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I want to find the first $4$ coefficients of the Maclaurin series of $\tan(z)$ by multiplying by $\cos(z)$ and using a Cauchy product.

Letting $\tan(z)=\sum\limits_{k=0}^\infty c_kz^k$ and multiplying by $\cos(z)=\sum\limits_{k=0}^\infty \frac{(-1)^k}{2k!}z^{2k}$ gives $$\sin(z)=\sum\limits_{k=0}^\infty c_kz^k \cdot \sum\limits_{k=0}^\infty \frac{(-1)^k}{2k!}z^{2k}$$ and the coefficients can be equated to the series representation of $\sin(z)$ to find $c_0,\ldots,c_4$.

However, I am not sure how to take the Cauchy product of this since the indices of $z$ don't match. Is there some manipulation that can be done to the limits of the sums or the powers of $z$ to let this product fit the definition of a Cauchy product?

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If you wanted the indices to match, you can use this series instead: $$\cos(x)=\sum_{k=0}^\infty\frac{\cos\left(\frac{\pi k}{2} \right)}{k!}x^k$$

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