Prove the inequality and limitation












5












$begingroup$



Suppose ${a_k}$ ,${b_k}$ and ${xi_k}$ are non negative, and for all $kge 0$ , we have $$a_{k+1}^2le(a_k+b_k)^2-xi_k^2$$
Prove

1.$$sum_{i=1}^{k}xi_i^2le(a_1+sum_{i=0}^kb_i)^2$$

2. If ${b_k}$ also satisfy $sum_{k=0}^{infty}b_k^2lt+infty $, then $$lim_{ktoinfty}frac{1}{k}sum_{i=1}^kxi_i^2=0$$




After doing this transformation:$$xi_k^2le(a_k+b_k)^2-a_{k+1}^2$$
$Rightarrow$ $$sum_{i=1}^{k}xi_k^2le(a_1+b_1-a_2)(a_1+b_1+a_2)+(a_2+b_2-a_3)(a_2+b_2+a_3)+cdots(a_k+b_k-a_{k+1})(a_k+b_k+a_{k+1})$$
then I can't move forward any more.






calculus inequality







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$endgroup$












  • $begingroup$
    Is $a_{k}$ decreasing?
    $endgroup$
    – Alex
    Sep 3 '18 at 5:24










  • $begingroup$
    @Alex The condition does not include this.
    $endgroup$
    – Jaqen Chou
    Sep 3 '18 at 5:36










  • $begingroup$
    Could you tell me something about the origin of the problem? I am curious about it!
    $endgroup$
    – Fyhswdsxjj
    Dec 5 '18 at 23:32










  • $begingroup$
    @gcfsjfcus It has been a long time. I just remember it's from some college's entrance examination.
    $endgroup$
    – Jaqen Chou
    Dec 6 '18 at 8:13
















5












$begingroup$



Suppose ${a_k}$ ,${b_k}$ and ${xi_k}$ are non negative, and for all $kge 0$ , we have $$a_{k+1}^2le(a_k+b_k)^2-xi_k^2$$
Prove

1.$$sum_{i=1}^{k}xi_i^2le(a_1+sum_{i=0}^kb_i)^2$$

2. If ${b_k}$ also satisfy $sum_{k=0}^{infty}b_k^2lt+infty $, then $$lim_{ktoinfty}frac{1}{k}sum_{i=1}^kxi_i^2=0$$




After doing this transformation:$$xi_k^2le(a_k+b_k)^2-a_{k+1}^2$$
$Rightarrow$ $$sum_{i=1}^{k}xi_k^2le(a_1+b_1-a_2)(a_1+b_1+a_2)+(a_2+b_2-a_3)(a_2+b_2+a_3)+cdots(a_k+b_k-a_{k+1})(a_k+b_k+a_{k+1})$$
then I can't move forward any more.






calculus inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is $a_{k}$ decreasing?
    $endgroup$
    – Alex
    Sep 3 '18 at 5:24










  • $begingroup$
    @Alex The condition does not include this.
    $endgroup$
    – Jaqen Chou
    Sep 3 '18 at 5:36










  • $begingroup$
    Could you tell me something about the origin of the problem? I am curious about it!
    $endgroup$
    – Fyhswdsxjj
    Dec 5 '18 at 23:32










  • $begingroup$
    @gcfsjfcus It has been a long time. I just remember it's from some college's entrance examination.
    $endgroup$
    – Jaqen Chou
    Dec 6 '18 at 8:13














5












5








5


3



$begingroup$



Suppose ${a_k}$ ,${b_k}$ and ${xi_k}$ are non negative, and for all $kge 0$ , we have $$a_{k+1}^2le(a_k+b_k)^2-xi_k^2$$
Prove

1.$$sum_{i=1}^{k}xi_i^2le(a_1+sum_{i=0}^kb_i)^2$$

2. If ${b_k}$ also satisfy $sum_{k=0}^{infty}b_k^2lt+infty $, then $$lim_{ktoinfty}frac{1}{k}sum_{i=1}^kxi_i^2=0$$




After doing this transformation:$$xi_k^2le(a_k+b_k)^2-a_{k+1}^2$$
$Rightarrow$ $$sum_{i=1}^{k}xi_k^2le(a_1+b_1-a_2)(a_1+b_1+a_2)+(a_2+b_2-a_3)(a_2+b_2+a_3)+cdots(a_k+b_k-a_{k+1})(a_k+b_k+a_{k+1})$$
then I can't move forward any more.






calculus inequality







share|cite|improve this question











$endgroup$





Suppose ${a_k}$ ,${b_k}$ and ${xi_k}$ are non negative, and for all $kge 0$ , we have $$a_{k+1}^2le(a_k+b_k)^2-xi_k^2$$
Prove

1.$$sum_{i=1}^{k}xi_i^2le(a_1+sum_{i=0}^kb_i)^2$$

2. If ${b_k}$ also satisfy $sum_{k=0}^{infty}b_k^2lt+infty $, then $$lim_{ktoinfty}frac{1}{k}sum_{i=1}^kxi_i^2=0$$




After doing this transformation:$$xi_k^2le(a_k+b_k)^2-a_{k+1}^2$$
$Rightarrow$ $$sum_{i=1}^{k}xi_k^2le(a_1+b_1-a_2)(a_1+b_1+a_2)+(a_2+b_2-a_3)(a_2+b_2+a_3)+cdots(a_k+b_k-a_{k+1})(a_k+b_k+a_{k+1})$$
then I can't move forward any more.






calculus inequality




calculus inequality






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 1 '18 at 4:20







Jaqen Chou

















asked Aug 31 '18 at 18:18









Jaqen ChouJaqen Chou

422110




422110












  • $begingroup$
    Is $a_{k}$ decreasing?
    $endgroup$
    – Alex
    Sep 3 '18 at 5:24










  • $begingroup$
    @Alex The condition does not include this.
    $endgroup$
    – Jaqen Chou
    Sep 3 '18 at 5:36










  • $begingroup$
    Could you tell me something about the origin of the problem? I am curious about it!
    $endgroup$
    – Fyhswdsxjj
    Dec 5 '18 at 23:32










  • $begingroup$
    @gcfsjfcus It has been a long time. I just remember it's from some college's entrance examination.
    $endgroup$
    – Jaqen Chou
    Dec 6 '18 at 8:13


















  • $begingroup$
    Is $a_{k}$ decreasing?
    $endgroup$
    – Alex
    Sep 3 '18 at 5:24










  • $begingroup$
    @Alex The condition does not include this.
    $endgroup$
    – Jaqen Chou
    Sep 3 '18 at 5:36










  • $begingroup$
    Could you tell me something about the origin of the problem? I am curious about it!
    $endgroup$
    – Fyhswdsxjj
    Dec 5 '18 at 23:32










  • $begingroup$
    @gcfsjfcus It has been a long time. I just remember it's from some college's entrance examination.
    $endgroup$
    – Jaqen Chou
    Dec 6 '18 at 8:13
















$begingroup$
Is $a_{k}$ decreasing?
$endgroup$
– Alex
Sep 3 '18 at 5:24




$begingroup$
Is $a_{k}$ decreasing?
$endgroup$
– Alex
Sep 3 '18 at 5:24












$begingroup$
@Alex The condition does not include this.
$endgroup$
– Jaqen Chou
Sep 3 '18 at 5:36




$begingroup$
@Alex The condition does not include this.
$endgroup$
– Jaqen Chou
Sep 3 '18 at 5:36












$begingroup$
Could you tell me something about the origin of the problem? I am curious about it!
$endgroup$
– Fyhswdsxjj
Dec 5 '18 at 23:32




$begingroup$
Could you tell me something about the origin of the problem? I am curious about it!
$endgroup$
– Fyhswdsxjj
Dec 5 '18 at 23:32












$begingroup$
@gcfsjfcus It has been a long time. I just remember it's from some college's entrance examination.
$endgroup$
– Jaqen Chou
Dec 6 '18 at 8:13




$begingroup$
@gcfsjfcus It has been a long time. I just remember it's from some college's entrance examination.
$endgroup$
– Jaqen Chou
Dec 6 '18 at 8:13










1 Answer
1






active

oldest

votes


















3





+50







$begingroup$

Let us first prove by induction that
$$tag{1} a_1 +sum_{i=0}^k b_i ge a_{k+1}.$$
For $k=0$ this inequality is trivial. Next assume that (1) is already shown for $0 le i le k$, then we have
$$a_1 + sum_{i=0}^{k+1} b_i ge a_{k+1} + b_{k+1} ge a_{k+1} + big(sqrt{a_{k+2}^2 + xi_{k+1}^2} - a_{k+1} big) ge a_{k+2}.$$
Now we are ready to show the stronger inequality
$$tag{2}sum_{i=1}^k xi_i^2 leq big(a_1 + sum_{i=0}^k b_i big)^2 - a_{k+1}^2.$$
Again, we prove this by induction. In fact, $k=1$ is a direct consequence of the initial condition:
$$a_2^2 le (a_1+b_1)^2-xi_1^2 le (a_1+b_0+b_1)^2 - xi_1^2.$$
Assume that (2) is true for $k$, then
begin{align}
big(a_1 + sum_{i=0}^{k+1} b_i big)^2-a_{k+2}^2 &ge sum_{i=1}^k xi_i^2 + 2 big( a_1 +sum_{i=0}^k b_k big) b_{k+1} + b_{k+1}^2 + (a_{k+1}^2 -a_{k+2}^2) \
&ge sum_{i=1}^k xi_i^2 + (2 a_{k+1} b_{k+1} +b_{k+1}^2) + (a_{k+1}^2 -a_{k+2}^2) \
&ge sum_{i=1}^k xi_i^2 + (a_{k+2}^2 +xi_{k+1}^2 - a_{k+1}^2) + (a_{k+1}^2 -a_{k+2}^2) \
&ge sum_{i=1}^{k+1} xi_i^2
end{align}
Note that I have used in (1) the second inequality and the initial condition in the third inequality.



Additionally assuming that $sum_{k=0}^infty b_k^2 < infty$, we can apply the previous result with shifted-indizes in order to conclude (by using $(a+b)^2 le 2 a^2+2b^2$) that
begin{align}
frac{1}{1+k-j} sum_{i=j}^k xi_i^2 &le frac{2}{1+k-j} Big(a_j^2 + (sum_{i=j}^k b_i)^2Big)\
&le frac{2a_j^2}{1+k-j} + 2sum_{i=j}^k b_i^2,
end{align}
where in the last step the Cauchy-Schwarz inequality was used.
Thus
begin{align}
limsup_{k rightarrow infty} frac{1}{k} sum_{i=1}^k xi_i^2 &le limsup_{k rightarrow infty} frac{1}{k} sum_{i=1}^{j-1} xi_i^2 + limsup_{k rightarrow infty} frac{1}{k} sum_{i=j}^k xi_i^2 \
&= limsup_{k rightarrow infty} frac{k+1-j}{k} frac{1}{k+1-j} sum_{i=j}^k xi_i^2 \
&le limsup_{k rightarrow infty} frac{2a_j^2}{1+k-j} + 2sum_{i=j}^k b_i^2 \
&= 2sum_{i=j}^infty b_i^2 rightarrow 0 quad text{for} quad j rightarrow infty
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is impressive! I just have a question about the last inquality. How does $sum_{i=1}^k xi_i^2 le 2(a_1^2+(sum_{i=0}^k b_i)^2)$ come to $sum_{i=1}^k xi_i^2 le 2(a_1^2+sum_{i=0}^k b_i^2)$
    $endgroup$
    – Jaqen Chou
    Sep 4 '18 at 4:24












  • $begingroup$
    My argument was not correct at all. I have added a complete answer.
    $endgroup$
    – p4sch
    Sep 4 '18 at 7:47










  • $begingroup$
    I think the application of Cauchy-Schwarz inequality is not true. We can check it by set $k=j+1$.
    $endgroup$
    – Jaqen Chou
    Sep 4 '18 at 9:00










  • $begingroup$
    By the Cauchy-Schwarz inequality we find $big( sum_{i=j}^k b_i)^2 le big( sum_{i=j}^k 1) big( sum_{i=j}^k b_i^2 big) = (k+1-j) big( sum_{i=j}^k b_i^2 big)$. The factor $(k+1-j)$ cancels with $frac{1}{k+1-j}$.
    $endgroup$
    – p4sch
    Sep 4 '18 at 9:48










  • $begingroup$
    That's right. I have no question any more. Also maybe you could edit $xi_k$ to $xi_i$ in the last formula series.
    $endgroup$
    – Jaqen Chou
    Sep 4 '18 at 13:28











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3





+50







$begingroup$

Let us first prove by induction that
$$tag{1} a_1 +sum_{i=0}^k b_i ge a_{k+1}.$$
For $k=0$ this inequality is trivial. Next assume that (1) is already shown for $0 le i le k$, then we have
$$a_1 + sum_{i=0}^{k+1} b_i ge a_{k+1} + b_{k+1} ge a_{k+1} + big(sqrt{a_{k+2}^2 + xi_{k+1}^2} - a_{k+1} big) ge a_{k+2}.$$
Now we are ready to show the stronger inequality
$$tag{2}sum_{i=1}^k xi_i^2 leq big(a_1 + sum_{i=0}^k b_i big)^2 - a_{k+1}^2.$$
Again, we prove this by induction. In fact, $k=1$ is a direct consequence of the initial condition:
$$a_2^2 le (a_1+b_1)^2-xi_1^2 le (a_1+b_0+b_1)^2 - xi_1^2.$$
Assume that (2) is true for $k$, then
begin{align}
big(a_1 + sum_{i=0}^{k+1} b_i big)^2-a_{k+2}^2 &ge sum_{i=1}^k xi_i^2 + 2 big( a_1 +sum_{i=0}^k b_k big) b_{k+1} + b_{k+1}^2 + (a_{k+1}^2 -a_{k+2}^2) \
&ge sum_{i=1}^k xi_i^2 + (2 a_{k+1} b_{k+1} +b_{k+1}^2) + (a_{k+1}^2 -a_{k+2}^2) \
&ge sum_{i=1}^k xi_i^2 + (a_{k+2}^2 +xi_{k+1}^2 - a_{k+1}^2) + (a_{k+1}^2 -a_{k+2}^2) \
&ge sum_{i=1}^{k+1} xi_i^2
end{align}
Note that I have used in (1) the second inequality and the initial condition in the third inequality.



Additionally assuming that $sum_{k=0}^infty b_k^2 < infty$, we can apply the previous result with shifted-indizes in order to conclude (by using $(a+b)^2 le 2 a^2+2b^2$) that
begin{align}
frac{1}{1+k-j} sum_{i=j}^k xi_i^2 &le frac{2}{1+k-j} Big(a_j^2 + (sum_{i=j}^k b_i)^2Big)\
&le frac{2a_j^2}{1+k-j} + 2sum_{i=j}^k b_i^2,
end{align}
where in the last step the Cauchy-Schwarz inequality was used.
Thus
begin{align}
limsup_{k rightarrow infty} frac{1}{k} sum_{i=1}^k xi_i^2 &le limsup_{k rightarrow infty} frac{1}{k} sum_{i=1}^{j-1} xi_i^2 + limsup_{k rightarrow infty} frac{1}{k} sum_{i=j}^k xi_i^2 \
&= limsup_{k rightarrow infty} frac{k+1-j}{k} frac{1}{k+1-j} sum_{i=j}^k xi_i^2 \
&le limsup_{k rightarrow infty} frac{2a_j^2}{1+k-j} + 2sum_{i=j}^k b_i^2 \
&= 2sum_{i=j}^infty b_i^2 rightarrow 0 quad text{for} quad j rightarrow infty
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is impressive! I just have a question about the last inquality. How does $sum_{i=1}^k xi_i^2 le 2(a_1^2+(sum_{i=0}^k b_i)^2)$ come to $sum_{i=1}^k xi_i^2 le 2(a_1^2+sum_{i=0}^k b_i^2)$
    $endgroup$
    – Jaqen Chou
    Sep 4 '18 at 4:24












  • $begingroup$
    My argument was not correct at all. I have added a complete answer.
    $endgroup$
    – p4sch
    Sep 4 '18 at 7:47










  • $begingroup$
    I think the application of Cauchy-Schwarz inequality is not true. We can check it by set $k=j+1$.
    $endgroup$
    – Jaqen Chou
    Sep 4 '18 at 9:00










  • $begingroup$
    By the Cauchy-Schwarz inequality we find $big( sum_{i=j}^k b_i)^2 le big( sum_{i=j}^k 1) big( sum_{i=j}^k b_i^2 big) = (k+1-j) big( sum_{i=j}^k b_i^2 big)$. The factor $(k+1-j)$ cancels with $frac{1}{k+1-j}$.
    $endgroup$
    – p4sch
    Sep 4 '18 at 9:48










  • $begingroup$
    That's right. I have no question any more. Also maybe you could edit $xi_k$ to $xi_i$ in the last formula series.
    $endgroup$
    – Jaqen Chou
    Sep 4 '18 at 13:28
















3





+50







$begingroup$

Let us first prove by induction that
$$tag{1} a_1 +sum_{i=0}^k b_i ge a_{k+1}.$$
For $k=0$ this inequality is trivial. Next assume that (1) is already shown for $0 le i le k$, then we have
$$a_1 + sum_{i=0}^{k+1} b_i ge a_{k+1} + b_{k+1} ge a_{k+1} + big(sqrt{a_{k+2}^2 + xi_{k+1}^2} - a_{k+1} big) ge a_{k+2}.$$
Now we are ready to show the stronger inequality
$$tag{2}sum_{i=1}^k xi_i^2 leq big(a_1 + sum_{i=0}^k b_i big)^2 - a_{k+1}^2.$$
Again, we prove this by induction. In fact, $k=1$ is a direct consequence of the initial condition:
$$a_2^2 le (a_1+b_1)^2-xi_1^2 le (a_1+b_0+b_1)^2 - xi_1^2.$$
Assume that (2) is true for $k$, then
begin{align}
big(a_1 + sum_{i=0}^{k+1} b_i big)^2-a_{k+2}^2 &ge sum_{i=1}^k xi_i^2 + 2 big( a_1 +sum_{i=0}^k b_k big) b_{k+1} + b_{k+1}^2 + (a_{k+1}^2 -a_{k+2}^2) \
&ge sum_{i=1}^k xi_i^2 + (2 a_{k+1} b_{k+1} +b_{k+1}^2) + (a_{k+1}^2 -a_{k+2}^2) \
&ge sum_{i=1}^k xi_i^2 + (a_{k+2}^2 +xi_{k+1}^2 - a_{k+1}^2) + (a_{k+1}^2 -a_{k+2}^2) \
&ge sum_{i=1}^{k+1} xi_i^2
end{align}
Note that I have used in (1) the second inequality and the initial condition in the third inequality.



Additionally assuming that $sum_{k=0}^infty b_k^2 < infty$, we can apply the previous result with shifted-indizes in order to conclude (by using $(a+b)^2 le 2 a^2+2b^2$) that
begin{align}
frac{1}{1+k-j} sum_{i=j}^k xi_i^2 &le frac{2}{1+k-j} Big(a_j^2 + (sum_{i=j}^k b_i)^2Big)\
&le frac{2a_j^2}{1+k-j} + 2sum_{i=j}^k b_i^2,
end{align}
where in the last step the Cauchy-Schwarz inequality was used.
Thus
begin{align}
limsup_{k rightarrow infty} frac{1}{k} sum_{i=1}^k xi_i^2 &le limsup_{k rightarrow infty} frac{1}{k} sum_{i=1}^{j-1} xi_i^2 + limsup_{k rightarrow infty} frac{1}{k} sum_{i=j}^k xi_i^2 \
&= limsup_{k rightarrow infty} frac{k+1-j}{k} frac{1}{k+1-j} sum_{i=j}^k xi_i^2 \
&le limsup_{k rightarrow infty} frac{2a_j^2}{1+k-j} + 2sum_{i=j}^k b_i^2 \
&= 2sum_{i=j}^infty b_i^2 rightarrow 0 quad text{for} quad j rightarrow infty
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is impressive! I just have a question about the last inquality. How does $sum_{i=1}^k xi_i^2 le 2(a_1^2+(sum_{i=0}^k b_i)^2)$ come to $sum_{i=1}^k xi_i^2 le 2(a_1^2+sum_{i=0}^k b_i^2)$
    $endgroup$
    – Jaqen Chou
    Sep 4 '18 at 4:24












  • $begingroup$
    My argument was not correct at all. I have added a complete answer.
    $endgroup$
    – p4sch
    Sep 4 '18 at 7:47










  • $begingroup$
    I think the application of Cauchy-Schwarz inequality is not true. We can check it by set $k=j+1$.
    $endgroup$
    – Jaqen Chou
    Sep 4 '18 at 9:00










  • $begingroup$
    By the Cauchy-Schwarz inequality we find $big( sum_{i=j}^k b_i)^2 le big( sum_{i=j}^k 1) big( sum_{i=j}^k b_i^2 big) = (k+1-j) big( sum_{i=j}^k b_i^2 big)$. The factor $(k+1-j)$ cancels with $frac{1}{k+1-j}$.
    $endgroup$
    – p4sch
    Sep 4 '18 at 9:48










  • $begingroup$
    That's right. I have no question any more. Also maybe you could edit $xi_k$ to $xi_i$ in the last formula series.
    $endgroup$
    – Jaqen Chou
    Sep 4 '18 at 13:28














3





+50







3





+50



3




+50



$begingroup$

Let us first prove by induction that
$$tag{1} a_1 +sum_{i=0}^k b_i ge a_{k+1}.$$
For $k=0$ this inequality is trivial. Next assume that (1) is already shown for $0 le i le k$, then we have
$$a_1 + sum_{i=0}^{k+1} b_i ge a_{k+1} + b_{k+1} ge a_{k+1} + big(sqrt{a_{k+2}^2 + xi_{k+1}^2} - a_{k+1} big) ge a_{k+2}.$$
Now we are ready to show the stronger inequality
$$tag{2}sum_{i=1}^k xi_i^2 leq big(a_1 + sum_{i=0}^k b_i big)^2 - a_{k+1}^2.$$
Again, we prove this by induction. In fact, $k=1$ is a direct consequence of the initial condition:
$$a_2^2 le (a_1+b_1)^2-xi_1^2 le (a_1+b_0+b_1)^2 - xi_1^2.$$
Assume that (2) is true for $k$, then
begin{align}
big(a_1 + sum_{i=0}^{k+1} b_i big)^2-a_{k+2}^2 &ge sum_{i=1}^k xi_i^2 + 2 big( a_1 +sum_{i=0}^k b_k big) b_{k+1} + b_{k+1}^2 + (a_{k+1}^2 -a_{k+2}^2) \
&ge sum_{i=1}^k xi_i^2 + (2 a_{k+1} b_{k+1} +b_{k+1}^2) + (a_{k+1}^2 -a_{k+2}^2) \
&ge sum_{i=1}^k xi_i^2 + (a_{k+2}^2 +xi_{k+1}^2 - a_{k+1}^2) + (a_{k+1}^2 -a_{k+2}^2) \
&ge sum_{i=1}^{k+1} xi_i^2
end{align}
Note that I have used in (1) the second inequality and the initial condition in the third inequality.



Additionally assuming that $sum_{k=0}^infty b_k^2 < infty$, we can apply the previous result with shifted-indizes in order to conclude (by using $(a+b)^2 le 2 a^2+2b^2$) that
begin{align}
frac{1}{1+k-j} sum_{i=j}^k xi_i^2 &le frac{2}{1+k-j} Big(a_j^2 + (sum_{i=j}^k b_i)^2Big)\
&le frac{2a_j^2}{1+k-j} + 2sum_{i=j}^k b_i^2,
end{align}
where in the last step the Cauchy-Schwarz inequality was used.
Thus
begin{align}
limsup_{k rightarrow infty} frac{1}{k} sum_{i=1}^k xi_i^2 &le limsup_{k rightarrow infty} frac{1}{k} sum_{i=1}^{j-1} xi_i^2 + limsup_{k rightarrow infty} frac{1}{k} sum_{i=j}^k xi_i^2 \
&= limsup_{k rightarrow infty} frac{k+1-j}{k} frac{1}{k+1-j} sum_{i=j}^k xi_i^2 \
&le limsup_{k rightarrow infty} frac{2a_j^2}{1+k-j} + 2sum_{i=j}^k b_i^2 \
&= 2sum_{i=j}^infty b_i^2 rightarrow 0 quad text{for} quad j rightarrow infty
end{align}






share|cite|improve this answer











$endgroup$



Let us first prove by induction that
$$tag{1} a_1 +sum_{i=0}^k b_i ge a_{k+1}.$$
For $k=0$ this inequality is trivial. Next assume that (1) is already shown for $0 le i le k$, then we have
$$a_1 + sum_{i=0}^{k+1} b_i ge a_{k+1} + b_{k+1} ge a_{k+1} + big(sqrt{a_{k+2}^2 + xi_{k+1}^2} - a_{k+1} big) ge a_{k+2}.$$
Now we are ready to show the stronger inequality
$$tag{2}sum_{i=1}^k xi_i^2 leq big(a_1 + sum_{i=0}^k b_i big)^2 - a_{k+1}^2.$$
Again, we prove this by induction. In fact, $k=1$ is a direct consequence of the initial condition:
$$a_2^2 le (a_1+b_1)^2-xi_1^2 le (a_1+b_0+b_1)^2 - xi_1^2.$$
Assume that (2) is true for $k$, then
begin{align}
big(a_1 + sum_{i=0}^{k+1} b_i big)^2-a_{k+2}^2 &ge sum_{i=1}^k xi_i^2 + 2 big( a_1 +sum_{i=0}^k b_k big) b_{k+1} + b_{k+1}^2 + (a_{k+1}^2 -a_{k+2}^2) \
&ge sum_{i=1}^k xi_i^2 + (2 a_{k+1} b_{k+1} +b_{k+1}^2) + (a_{k+1}^2 -a_{k+2}^2) \
&ge sum_{i=1}^k xi_i^2 + (a_{k+2}^2 +xi_{k+1}^2 - a_{k+1}^2) + (a_{k+1}^2 -a_{k+2}^2) \
&ge sum_{i=1}^{k+1} xi_i^2
end{align}
Note that I have used in (1) the second inequality and the initial condition in the third inequality.



Additionally assuming that $sum_{k=0}^infty b_k^2 < infty$, we can apply the previous result with shifted-indizes in order to conclude (by using $(a+b)^2 le 2 a^2+2b^2$) that
begin{align}
frac{1}{1+k-j} sum_{i=j}^k xi_i^2 &le frac{2}{1+k-j} Big(a_j^2 + (sum_{i=j}^k b_i)^2Big)\
&le frac{2a_j^2}{1+k-j} + 2sum_{i=j}^k b_i^2,
end{align}
where in the last step the Cauchy-Schwarz inequality was used.
Thus
begin{align}
limsup_{k rightarrow infty} frac{1}{k} sum_{i=1}^k xi_i^2 &le limsup_{k rightarrow infty} frac{1}{k} sum_{i=1}^{j-1} xi_i^2 + limsup_{k rightarrow infty} frac{1}{k} sum_{i=j}^k xi_i^2 \
&= limsup_{k rightarrow infty} frac{k+1-j}{k} frac{1}{k+1-j} sum_{i=j}^k xi_i^2 \
&le limsup_{k rightarrow infty} frac{2a_j^2}{1+k-j} + 2sum_{i=j}^k b_i^2 \
&= 2sum_{i=j}^infty b_i^2 rightarrow 0 quad text{for} quad j rightarrow infty
end{align}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 4 '18 at 13:37

























answered Sep 3 '18 at 13:01









p4schp4sch

4,995217




4,995217












  • $begingroup$
    This is impressive! I just have a question about the last inquality. How does $sum_{i=1}^k xi_i^2 le 2(a_1^2+(sum_{i=0}^k b_i)^2)$ come to $sum_{i=1}^k xi_i^2 le 2(a_1^2+sum_{i=0}^k b_i^2)$
    $endgroup$
    – Jaqen Chou
    Sep 4 '18 at 4:24












  • $begingroup$
    My argument was not correct at all. I have added a complete answer.
    $endgroup$
    – p4sch
    Sep 4 '18 at 7:47










  • $begingroup$
    I think the application of Cauchy-Schwarz inequality is not true. We can check it by set $k=j+1$.
    $endgroup$
    – Jaqen Chou
    Sep 4 '18 at 9:00










  • $begingroup$
    By the Cauchy-Schwarz inequality we find $big( sum_{i=j}^k b_i)^2 le big( sum_{i=j}^k 1) big( sum_{i=j}^k b_i^2 big) = (k+1-j) big( sum_{i=j}^k b_i^2 big)$. The factor $(k+1-j)$ cancels with $frac{1}{k+1-j}$.
    $endgroup$
    – p4sch
    Sep 4 '18 at 9:48










  • $begingroup$
    That's right. I have no question any more. Also maybe you could edit $xi_k$ to $xi_i$ in the last formula series.
    $endgroup$
    – Jaqen Chou
    Sep 4 '18 at 13:28


















  • $begingroup$
    This is impressive! I just have a question about the last inquality. How does $sum_{i=1}^k xi_i^2 le 2(a_1^2+(sum_{i=0}^k b_i)^2)$ come to $sum_{i=1}^k xi_i^2 le 2(a_1^2+sum_{i=0}^k b_i^2)$
    $endgroup$
    – Jaqen Chou
    Sep 4 '18 at 4:24












  • $begingroup$
    My argument was not correct at all. I have added a complete answer.
    $endgroup$
    – p4sch
    Sep 4 '18 at 7:47










  • $begingroup$
    I think the application of Cauchy-Schwarz inequality is not true. We can check it by set $k=j+1$.
    $endgroup$
    – Jaqen Chou
    Sep 4 '18 at 9:00










  • $begingroup$
    By the Cauchy-Schwarz inequality we find $big( sum_{i=j}^k b_i)^2 le big( sum_{i=j}^k 1) big( sum_{i=j}^k b_i^2 big) = (k+1-j) big( sum_{i=j}^k b_i^2 big)$. The factor $(k+1-j)$ cancels with $frac{1}{k+1-j}$.
    $endgroup$
    – p4sch
    Sep 4 '18 at 9:48










  • $begingroup$
    That's right. I have no question any more. Also maybe you could edit $xi_k$ to $xi_i$ in the last formula series.
    $endgroup$
    – Jaqen Chou
    Sep 4 '18 at 13:28
















$begingroup$
This is impressive! I just have a question about the last inquality. How does $sum_{i=1}^k xi_i^2 le 2(a_1^2+(sum_{i=0}^k b_i)^2)$ come to $sum_{i=1}^k xi_i^2 le 2(a_1^2+sum_{i=0}^k b_i^2)$
$endgroup$
– Jaqen Chou
Sep 4 '18 at 4:24






$begingroup$
This is impressive! I just have a question about the last inquality. How does $sum_{i=1}^k xi_i^2 le 2(a_1^2+(sum_{i=0}^k b_i)^2)$ come to $sum_{i=1}^k xi_i^2 le 2(a_1^2+sum_{i=0}^k b_i^2)$
$endgroup$
– Jaqen Chou
Sep 4 '18 at 4:24














$begingroup$
My argument was not correct at all. I have added a complete answer.
$endgroup$
– p4sch
Sep 4 '18 at 7:47




$begingroup$
My argument was not correct at all. I have added a complete answer.
$endgroup$
– p4sch
Sep 4 '18 at 7:47












$begingroup$
I think the application of Cauchy-Schwarz inequality is not true. We can check it by set $k=j+1$.
$endgroup$
– Jaqen Chou
Sep 4 '18 at 9:00




$begingroup$
I think the application of Cauchy-Schwarz inequality is not true. We can check it by set $k=j+1$.
$endgroup$
– Jaqen Chou
Sep 4 '18 at 9:00












$begingroup$
By the Cauchy-Schwarz inequality we find $big( sum_{i=j}^k b_i)^2 le big( sum_{i=j}^k 1) big( sum_{i=j}^k b_i^2 big) = (k+1-j) big( sum_{i=j}^k b_i^2 big)$. The factor $(k+1-j)$ cancels with $frac{1}{k+1-j}$.
$endgroup$
– p4sch
Sep 4 '18 at 9:48




$begingroup$
By the Cauchy-Schwarz inequality we find $big( sum_{i=j}^k b_i)^2 le big( sum_{i=j}^k 1) big( sum_{i=j}^k b_i^2 big) = (k+1-j) big( sum_{i=j}^k b_i^2 big)$. The factor $(k+1-j)$ cancels with $frac{1}{k+1-j}$.
$endgroup$
– p4sch
Sep 4 '18 at 9:48












$begingroup$
That's right. I have no question any more. Also maybe you could edit $xi_k$ to $xi_i$ in the last formula series.
$endgroup$
– Jaqen Chou
Sep 4 '18 at 13:28




$begingroup$
That's right. I have no question any more. Also maybe you could edit $xi_k$ to $xi_i$ in the last formula series.
$endgroup$
– Jaqen Chou
Sep 4 '18 at 13:28


















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