1

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 $(x_j)$ 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 $\lim y_i=\lim x_i$

I don't really know how to approach the problem, thanks in advance

Jeff
  • 1,199
  • The $n$ made all the difference. The third bullet looks fine now. You still to make 1 more edit: $y_i=\sum_{j=1}^i P_{ij}x_j$. – Kim Jong Un Oct 06 '14 at 04:32
  • @KimJongUn thanks, sometimes typing here is tedious to get it all right, mistakes fly under the radar all the time – Jeff Oct 06 '14 at 04:33

1 Answers1

4

Firstly, we may assume that $\lim x_i=0$. We may assume that $|x_i|<M$ holds.

Fix arbitrary $\epsilon>0$, there exists $N$ such that $|x_i|<\epsilon $ for all $i>N$.

Then you see that

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

Note that $\lim_i\sum_{j\leq N} P_{ij}M=0$ since it is a finite sum.

That means there exist $N'$ such that $|y_i|<2\epsilon$ for $i>N'$.

QED

Chen Jiang
  • 1,393