When does this sequence converge? I have an infinite sequence in the following form:
$$
\left(\sum_{i=0}^{k}f(i)x^{k-i}\right)_{k\in \Bbb N}
$$
where $0<x<1$.
I am trying to show that the sequence converges, but $f(i)$ does not have a very tractable form. Are there general properties for $f(i)$ that make the sequence convergent?
 A: Let $\displaystyle\quad F_k = \sum_{i=0}^k f(i)x^{k-i},\quad$ then as a sequence, 
$$F_k \text{ converges } \quad\iff\quad f(i) \text{ converges.}\tag{*1}$$
Furthermore, when they converge, one have:
$$\lim_{k\to\infty} F_k = \frac{\lim_{i\to\infty}f(i)}{1-x}\tag{*2}$$
Notice $F_k$ satisfies the recurrence relation:
$$F_{k+1} = \sum_{i=0}^{k+1} f(i) x^{k+1-i} = f(k+1) + x\sum_{k=0}^{k}f(i)x^{k-i} = f(k+1) + x F_{k}$$
The $\implies$ part of $(*1)$ and $(*2)$ is trivial. 
To prove the $\Longleftarrow$ part of $(*1)$, let $L$ be $\displaystyle \lim_{i\to\infty} f(i)$.
For any $\epsilon > 0$, pick a $N$ such that:
$$
|f(i) - L| < \frac{\epsilon}{3} (1-x) \quad\text{ for } i > N
\quad\quad\text{ and }\quad\quad
|L| x^N    < \frac{\epsilon}{3} (1-x)
$$
For any $k > N$, we have the estimate:
$$
\begin{align}
\left| F_k - \frac{L}{1-x}\right|
\le &
\left| F_k - L \sum_{i=0}^k x^{k-i} \right| + \frac{|L| x^{k+1}}{1-x}\\
\le & \left|\sum_{i=0}^k ( f(i) - L ) x^{k-i}\right| + \frac{\epsilon}{3}\\
\le & \frac{\epsilon}{3}\left( 1 + (1 - x) \sum_{i=N+1}^k x^{k-i}\right) + x^{k-N}\left|\sum_{i=0}^N (f(i) - L ) x^{N-i}\right| \\
\le & \frac{2\epsilon}{3} + x^{k-N}\left|\sum_{i=0}^N (f(i) - L ) x^{N-i}\right|
\end{align}$$
Now pick another $N' > N$ such that
$$x^{N'-N} \left|\sum_{i=0}^N (f(i) - L ) x^{N-i}\right| < \frac{\epsilon}{3},$$
It is easy to see for any $k \ge N'$, we will have
$$\left|F_k - \frac{L}{1-x}\right| < \epsilon$$
From this we can conclude $\displaystyle \lim_{k\to\infty} F_k$ exists and equal to $\displaystyle \frac{L}{1-x}$.
