How do I prove it formally? If $f(z)=\sum_{k=0}^\infty a_k z^k$ and $g(z)=\sum_{k=0}^\infty b_k z^k$ are two power series whose radii of convergence are $1$. Let the product of $f.g$ has a power series representation $\sum_{k=0}^\infty c_k z^k$ on the open disk. Is the statement true or false?
My attempt:-
I took $a_k=1$ and $b_k=1$ , By Cauchy's product $c_k=k+1$. $\sum_{k=0}^\infty c_k z^k$. So, it is convergence in the unit disk. The statement is true. How do I prove it formally? 
 A: The Cauchy-Hadamard Radius Formula implies $\lim \sup_{k\to \infty}|a_k|^{1/k}\le 1\ge \lim \sup_{k\to \infty}|b_k|^{1/k}.$
For $r>0$ let $S(r)=\{0\}\cup \{k>0: |a_k|^{1/k}\ge 1+r \lor |b_k|^{1/k}\ge 1+r\}.$ Then $S(r)$ is finite. Let $M(r)=\max S(r).$ Let $P(r)=\max \{\max \{|a_k|,|b_k|\}: k\le M(r)\}.$
Let $c_k=\sum_{i=0}^k a_i b_{k-j}.$ 
Given $r>0,$ consider any $k$ large enough that $k>3M(r)$ and $(1+r)^k>P(r).$ If $i+j=k$ then 
......(I). If  $M(r)< i<k-M(r)$  then $j>M(r)$ so $|a_ib_j|<(1+r)^i(1+r)^j=(1+r)^k<(1+r)^{2k}.$
.....(II). If $i\le M(r)$ then $j>M(r)$ so $|a_ib_j|\le P(r)(1+r)^j<(1+r)^k(1+r)^j\le (1+r)^{2k}.$ 
....(III). If $i\ge k-M(r)$ then $j<M(r)<i$ so $|a_ib_j|\le (1+r)^iP(r)<(1+r)^i(1+r)^k\le(1+r)^{2k}.$
So for all sufficiently large $k$ we have $|c_k|^{1/k}\le (\,\sum_{i=0}^k|a_ib_{k-j}|\,)^{1/k}\le$ $ (\,\sum_{i=0}^k(1+r)^{2k}\,)^{1/k}=$ $=(1+r)^2(1+k)^{1/k}.$ 
Since $\lim_{k\to \infty}(1+k)^{1/k}=1,$ we infer that $\lim \sup_{k\to \infty}|c_k|^{1/k}\le (1+r)^2 $ for every $r>0,$ and hence $\lim \sup_{k\to \infty}|c_k|^{1/k}\le 1.$ The C-H Radius Formula then implies that $\sum_kc_kx^k$ converges whenever $|x|<1.$
