I have a question about my attempt to demonstrate the following exercise:
exercise:
Let $f$ be analytic in a domain $\Omega$ and for $a \in \Omega$ let
$$ f_n = \sum_{k=1}^n f^{(k)}(a)(z-a)^k, z \in \Omega $$
so $f_n(z)$ is the nth partial sum of the Taylor expansion around $a$. Show that if $\overline{D}(a,R) \subset \Omega$, then
$$f_n(z) = \frac{1}{2\pi i} \int_\alpha \frac{f(\epsilon)}{(\epsilon-a)^{n+1}} \frac{(\epsilon-a)^{n+1}-(z-a)^{n+1}}{\epsilon-z}d\epsilon$$ $z\in D(a,R)$, where $\alpha(t) = a+Re^{2\pi i t}$
My proof:
Note that: $$ \frac{1}{2\pi i }\int_\alpha \frac{f(\epsilon)}{(\epsilon-a)^{n+1}}\frac{(\epsilon-a)^{n +1}-(z-a)^{n+1}}{\epsilon-z}d\epsilon$$ $$ = \frac{1}{2\pi i} \int_\alpha \frac{f(\epsilon)}{\epsilon-z} d\epsilon - \frac{1}{2\pi i} \int_\alpha \frac{f(\epsilon)(z-a)^{n+1}}{(\epsilon-a)^{n+1}(\epsilon-z)}d\epsilon $$ but by the Cauchy integral, $$ = f(z) - \frac{(z-a)^{n+1}}{2\pi i} \int_\alpha \frac{f(\epsilon)}{(\epsilon-a)^{n+1}(\epsilon-a)} $$ Since $f$ is analytic, consider the taylor sum of $f$ centered at $a$, hence $$ = f(z) - \frac{(z-a)^{n+1}}{2\pi i} \sum_{k = 0}^{\infty} \frac{1}{k!} f^{( k)}(a) \int_\alpha \frac{(\epsilon-a)^{k-(n+1)}}{\epsilon-a}d\epsilon $$ considering $g(z) = (z-a)^{k-(n+1)}$, then $g(z)$ is analytic in $\mathbb{C}$, therefore I apply the cauchy integral and I would stay: $$ = f(z) - (z-a)^{n+1}\sum_{k = 0}^{\infty} \frac{1}{k!} f^{(k)}(a)(z-a)^ {k-(n+1)} = f(z)-f(z) = 0 $$
my question: where did i get wrong? or why did I get to 0?