Convergence of $b_n = frac{sum_{i=1}^na_i}{4}$
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Question:
Let ${a_n}_{nin mathbb N}$ be a sequence of real numbers, and for each $nin mathbb N$ define
$$b_n= frac{sum_{i=1}^na_i}{4}.$$
Prove that if ${a_n}$ converges to $A$, then so does ${b_n}$.
Attempt:
Since ${a_n}$ converges to $A$, for each $epsilon>0$, there exists $N>0$ such that whenever $ngeq N$, $A-epsilon<a_n<A+epsilon$. We can write $b_n$ as
$$ b_n=frac14left(sum_{i=1}^Na_i + sum_{i=N+1}^na_iright),$$
and by the convergence of ${a_n}$, we can write
$$(n-N)(A-epsilon)<sum_{i=N+1}^na_i<(n-N)(A+epsilon).$$ If we write $C=sum_{i=1}^Na_i$, then
$$frac{C+(n-N)(A-epsilon)}{4}<b_n<frac{C+(n-N)(A+epsilon)}{4}.$$
We did an example where $b_n$ was the arithmetic mean and it worked because the denominator was $n$ and we could see that in the large $n$ limit, $|b_n-A|<epsilon$. But here, I'm not sure how to take it home, because $(n-N)$ can grow without bound. Am I even on the right path?
Edit: The proposition is false :(
real-analysis sequences-and-series convergence
asked Dec 9 '18 at 19:55
Wyatt KuehsterWyatt Kuehster
637
$endgroup$
|
show 4 more comments
$begingroup$
Question:
Let ${a_n}_{nin mathbb N}$ be a sequence of real numbers, and for each $nin mathbb N$ define
$$b_n= frac{sum_{i=1}^na_i}{4}.$$
Prove that if ${a_n}$ converges to $A$, then so does ${b_n}$.
Attempt:
Since ${a_n}$ converges to $A$, for each $epsilon>0$, there exists $N>0$ such that whenever $ngeq N$, $A-epsilon<a_n<A+epsilon$. We can write $b_n$ as
$$ b_n=frac14left(sum_{i=1}^Na_i + sum_{i=N+1}^na_iright),$$
and by the convergence of ${a_n}$, we can write
$$(n-N)(A-epsilon)<sum_{i=N+1}^na_i<(n-N)(A+epsilon).$$ If we write $C=sum_{i=1}^Na_i$, then
$$frac{C+(n-N)(A-epsilon)}{4}<b_n<frac{C+(n-N)(A+epsilon)}{4}.$$
We did an example where $b_n$ was the arithmetic mean and it worked because the denominator was $n$ and we could see that in the large $n$ limit, $|b_n-A|<epsilon$. But here, I'm not sure how to take it home, because $(n-N)$ can grow without bound. Am I even on the right path?
Edit: The proposition is false :(
real-analysis sequences-and-series convergence
asked Dec 9 '18 at 19:55
Wyatt KuehsterWyatt Kuehster
637
$endgroup$
2
$begingroup$
Take ${a_n}$ to be the constant sequence $A$. $b_n = cfrac{nA}{4} to +infty$
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 19:58
1
$begingroup$
Perhaps there is a typo, and this math.stackexchange.com/q/155839/42969 is what you really mean?
$endgroup$
– Martin R
Dec 9 '18 at 20:00
$begingroup$
@MartinR which is solved right away with Stolz-Ces$mathrm{grave{a}}$ro !!!!.
$endgroup$
– Felix Marin
Dec 9 '18 at 21:03
1
$begingroup$
@WyattKuehster: Then your statement is obviously wrong, as pointed out above. More generally, if $a_n to A > 0$ then $b_n to infty$.
$endgroup$
– Martin R
Dec 9 '18 at 22:26
1
$begingroup$
Or take any non-negative sequence such that $sum a_n$ diverges, such as $a_n = 1/n$.
$endgroup$
– Martin R
Dec 9 '18 at 22:34
|
show 4 more comments
$begingroup$
Question:
Let ${a_n}_{nin mathbb N}$ be a sequence of real numbers, and for each $nin mathbb N$ define
$$b_n= frac{sum_{i=1}^na_i}{4}.$$
Prove that if ${a_n}$ converges to $A$, then so does ${b_n}$.
Attempt:
Since ${a_n}$ converges to $A$, for each $epsilon>0$, there exists $N>0$ such that whenever $ngeq N$, $A-epsilon<a_n<A+epsilon$. We can write $b_n$ as
$$ b_n=frac14left(sum_{i=1}^Na_i + sum_{i=N+1}^na_iright),$$
and by the convergence of ${a_n}$, we can write
$$(n-N)(A-epsilon)<sum_{i=N+1}^na_i<(n-N)(A+epsilon).$$ If we write $C=sum_{i=1}^Na_i$, then
$$frac{C+(n-N)(A-epsilon)}{4}<b_n<frac{C+(n-N)(A+epsilon)}{4}.$$
We did an example where $b_n$ was the arithmetic mean and it worked because the denominator was $n$ and we could see that in the large $n$ limit, $|b_n-A|<epsilon$. But here, I'm not sure how to take it home, because $(n-N)$ can grow without bound. Am I even on the right path?
Edit: The proposition is false :(
real-analysis sequences-and-series convergence
asked Dec 9 '18 at 19:55
Wyatt KuehsterWyatt Kuehster
637
$endgroup$
Question:
Let ${a_n}_{nin mathbb N}$ be a sequence of real numbers, and for each $nin mathbb N$ define
$$b_n= frac{sum_{i=1}^na_i}{4}.$$
Prove that if ${a_n}$ converges to $A$, then so does ${b_n}$.
Attempt:
Since ${a_n}$ converges to $A$, for each $epsilon>0$, there exists $N>0$ such that whenever $ngeq N$, $A-epsilon<a_n<A+epsilon$. We can write $b_n$ as
$$ b_n=frac14left(sum_{i=1}^Na_i + sum_{i=N+1}^na_iright),$$
and by the convergence of ${a_n}$, we can write
$$(n-N)(A-epsilon)<sum_{i=N+1}^na_i<(n-N)(A+epsilon).$$ If we write $C=sum_{i=1}^Na_i$, then
$$frac{C+(n-N)(A-epsilon)}{4}<b_n<frac{C+(n-N)(A+epsilon)}{4}.$$
We did an example where $b_n$ was the arithmetic mean and it worked because the denominator was $n$ and we could see that in the large $n$ limit, $|b_n-A|<epsilon$. But here, I'm not sure how to take it home, because $(n-N)$ can grow without bound. Am I even on the right path?
Edit: The proposition is false :(
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
asked Dec 9 '18 at 19:55
Wyatt KuehsterWyatt Kuehster
637
asked Dec 9 '18 at 19:55
Wyatt KuehsterWyatt Kuehster
637
edited Dec 9 '18 at 22:39
Wyatt Kuehster
asked Dec 9 '18 at 19:55
Wyatt KuehsterWyatt Kuehster
637
asked Dec 9 '18 at 19:55
Wyatt KuehsterWyatt Kuehster
637
asked Dec 9 '18 at 19:55
Wyatt KuehsterWyatt Kuehster
637
637
2
$begingroup$
Take ${a_n}$ to be the constant sequence $A$. $b_n = cfrac{nA}{4} to +infty$
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 19:58
1
$begingroup$
Perhaps there is a typo, and this math.stackexchange.com/q/155839/42969 is what you really mean?
$endgroup$
– Martin R
Dec 9 '18 at 20:00
$begingroup$
@MartinR which is solved right away with Stolz-Ces$mathrm{grave{a}}$ro !!!!.
$endgroup$
– Felix Marin
Dec 9 '18 at 21:03
1
$begingroup$
@WyattKuehster: Then your statement is obviously wrong, as pointed out above. More generally, if $a_n to A > 0$ then $b_n to infty$.
$endgroup$
– Martin R
Dec 9 '18 at 22:26
1
$begingroup$
Or take any non-negative sequence such that $sum a_n$ diverges, such as $a_n = 1/n$.
$endgroup$
– Martin R
Dec 9 '18 at 22:34
|
show 4 more comments
2
$begingroup$
Take ${a_n}$ to be the constant sequence $A$. $b_n = cfrac{nA}{4} to +infty$
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 19:58
1
$begingroup$
Perhaps there is a typo, and this math.stackexchange.com/q/155839/42969 is what you really mean?
$endgroup$
– Martin R
Dec 9 '18 at 20:00
$begingroup$
@MartinR which is solved right away with Stolz-Ces$mathrm{grave{a}}$ro !!!!.
$endgroup$
– Felix Marin
Dec 9 '18 at 21:03
1
$begingroup$
@WyattKuehster: Then your statement is obviously wrong, as pointed out above. More generally, if $a_n to A > 0$ then $b_n to infty$.
$endgroup$
– Martin R
Dec 9 '18 at 22:26
1
$begingroup$
Or take any non-negative sequence such that $sum a_n$ diverges, such as $a_n = 1/n$.
$endgroup$
– Martin R
Dec 9 '18 at 22:34
2
2
$begingroup$
Take ${a_n}$ to be the constant sequence $A$. $b_n = cfrac{nA}{4} to +infty$
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 19:58
$begingroup$
Take ${a_n}$ to be the constant sequence $A$. $b_n = cfrac{nA}{4} to +infty$
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 19:58
1
1
$begingroup$
Perhaps there is a typo, and this math.stackexchange.com/q/155839/42969 is what you really mean?
$endgroup$
– Martin R
Dec 9 '18 at 20:00
$begingroup$
Perhaps there is a typo, and this math.stackexchange.com/q/155839/42969 is what you really mean?
$endgroup$
– Martin R
Dec 9 '18 at 20:00
$begingroup$
@MartinR which is solved right away with Stolz-Ces$mathrm{grave{a}}$ro !!!!.
$endgroup$
– Felix Marin
Dec 9 '18 at 21:03
$begingroup$
@MartinR which is solved right away with Stolz-Ces$mathrm{grave{a}}$ro !!!!.
$endgroup$
– Felix Marin
Dec 9 '18 at 21:03
1
1
$begingroup$
@WyattKuehster: Then your statement is obviously wrong, as pointed out above. More generally, if $a_n to A > 0$ then $b_n to infty$.
$endgroup$
– Martin R
Dec 9 '18 at 22:26
$begingroup$
@WyattKuehster: Then your statement is obviously wrong, as pointed out above. More generally, if $a_n to A > 0$ then $b_n to infty$.
$endgroup$
– Martin R
Dec 9 '18 at 22:26
1
1
$begingroup$
Or take any non-negative sequence such that $sum a_n$ diverges, such as $a_n = 1/n$.
$endgroup$
– Martin R
Dec 9 '18 at 22:34
$begingroup$
Or take any non-negative sequence such that $sum a_n$ diverges, such as $a_n = 1/n$.
$endgroup$
– Martin R
Dec 9 '18 at 22:34
|
show 4 more comments
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2
$begingroup$
Take ${a_n}$ to be the constant sequence $A$. $b_n = cfrac{nA}{4} to +infty$
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 19:58
1
$begingroup$
Perhaps there is a typo, and this math.stackexchange.com/q/155839/42969 is what you really mean?
$endgroup$
– Martin R
Dec 9 '18 at 20:00
$begingroup$
@MartinR which is solved right away with Stolz-Ces$mathrm{grave{a}}$ro !!!!.
$endgroup$
– Felix Marin
Dec 9 '18 at 21:03
1
$begingroup$
@WyattKuehster: Then your statement is obviously wrong, as pointed out above. More generally, if $a_n to A > 0$ then $b_n to infty$.
$endgroup$
– Martin R
Dec 9 '18 at 22:26
1
$begingroup$
Or take any non-negative sequence such that $sum a_n$ diverges, such as $a_n = 1/n$.
$endgroup$
– Martin R
Dec 9 '18 at 22:34