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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 $|x_i-x|<M$ holds always, for a constant M.

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.

Community
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Jeff
  • 1,199

1 Answers1

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The sum $\sum_{j\leq N} P_{ij}$ has a fixed number of terms. Each of them tends to zero as $i$ tends to infinity. Thus the inequality $\sum_{j\leq N} P_{ij} M \leq \epsilon$ holds for every sufficiently large $i$.

mandourin
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  • so would I be allowed to say that that sum is at most $\epsilon$, so then $|y_n-x|\le 2\epsilon$, completing the proof? And is the rest good? – Jeff Oct 06 '14 at 17:05
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    It looks good to me.

    On a homework paper, I would end with a recapitulation of what has been proven with all due quantification to match the definition of a convergent sequence.

     – mandourin Oct 06 '14 at 17:10