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Cesaro limits

  • #1
1,456
44

Homework Statement


For each ##n\in\mathbb{N}##, let the finite sequence ##\{b_{n,m}\}_{m=1}^n\subset(0,\infty)## be given. Assume, for each ##n\in\mathbb{N}##, that ##b_{n,1}+b_{n,2}+\cdots+b_{n,n}=1##.

Show that ##\lim_{n\to\infty}( b_{n,1}\cdot a_1+b_{n,2}\cdot a_2+\cdots+b_{n,n}\cdot a_n) = \lim_{n\to\infty}a_n##, for every convergent sequence ##\{a_n\}_{n=1}^\infty\subset\mathbb{R}## if and only if, for each ##m\in\mathbb{N}##, ##\lim_{n\to\infty}b_{n,m}=0##.

Homework Equations




The Attempt at a Solution


I really need at least one hint for this one. Which direction of the proof should I start with? Which one is easier?

My idea for the <---- direction is that since for each ##m\in\mathbb{N}##, ##\lim_{n\to\infty}b_{n,m}=0##, we see that ## \lim_{n\to\infty}b_{n,1}\cdot a_1+\lim_{n\to\infty}b_{n,2}\cdot a_2+\cdots+\lim_{n\to\infty}b_{n,n}\cdot a_n) = 0 + 0 + \cdots + 0##... But I'm not sure where this gets me.
 

Answers and Replies

  • #2
1,456
44

Homework Statement


For each ##n\in\mathbb{N}##, let the finite sequence ##\{b_{n,m}\}_{m=1}^n\subset(0,\infty)## be given. Assume, for each ##n\in\mathbb{N}##, that ##b_{n,1}+b_{n,2}+\cdots+b_{n,n}=1##.

Show that ##\lim_{n\to\infty}( b_{n,1}\cdot a_1+b_{n,2}\cdot a_2+\cdots+b_{n,n}\cdot a_n) = \lim_{n\to\infty}a_n##, for every convergent sequence ##\{a_n\}_{n=1}^\infty\subset\mathbb{R}## if and only if, for each ##m\in\mathbb{N}##, ##\lim_{n\to\infty}b_{n,m}=0##.

Homework Equations




The Attempt at a Solution


I really need at least one hint for this one. Which direction of the proof should I start with? Which one is easier?

My idea for the <---- direction is that since for each ##m\in\mathbb{N}##, ##\lim_{n\to\infty}b_{n,m}=0##, we see that ## \lim_{n\to\infty}b_{n,1}\cdot a_1+\lim_{n\to\infty}b_{n,2}\cdot a_2+\cdots+\lim_{n\to\infty}b_{n,n}\cdot a_n) = 0 + 0 + \cdots + 0##... But I'm not sure where this gets me.
Here is my stab at a solution for the <--- direction.

Let ##(a_n)## be an arbitrary convergent sequence that converges to ##c##. We want to show that ##b_{n,1}a_n + \cdots b_{n,n}a_n## also converges to ##c##. To this end, given ##\epsilon > 0##, there exists ##N \in \mathbb{N}## such that if ##n \ge N## we have ##|a_n - c| \le \epsilon##. ##|b_{n,1} a_1 + \cdots b_{n,n}a_n - c| = |b_{n,1} a_1 + \cdots + b_{n,n}a_n - c(b_{n,1}+ \cdots + b_{n,n})| = |b_{n,1}(a_1 - c) + \cdots b_{n,n}(a_n - c)|##... This is kind of where I get stuck. I'm not sure what to do next.
 
  • #3
34,043
9,891
My idea for the <---- direction is that since for each ##m\in\mathbb{N}##, ##\lim_{n\to\infty}b_{n,m}=0##, we see that ## \lim_{n\to\infty}b_{n,1}\cdot a_1+\lim_{n\to\infty}b_{n,2}\cdot a_2+\cdots+\lim_{n\to\infty}b_{n,n}\cdot a_n) = 0 + 0 + \cdots + 0##... But I'm not sure where this gets me.
You cannot split a limit like that.

The other direction is easier if you make a proof by contradiction. Assume there is an m such that ##\lim_{n\to\infty}b_{n,m} \neq 0## and show that the original statement is no longer true. That should also give an idea how the overall structure works.
 

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