Convergence of $b_n = frac{sum_{i=1}^na_i}{4}$












1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 22:39







Wyatt Kuehster

















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














  • 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










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