I have a question for my class that asks to find the product of two infinite series, namely $a_n = b_n = \sum\limits_{k=0}^\infty r^n$ for $r\in(0,1)$. That is find $c_n = \sum\limits_{k=0}^n a_k b_{n-k}$. Then it asks to show without appealing to the "product rule" that the product is absolutely convergent. Lastly, find its sum.
My thought process is the following. Let, $a_n = b_n = \sum\limits_{k=0}^\infty r^n$. Then $c_n = \sum\limits_{k=0}^nr^kr^{n-k} = \sum\limits_{k=0}^nr^n.$
Now, if that is the cauchy product, then I figure that if we look at the partial sums of $c_n$, then as $n$ approaches $+\infty$ then $c_n$ approaches zero. Since we're dealing with strictly positive terms then it's also absolutely convergent.
Last thing is to find a sum of the series, so if $c_n = \sum\limits_{k=0}^nr^n$, then I believe that the sum would be like any other geometric sum, and since $a=1$, then $c_n$ converges to $\frac{1}{1-r}$.
Best Answer
You have, correctly, that $$c_n=\sum_{k=0}^n r^n.$$ That's like $\sum_{k=0}^n b$, the sum of $n+1$ copies of $b$, since $r^n$ does not depend on $k$. That means that $$c_n=(n+1)r^n.$$
Now that you have the correct $c_n$, perhaps you can complete things.
Edit: So we want $\sum_0^\infty (n+1)r^n$. Let's change letters to $\sum_0^\infty (n+1)x^n$. By the Ratio Test, or the Root Test, you can prove that we have convergence if $|x|\lt 1$.
There are many ways to show that the sum is $\frac{1}{(1-r)^2}$. It depends on how much marchinery you want to use. For minimal machinery (but not much fun) one can find an explicit formula for $\sum_0^n (k+1)r^k$. For freewheeling use of machinery, note that $\frac{1}{1-x}=1+x+x^2+x^3+\cdots$, and differentiate term by term.