If $f$ is the sum of a convergent power series on a disk $D(0;R)$ prove that the integral of $f$ over any closed path $\gamma$ in $D(0;R)$ is zero If $f$ is the sum of a convergent power series on a disk $D(0;R)$ prove that the integral of $f$ over any closed path $\gamma$ in $D(0;R)$ is zero.

How can I able to prove the above problem without using any form of cauchys theorem
 A: I guess what you want is Method $3$ below. I had not seen your Cauchy-free request. Note that I assume the closed path $\gamma$ to be piecewise $C^1$ for this method $3$ to work.
Method 1: By the Residue Theorem, such an integral is essentially adding, modulo $2i\pi$, the residues of $f$ at its singularities. If there are no singularities or only removable singularities, this makes $0$.
Method 2: Now let us check this with Cauchy's integral formula. Consider the holomorphic function $g(z)=zf(z)$ on $D$. Let $\gamma $ be a closed rectifiable curve in $D$ with winding number $1$ about $0$. Then
$$
g(0)=\frac{1}{2i\pi}\int_\gamma\frac{g(z)}{z}dz\qquad\Rightarrow\qquad0=\int_\gamma f(z)dz.
$$
Method 3: Finally, let us do it with the power series representation of $f$, $f(z)=\sum_{n\geq 0}a_nz^n$ whose radius of convergence is not smaller than $R$ by assumption. Let $\gamma:[0,1]\longrightarrow D$ be a closed and piecewise $C^1$ path. By compactness and continuity, there exists $r<R$ such that $|\gamma(t)|\leq r$ on $[0,1]$. Then the series defining $f$ converges absolutely at $r$. Also, we denote $M=\sup_{[0,1]}|\gamma'|<\infty$. Then
$$
\sum_{n\geq 0}\sup_{t\in[0,1]}|a_n||\gamma(t)|^n|\gamma'(t)|\leq M\sum_{n\geq 0}|a_n|r^n<\infty.
$$
So the series defining $f(\gamma(t))\gamma'(t)$ converges normally on $[0,1]$, or uniformly by Weierstrass M-test, if you prefer. Hence we can swap sum and integral:
$$
\int_0^1f(\gamma(t))\gamma'(t)dt=\sum_{n\geq 0}a_n\int_0^1\gamma(t)^n\gamma'(t)dt.
$$
Now for every $n\geq 0$, we have
$$
\int_0^1\gamma(t)^n\gamma'(t)dt=\left.\frac{\gamma(t)^{n+1}}{n+1}\right|_0^1=\frac{\gamma(1)^{n+1}-\gamma(0)^{n+1}}{n+1}=0.
$$
Therefore
$$\int_\gamma f(z)dz=\int_0^1f(\gamma(t))\gamma'(t)dt=\sum_{n\geq 0}a_n\cdot0=0.
$$
A: First, let us assume your closed path $\gamma : [0,1] \to D(0,R)$ is $C^1$.
Since $[0,1]$ is compact and the function $|\gamma(t)|$ is continuous on $[0,1]$, 
$|\gamma(t)|$ achieves maximum at some $t_0 \in [0,1]$. This implies the whole $\gamma([0,1])$
is contained in the closed ball $\bar{B}(0,R')$ where $R' = |\gamma(t_0)| < R$.
Let $f = \sum_{k=0}^{\infty} a_k z^k$ be your converging power series. $f$ converges on $D(0,R)$ implies it converges uniformly over any smaller closed disk, in particular on $D(0,R')$. You can then integrate the power series term by term:
$$\begin{align}\oint_\gamma f(z) dz &= \oint_\gamma \sum_{k=0}^{\infty} a_k z^k dz= \sum_{k=0}^{\infty} a_k \oint_{\gamma} z^k dz = \sum_{k=0}^{\infty}a_k\int_0^1 \gamma^k(t) \gamma'(t) dt \\&= \sum_{k=0}^{\infty} \frac{a_k}{k+1}\left[\gamma^{k+1}(t)\right]_0^1 = \sum_{k=0}^{\infty}\frac{a_k}{k+1} 0 = 0\end{align}$$
Same arguments work when $\gamma$ is only piecewise $C^1$ or only just rectifiable (i.e. path with finite length). For the most general case where $\gamma$ is only assumed to be $C^0$. We can still define contour integral and it is still true that:
$$\oint_{\gamma} f(z) dz = 0\tag{*}$$
However, I believe when one generalize contour integral from $C^1$ to $C^0$ path, 
some form of Cauchy's theorem has been used implicitly in the definition. If one completely avoid Cauchy's theorem, I have no idea how to prove $(*)$ from first principle.
