x \in (-1,\infty)$ by a constant $M_n$ so that $\sum M_n < \infty$.

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  • $\begingroup$ can you show me the part with the uniformly converge of the derivatives? I tried it but got stuck. $\endgroup$ Jul 17, 2018 at 16:30

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