real analysis - Proving Two Limits Are Equal to Each Other

Let $P_{ij}$, where $P_{ij}$ is a real number and $i, j$ natural number, allow the three statements to hold:

• $P_{ij}\ge 0$ for all $i, j$




• $\lim_{i\to\infty}P_{ij}=0$ for all $j$




• $\sum_{j=1}^i P_{ij}$ =1 for all $i$

Let $(x_i)$ be a convergent sequence and let a sequence $(y_i)$ be defined by

$$y_i=\sum_{j=1}^i P_{ij}x_j$$

Prove $(y_i)$ is a convergent sequence, and prove $\lim y_i=\lim x_i$

Ok so here's what I have:

Fix arbitrary $\epsilon>0$, there exists $N$ such that $|x_i-x|<\epsilon $ for all $i>N$, so $\lim x_n = x$, which exists since ($x_n$) is a convergent sequence. We may assume that $|x_i-x|

Then
$y_i=\sum_{j=1}^i P_{ij}x_j $, so by triangle inequality and since $\sum_{j=1}^i P_{ij} =1$ for all $i$, the following occurs:

\begin{align*}
|y_i-x|\leq {}&\sum_{j=1}^i P_{ij}|x_j- x|\\
={}&\sum_{j\leq N} P_{ij}|x_j- x| +\sum_{j> N} P_{ij}|x_j- x|\\
<{}&\sum_{j\leq N} P_{ij}M +\sum_{j> N} P_{ij}\epsilon\\
\leq{}&\sum_{j\leq N} P_{ij}M +\epsilon
\end{align*}

Now I don't know what to do with the sum that remains, if I factor out the $M$ and make the sum that remains equal to $1$, then I don't really get anywhere. Thanks in advance.