Relating $\sum_{n = 0}^{\infty} (-1)^{n}/(1 + np)$ to $\int_{0}^{1} dx/(1 + x^{p})$

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This is an application of the lemma that if $\{f_{n}\}$ is a sequence of nonnegative measurable functions, $\int \sum_{n = 0}^{\infty} f_{n} = \sum_{n = 0}^{\infty} \int f_{n}.$ I'm supposed to deduce that $\sum_{n = 0}^{\infty} (-1)^{n}/(1 + np) = \int_{0}^{1} dx/(1 + x^{p})$.

So the expression $(-1)^{n}/(1 + np) = \int_{0}^{1} f_{n}$, and $1/(1 + x^{p})$ is the sum of our $f_{n}$'s... I have a feeling there's a series formula I'm forgetting here, but this isn't quite clicking.

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Hint: By using Taylor series you can write $$\dfrac{1}{1+x^p}=1-x^p+x^{2p}-x^{3p}+\cdots$$can you finish now?