Show $limlimits_{n to infty} x_n=a$
$begingroup$
Given $n,k in mathbb{N}$, $t_{n,k} geq 0$, $sumlimits_{k=1}^n t_{n,k}=1$, $limlimits_{n to infty}t_{n,k}=0$. $limlimits_{n to infty}a_n=a$ and let $x_n := sumlimits_{k=1}^n t_{n,k}a_k$. Show $limlimits_{n to infty} x_n=a$.
I think I can see why intuitively. For large $n$, in $sumlimits_{k=1}^n t_{n,k}a_k$, we are summing big portion of term of the form $epsilon_j (apmepsilon_i)$, so it is approximately $(1-epsilon)a+epsilon*First_few_term_of_a_n approx a$.
Is there clever ways to prove it?
analysis
asked Dec 11 '18 at 21:21
mathpadawanmathpadawan
1,890422
$endgroup$
add a comment |
$begingroup$
Given $n,k in mathbb{N}$, $t_{n,k} geq 0$, $sumlimits_{k=1}^n t_{n,k}=1$, $limlimits_{n to infty}t_{n,k}=0$. $limlimits_{n to infty}a_n=a$ and let $x_n := sumlimits_{k=1}^n t_{n,k}a_k$. Show $limlimits_{n to infty} x_n=a$.
I think I can see why intuitively. For large $n$, in $sumlimits_{k=1}^n t_{n,k}a_k$, we are summing big portion of term of the form $epsilon_j (apmepsilon_i)$, so it is approximately $(1-epsilon)a+epsilon*First_few_term_of_a_n approx a$.
Is there clever ways to prove it?
analysis
asked Dec 11 '18 at 21:21
mathpadawanmathpadawan
1,890422
$endgroup$
add a comment |
$begingroup$
Given $n,k in mathbb{N}$, $t_{n,k} geq 0$, $sumlimits_{k=1}^n t_{n,k}=1$, $limlimits_{n to infty}t_{n,k}=0$. $limlimits_{n to infty}a_n=a$ and let $x_n := sumlimits_{k=1}^n t_{n,k}a_k$. Show $limlimits_{n to infty} x_n=a$.
I think I can see why intuitively. For large $n$, in $sumlimits_{k=1}^n t_{n,k}a_k$, we are summing big portion of term of the form $epsilon_j (apmepsilon_i)$, so it is approximately $(1-epsilon)a+epsilon*First_few_term_of_a_n approx a$.
Is there clever ways to prove it?
analysis
asked Dec 11 '18 at 21:21
mathpadawanmathpadawan
1,890422
$endgroup$
Given $n,k in mathbb{N}$, $t_{n,k} geq 0$, $sumlimits_{k=1}^n t_{n,k}=1$, $limlimits_{n to infty}t_{n,k}=0$. $limlimits_{n to infty}a_n=a$ and let $x_n := sumlimits_{k=1}^n t_{n,k}a_k$. Show $limlimits_{n to infty} x_n=a$.
I think I can see why intuitively. For large $n$, in $sumlimits_{k=1}^n t_{n,k}a_k$, we are summing big portion of term of the form $epsilon_j (apmepsilon_i)$, so it is approximately $(1-epsilon)a+epsilon*First_few_term_of_a_n approx a$.
Is there clever ways to prove it?
analysis
analysis
asked Dec 11 '18 at 21:21
mathpadawanmathpadawan
1,890422
asked Dec 11 '18 at 21:21
mathpadawanmathpadawan
1,890422
edited Dec 11 '18 at 21:45
mathpadawan
asked Dec 11 '18 at 21:21
mathpadawanmathpadawan
1,890422
asked Dec 11 '18 at 21:21
mathpadawanmathpadawan
1,890422
asked Dec 11 '18 at 21:21
mathpadawanmathpadawan
1,890422
1,890422
add a comment |
add a comment |
1 Answer
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$begingroup$
Note that the sequence $x_n$ forgets the 'past' values of $a_k$ at the infinity by observing
$$
lim_{ntoinfty}sum_{k=1}^M t_{n,k}a_k = 0
$$ for all fixed $M>0$. This enables us to assume without loss of generality, $|a_n-a|<epsilon$ for all $n$, giving the desired result. So the formal proof goes like this.
Choose $M>0$ such that
$$
|a_n-a|<epsilon,quad forall n>M.
$$ Then we have
$$
|x_n -a| leq sum_{k=1}^M t_{n,k}|a_k-a| + epsilonsum_{k=M+1}^n t_{n,k}<sum_{k=1}^M t_{n,k}|a_k-a|+ epsilon,quadforall n>M.
$$ Take $nto infty$ to get
$$
limsup_{ntoinfty} |x_n -a|leq epsilon,
$$ for arbitrary $epsilon>0$. Conclude from this that $lim_{ntoinfty} x_n =a$.
answered Dec 11 '18 at 21:54
SongSong
18.5k21651
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1 Answer
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1 Answer
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$begingroup$
Note that the sequence $x_n$ forgets the 'past' values of $a_k$ at the infinity by observing
$$
lim_{ntoinfty}sum_{k=1}^M t_{n,k}a_k = 0
$$ for all fixed $M>0$. This enables us to assume without loss of generality, $|a_n-a|<epsilon$ for all $n$, giving the desired result. So the formal proof goes like this.
Choose $M>0$ such that
$$
|a_n-a|<epsilon,quad forall n>M.
$$ Then we have
$$
|x_n -a| leq sum_{k=1}^M t_{n,k}|a_k-a| + epsilonsum_{k=M+1}^n t_{n,k}<sum_{k=1}^M t_{n,k}|a_k-a|+ epsilon,quadforall n>M.
$$ Take $nto infty$ to get
$$
limsup_{ntoinfty} |x_n -a|leq epsilon,
$$ for arbitrary $epsilon>0$. Conclude from this that $lim_{ntoinfty} x_n =a$.
answered Dec 11 '18 at 21:54
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
Note that the sequence $x_n$ forgets the 'past' values of $a_k$ at the infinity by observing
$$
lim_{ntoinfty}sum_{k=1}^M t_{n,k}a_k = 0
$$ for all fixed $M>0$. This enables us to assume without loss of generality, $|a_n-a|<epsilon$ for all $n$, giving the desired result. So the formal proof goes like this.
Choose $M>0$ such that
$$
|a_n-a|<epsilon,quad forall n>M.
$$ Then we have
$$
|x_n -a| leq sum_{k=1}^M t_{n,k}|a_k-a| + epsilonsum_{k=M+1}^n t_{n,k}<sum_{k=1}^M t_{n,k}|a_k-a|+ epsilon,quadforall n>M.
$$ Take $nto infty$ to get
$$
limsup_{ntoinfty} |x_n -a|leq epsilon,
$$ for arbitrary $epsilon>0$. Conclude from this that $lim_{ntoinfty} x_n =a$.
answered Dec 11 '18 at 21:54
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
Note that the sequence $x_n$ forgets the 'past' values of $a_k$ at the infinity by observing
$$
lim_{ntoinfty}sum_{k=1}^M t_{n,k}a_k = 0
$$ for all fixed $M>0$. This enables us to assume without loss of generality, $|a_n-a|<epsilon$ for all $n$, giving the desired result. So the formal proof goes like this.
Choose $M>0$ such that
$$
|a_n-a|<epsilon,quad forall n>M.
$$ Then we have
$$
|x_n -a| leq sum_{k=1}^M t_{n,k}|a_k-a| + epsilonsum_{k=M+1}^n t_{n,k}<sum_{k=1}^M t_{n,k}|a_k-a|+ epsilon,quadforall n>M.
$$ Take $nto infty$ to get
$$
limsup_{ntoinfty} |x_n -a|leq epsilon,
$$ for arbitrary $epsilon>0$. Conclude from this that $lim_{ntoinfty} x_n =a$.
answered Dec 11 '18 at 21:54
SongSong
18.5k21651
$endgroup$
Note that the sequence $x_n$ forgets the 'past' values of $a_k$ at the infinity by observing
$$
lim_{ntoinfty}sum_{k=1}^M t_{n,k}a_k = 0
$$ for all fixed $M>0$. This enables us to assume without loss of generality, $|a_n-a|<epsilon$ for all $n$, giving the desired result. So the formal proof goes like this.
Choose $M>0$ such that
$$
|a_n-a|<epsilon,quad forall n>M.
$$ Then we have
$$
|x_n -a| leq sum_{k=1}^M t_{n,k}|a_k-a| + epsilonsum_{k=M+1}^n t_{n,k}<sum_{k=1}^M t_{n,k}|a_k-a|+ epsilon,quadforall n>M.
$$ Take $nto infty$ to get
$$
limsup_{ntoinfty} |x_n -a|leq epsilon,
$$ for arbitrary $epsilon>0$. Conclude from this that $lim_{ntoinfty} x_n =a$.
answered Dec 11 '18 at 21:54
SongSong
18.5k21651
answered Dec 11 '18 at 21:54
SongSong
18.5k21651
answered Dec 11 '18 at 21:54
SongSong
18.5k21651
answered Dec 11 '18 at 21:54
SongSong
18.5k21651
18.5k21651
add a comment |
add a comment |
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