$(x_n)$ real, $(x_n) \to 0$ and $0So, I'm supposed to show that:
Given $(x_n)$ a sequence of real numbers, with $(x_n) \to 0$, and given $0<c<1$, then $(y_n) \to 0$, where $y_n = c^n x_0 + c^{n-1} x_1 + ... + c^0 x_n$.
Here's my attempt, and I'd appreciate any corrections:
Since $|y_n| < c^n |x_0| + c^{n-1} |x_1| + ... + c^0 |x_n|$, I'll assume the $x_n$ are all positive and try to show that $y_n$ still goes to zero.
So suppose that, for a given $d > 0$, there exist arbitrarily large $n$, such that $y_n > d$. Then, since $\forall n$, 
$$y_{n+1} = c (y_n) + x_{n+1} \implies y_{n+1} - y_n = (c-1) y_n + x_{n+1}$$, 
we would have 
$$y_{n+1} - y_n = (c-1) y_n + x_{n+1} < (c-1) d + x_{n+1}$$,
 and since $\exists N_0$ such that $n > N_0 \implies x_n < \frac{(1-c)}{2}d$, we would have, for $n > N_0$, 
$$y_{n+1} - y_n < (c-1) d + x_{n+1} < (c-1) d + \frac{(1-c)}{2}d = \frac{(c-1)}{2}d < 0$$.
Assuming $y_{n+1} > d$ and continuing in this fashion, we would eventually get an $N$ such that $y_N \leq d$. Then, for the subsequent term, 
$$y_{N+1} = c \cdot y_{N} + x_{N+1} \leq c \cdot d + \frac{(1-c)}{2}d = \frac{(c+1)}{2}d$$ . 
If $y_{N+1} > d$, then
$$y_{N+2} - y_{N+1} < \frac{(c-1)}{2}d \implies y_{N+2} < y_{N+1} + \frac{(c-1)}{2}d < \frac{(c+1)}{2}d + \frac{(c-1)}{2}d = c \cdot d < d$$.
Therefore, for sufficiently large $N$, $n > N \implies y_n < 2 d$.
Since $d$ was arbitrary, $(y_n) \to 0$.

Let me suggest a totally different approach.
Suppose that
$$
f(z)=\sum_{n=0}^\infty x_n z^n, \quad g(z)=\sum_{n=0}
^\infty c^nz^n, \quad\text{and then}\quad h(z)=f(z)g(z)=\sum_{n=0}^\infty y_nz^n,
$$
where $\,y_n=x_n+cx_{n-1}+\cdots+ c^n x_0$.
Then, the radius of converge of $f$ and $g$ is at least 1, and hence at least 1 is the radius of convergence of $h$. Thus $y_n\to 0$.
