I just got my homework back and I don't really know why my proof wasn't enough, it's about rearranging the alternating harmonic series.
$(a_k)_{a \in \mathbb{N}}$ is a sequence with $a_k= (-1)^k/k$. $g_k =2k$ and $u_k=2k-1$, define a new sequence $(b_k)_{b \in \mathbb{N}}$: $a_{u_1}, a_{g_1}, a_{g_2}; a_{u_2}, a_{g_3}, a_{g_4};...$.
We already know that the limit $a:=\sum\limits_{k=1}^\infty a_k$ exists. Show that $\sum\limits_{k=1}^\infty b_k$ converges as well and $\sum\limits_{k=1}^\infty b_k=\frac{1}{2}a$.
$\sum\limits_{k=1}^\infty a_k=-1 + \frac{1}{2} -\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}+\frac{1}{8}-\frac{1}{9}+\frac{1}{10}-...$
$\sum\limits_{k=1}^\infty b_k=-1 + \frac{1}{2} +\frac{1}{4}-\frac{1}{3}+\frac{1}{6}+\frac{1}{8}-\frac{1}{5}+\frac{1}{10}+\frac{1}{12}-...$
$=(-1 + \frac{1}{2}) +\frac{1}{4}+(-\frac{1}{3}+\frac{1}{6})+\frac{1}{8}+(-\frac{1}{5}+\frac{1}{10})+\frac{1}{12}-...$
$= \frac{1}{2}(-1 + \frac{1}{2} -\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-...)=\frac{1}{2}a$
I got 2/3 points for this and he said that I can't use $...$ notation here. But I don't understand why, I did some other proofs with $exp$ $sin$ and $cos$ and it was fine there. For example:
Show that $cos(z)=\sum\limits_{k=1}^\infty (-1)^k\frac{z^{2k}}{(2k)!}=1-\frac{z^2}{2!}+\frac{z^4}{4!}...$.
$cos(z) = \frac{1}{2}(exp(iz)+exp(-iz))=\sum\limits_{k=1}^\infty\frac{(iz)^{k}}{(2k)!}+\sum\limits_{k=1}^\infty\frac{(-iz)^{k}}{(2k)!}$
$=\frac{1}{2}+\frac{iz}{2}-\frac{z^2}{2(2!)}-\frac{iz^3}{2(3!)}+\frac{z^4}{2(4!)}+...$
$+\frac{1}{2}-\frac{iz}{2}-\frac{z^2}{2(2!)}+\frac{iz^3}{2(3!)}+\frac{z^4}{2(4!)}-...$
$=1-\frac{z^2}{2!}+\frac{z^4}{4!}-...$
I got 2/2 points for this, why is using $...$ fine here and not in the other exercise? When am I allowed to use $...$ notation and when do I have to use $\sum\limits_{k=1}^\infty$?
You did not prove that your series converges. All that you proved was that if it converges, then its sum is half the sum of the harmonic series.