I follows these steps:

  1. Define $e^z:=\sum_{k=0}^{\infty}\frac{z^k}{k!}$.
  2. Show thatthe series is absolutely convergent.
  3. Define $\sin(z):=\frac{e^{iz}-e^{-iz}}{2i}$, and $\cos(z):=\frac{e^{iz}+e^{-iz}}{2}.$
  4. Derive $$ \sin(z)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}z^{2n+1}, \quad \cos(z)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}z^{2n}. $$ just by sum manipulation.

Can I now trust that $\frac{e^{iz}+e^{-iz}}{2}=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}z^{2n+1}$? Or: is this question equivalent to the following lemma?

Lemma If $f(x)=\sum a_k$ and $g(x)=\sum b_k$ are abs. conv. then (a) $\sum ca_k$ is abs. conv. and $cf=\sum ca_k$; (b) $\sum (a_k+b_k)$ is abs. conv. and $f+g=\sum (a_k+b_k)$.

Is the given lemma treated in a standard textbook or is assumed as evident?


Right now, I can't provide an example of a textbook proving that lemma. But it's quite simple:

(a) If $\sum_{n=0}^\infty\lvert a_k\rvert$ converges, then $\sum_{n=0}^\infty\lvert ca_k\rvert$ converges too, since it is equal to $\sum_{n=0}^\infty\lvert c\rvert\lvert a_k\rvert$.

(b) If both series $\sum_{n=0}^\infty\lvert a_k\rvert$ and $\sum_{n=0}^\infty\lvert b_k\rvert$ converge, then $\sum_{n=0}^\infty\lvert a_k+b_k\rvert$ converges too, since$$(\forall k\in\mathbb Z^+):\lvert a_k+b_k\rvert\leqslant\lvert a_k\rvert+\lvert b_k\rvert$$and $\sum_{n=0}^\infty\bigl(\lvert a_k\rvert+\lvert b_k\rvert\bigr)$ converges.

  • $\begingroup$ If I understand your answer, you mean that the Lemma is needed, and I'm not making my life unnecessarily difficult (that was the spirit of my question). $\endgroup$ – Aaron Lenz May 14 at 8:12
  • $\begingroup$ Yes, the lemma is needed. No doubt about that. $\endgroup$ – José Carlos Santos May 14 at 8:14

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