Prove $fracb-a+1ab leq sum_i=a^b frac1i^2 leq fracb-a+1(a-1)b$ [closed]
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Given $b>a>1; a,b in Z^+$. Prove $fracb-a+1ab leq sum_i=a^b frac1i^2 leq fracb-a+1(a-1)b. $
I mostly care about the second inequality. Thank you!
real-analysis algebra-precalculus discrete-mathematics real-numbers
asked Sep 3 at 12:47
Zichong Li
133
closed as off-topic by amWhy, Xander Henderson, rtybase, user91500, Jendrik Stelzner Sep 6 at 10:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, rtybase, user91500, Jendrik Stelzner
add a comment |Â
up vote
1
down vote
favorite
Given $b>a>1; a,b in Z^+$. Prove $fracb-a+1ab leq sum_i=a^b frac1i^2 leq fracb-a+1(a-1)b. $
I mostly care about the second inequality. Thank you!
real-analysis algebra-precalculus discrete-mathematics real-numbers
asked Sep 3 at 12:47
Zichong Li
133
closed as off-topic by amWhy, Xander Henderson, rtybase, user91500, Jendrik Stelzner Sep 6 at 10:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, rtybase, user91500, Jendrik Stelzner
Welcome to MSE. Your tags are in conflict. How can "real analysis" and "algebra precalculus" both apply, for example? What are you studying? What techniques do you know about? You have to give some context so we know hat kinds of answers are likely to be helpful to you. Also, you should show us what you have done so far, and where you are getting stuck, so we know where you need help.
– saulspatz
Sep 3 at 12:52
As you can see, this simple-looking inequality fits both tags. Also I made many failed attempts only, I have no idea if I made any "progress"
– Zichong Li
Sep 3 at 12:59
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given $b>a>1; a,b in Z^+$. Prove $fracb-a+1ab leq sum_i=a^b frac1i^2 leq fracb-a+1(a-1)b. $
I mostly care about the second inequality. Thank you!
real-analysis algebra-precalculus discrete-mathematics real-numbers
asked Sep 3 at 12:47
Zichong Li
133
Given $b>a>1; a,b in Z^+$. Prove $fracb-a+1ab leq sum_i=a^b frac1i^2 leq fracb-a+1(a-1)b. $
I mostly care about the second inequality. Thank you!
real-analysis algebra-precalculus discrete-mathematics real-numbers
real-analysis algebra-precalculus discrete-mathematics real-numbers
asked Sep 3 at 12:47
Zichong Li
133
asked Sep 3 at 12:47
Zichong Li
133
asked Sep 3 at 12:47
Zichong Li
133
asked Sep 3 at 12:47
Zichong Li
133
asked Sep 3 at 12:47
Zichong Li
133
133
closed as off-topic by amWhy, Xander Henderson, rtybase, user91500, Jendrik Stelzner Sep 6 at 10:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, rtybase, user91500, Jendrik Stelzner
closed as off-topic by amWhy, Xander Henderson, rtybase, user91500, Jendrik Stelzner Sep 6 at 10:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, rtybase, user91500, Jendrik Stelzner
Welcome to MSE. Your tags are in conflict. How can "real analysis" and "algebra precalculus" both apply, for example? What are you studying? What techniques do you know about? You have to give some context so we know hat kinds of answers are likely to be helpful to you. Also, you should show us what you have done so far, and where you are getting stuck, so we know where you need help.
– saulspatz
Sep 3 at 12:52
As you can see, this simple-looking inequality fits both tags. Also I made many failed attempts only, I have no idea if I made any "progress"
– Zichong Li
Sep 3 at 12:59
add a comment |Â
Welcome to MSE. Your tags are in conflict. How can "real analysis" and "algebra precalculus" both apply, for example? What are you studying? What techniques do you know about? You have to give some context so we know hat kinds of answers are likely to be helpful to you. Also, you should show us what you have done so far, and where you are getting stuck, so we know where you need help.
– saulspatz
Sep 3 at 12:52
As you can see, this simple-looking inequality fits both tags. Also I made many failed attempts only, I have no idea if I made any "progress"
– Zichong Li
Sep 3 at 12:59
Welcome to MSE. Your tags are in conflict. How can "real analysis" and "algebra precalculus" both apply, for example? What are you studying? What techniques do you know about? You have to give some context so we know hat kinds of answers are likely to be helpful to you. Also, you should show us what you have done so far, and where you are getting stuck, so we know where you need help.
– saulspatz
Sep 3 at 12:52
Welcome to MSE. Your tags are in conflict. How can "real analysis" and "algebra precalculus" both apply, for example? What are you studying? What techniques do you know about? You have to give some context so we know hat kinds of answers are likely to be helpful to you. Also, you should show us what you have done so far, and where you are getting stuck, so we know where you need help.
– saulspatz
Sep 3 at 12:52
As you can see, this simple-looking inequality fits both tags. Also I made many failed attempts only, I have no idea if I made any "progress"
– Zichong Li
Sep 3 at 12:59
As you can see, this simple-looking inequality fits both tags. Also I made many failed attempts only, I have no idea if I made any "progress"
– Zichong Li
Sep 3 at 12:59
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
For the second inequality just observe that
$$
frac1i^2 leq frac1i(i-1) = frac1i-1 - frac1i,
$$
hence
$$
sum_i=a^b frac1i^2 leq
sum_i=a^bleft(frac1i-1 - frac1iright)
= frac1a-1 - frac1b = fracb-a+1(a-1)b,.
$$
The first inequality can be proved by induction on $b$.
Let $ainmathbbZ^+$ be fixed, and let us prove that
$$
S(b) := sum_i=a^b frac1i^2 - fracb-a+1abgeq 0
qquad
forall bgeq a.
$$
If $b = a$ then $S(b) = 0$.
Assume now that the inequality holds for some $bgeq a$, and let us prove that it holds for $b+1$.
Using the induction assumption $S(b) geq 0$, we have that:
$$
beginsplit
S(b+1) & = S(b) + frac1(b+1)^2 -fracb-a+2a(b+1) + fracb-a+1ab
\ & geq frac1(b+1)^2 -fracb-a+2a(b+1) + fracb-a+1ab
\ & =
fracb-a+1ab(b+1)^2 > 0.
endsplit
$$
answered Sep 3 at 12:54
Rigel
10.4k11319
Thank you so much sir! The second inequality indeed works. But using the same technique in the first one, it seems we can only get a weaker result. Can you elaborate on how to prove the first inequality please?
– Zichong Li
Sep 3 at 13:09
You are right. Using the same argument, at the denominator you obtain $a(b+1)$.
– Rigel
Sep 3 at 19:04
Right, I'm just curious how to prove the stronger ($ab$ on denominator) inequality.
– Zichong Li
Sep 3 at 21:38
I have added a proof of the first inequality.
– Rigel
Sep 4 at 10:06
Wow that works. Thank you!
– Zichong Li
Sep 5 at 14:12
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
For the second inequality just observe that
$$
frac1i^2 leq frac1i(i-1) = frac1i-1 - frac1i,
$$
hence
$$
sum_i=a^b frac1i^2 leq
sum_i=a^bleft(frac1i-1 - frac1iright)
= frac1a-1 - frac1b = fracb-a+1(a-1)b,.
$$
The first inequality can be proved by induction on $b$.
Let $ainmathbbZ^+$ be fixed, and let us prove that
$$
S(b) := sum_i=a^b frac1i^2 - fracb-a+1abgeq 0
qquad
forall bgeq a.
$$
If $b = a$ then $S(b) = 0$.
Assume now that the inequality holds for some $bgeq a$, and let us prove that it holds for $b+1$.
Using the induction assumption $S(b) geq 0$, we have that:
$$
beginsplit
S(b+1) & = S(b) + frac1(b+1)^2 -fracb-a+2a(b+1) + fracb-a+1ab
\ & geq frac1(b+1)^2 -fracb-a+2a(b+1) + fracb-a+1ab
\ & =
fracb-a+1ab(b+1)^2 > 0.
endsplit
$$
answered Sep 3 at 12:54
Rigel
10.4k11319
Thank you so much sir! The second inequality indeed works. But using the same technique in the first one, it seems we can only get a weaker result. Can you elaborate on how to prove the first inequality please?
– Zichong Li
Sep 3 at 13:09
You are right. Using the same argument, at the denominator you obtain $a(b+1)$.
– Rigel
Sep 3 at 19:04
Right, I'm just curious how to prove the stronger ($ab$ on denominator) inequality.
– Zichong Li
Sep 3 at 21:38
I have added a proof of the first inequality.
– Rigel
Sep 4 at 10:06
Wow that works. Thank you!
– Zichong Li
Sep 5 at 14:12
add a comment |Â
up vote
4
down vote
accepted
For the second inequality just observe that
$$
frac1i^2 leq frac1i(i-1) = frac1i-1 - frac1i,
$$
hence
$$
sum_i=a^b frac1i^2 leq
sum_i=a^bleft(frac1i-1 - frac1iright)
= frac1a-1 - frac1b = fracb-a+1(a-1)b,.
$$
The first inequality can be proved by induction on $b$.
Let $ainmathbbZ^+$ be fixed, and let us prove that
$$
S(b) := sum_i=a^b frac1i^2 - fracb-a+1abgeq 0
qquad
forall bgeq a.
$$
If $b = a$ then $S(b) = 0$.
Assume now that the inequality holds for some $bgeq a$, and let us prove that it holds for $b+1$.
Using the induction assumption $S(b) geq 0$, we have that:
$$
beginsplit
S(b+1) & = S(b) + frac1(b+1)^2 -fracb-a+2a(b+1) + fracb-a+1ab
\ & geq frac1(b+1)^2 -fracb-a+2a(b+1) + fracb-a+1ab
\ & =
fracb-a+1ab(b+1)^2 > 0.
endsplit
$$
answered Sep 3 at 12:54
Rigel
10.4k11319
Thank you so much sir! The second inequality indeed works. But using the same technique in the first one, it seems we can only get a weaker result. Can you elaborate on how to prove the first inequality please?
– Zichong Li
Sep 3 at 13:09
You are right. Using the same argument, at the denominator you obtain $a(b+1)$.
– Rigel
Sep 3 at 19:04
Right, I'm just curious how to prove the stronger ($ab$ on denominator) inequality.
– Zichong Li
Sep 3 at 21:38
I have added a proof of the first inequality.
– Rigel
Sep 4 at 10:06
Wow that works. Thank you!
– Zichong Li
Sep 5 at 14:12
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
For the second inequality just observe that
$$
frac1i^2 leq frac1i(i-1) = frac1i-1 - frac1i,
$$
hence
$$
sum_i=a^b frac1i^2 leq
sum_i=a^bleft(frac1i-1 - frac1iright)
= frac1a-1 - frac1b = fracb-a+1(a-1)b,.
$$
The first inequality can be proved by induction on $b$.
Let $ainmathbbZ^+$ be fixed, and let us prove that
$$
S(b) := sum_i=a^b frac1i^2 - fracb-a+1abgeq 0
qquad
forall bgeq a.
$$
If $b = a$ then $S(b) = 0$.
Assume now that the inequality holds for some $bgeq a$, and let us prove that it holds for $b+1$.
Using the induction assumption $S(b) geq 0$, we have that:
$$
beginsplit
S(b+1) & = S(b) + frac1(b+1)^2 -fracb-a+2a(b+1) + fracb-a+1ab
\ & geq frac1(b+1)^2 -fracb-a+2a(b+1) + fracb-a+1ab
\ & =
fracb-a+1ab(b+1)^2 > 0.
endsplit
$$
answered Sep 3 at 12:54
Rigel
10.4k11319
For the second inequality just observe that
$$
frac1i^2 leq frac1i(i-1) = frac1i-1 - frac1i,
$$
hence
$$
sum_i=a^b frac1i^2 leq
sum_i=a^bleft(frac1i-1 - frac1iright)
= frac1a-1 - frac1b = fracb-a+1(a-1)b,.
$$
The first inequality can be proved by induction on $b$.
Let $ainmathbbZ^+$ be fixed, and let us prove that
$$
S(b) := sum_i=a^b frac1i^2 - fracb-a+1abgeq 0
qquad
forall bgeq a.
$$
If $b = a$ then $S(b) = 0$.
Assume now that the inequality holds for some $bgeq a$, and let us prove that it holds for $b+1$.
Using the induction assumption $S(b) geq 0$, we have that:
$$
beginsplit
S(b+1) & = S(b) + frac1(b+1)^2 -fracb-a+2a(b+1) + fracb-a+1ab
\ & geq frac1(b+1)^2 -fracb-a+2a(b+1) + fracb-a+1ab
\ & =
fracb-a+1ab(b+1)^2 > 0.
endsplit
$$
answered Sep 3 at 12:54
Rigel
10.4k11319
edited Sep 4 at 10:06
answered Sep 3 at 12:54
Rigel
10.4k11319
answered Sep 3 at 12:54
Rigel
10.4k11319
answered Sep 3 at 12:54
Rigel
10.4k11319
10.4k11319
Thank you so much sir! The second inequality indeed works. But using the same technique in the first one, it seems we can only get a weaker result. Can you elaborate on how to prove the first inequality please?
– Zichong Li
Sep 3 at 13:09
You are right. Using the same argument, at the denominator you obtain $a(b+1)$.
– Rigel
Sep 3 at 19:04
Right, I'm just curious how to prove the stronger ($ab$ on denominator) inequality.
– Zichong Li
Sep 3 at 21:38
I have added a proof of the first inequality.
– Rigel
Sep 4 at 10:06
Wow that works. Thank you!
– Zichong Li
Sep 5 at 14:12
add a comment |Â
Thank you so much sir! The second inequality indeed works. But using the same technique in the first one, it seems we can only get a weaker result. Can you elaborate on how to prove the first inequality please?
– Zichong Li
Sep 3 at 13:09
You are right. Using the same argument, at the denominator you obtain $a(b+1)$.
– Rigel
Sep 3 at 19:04
Right, I'm just curious how to prove the stronger ($ab$ on denominator) inequality.
– Zichong Li
Sep 3 at 21:38
I have added a proof of the first inequality.
– Rigel
Sep 4 at 10:06
Wow that works. Thank you!
– Zichong Li
Sep 5 at 14:12
Thank you so much sir! The second inequality indeed works. But using the same technique in the first one, it seems we can only get a weaker result. Can you elaborate on how to prove the first inequality please?
– Zichong Li
Sep 3 at 13:09
Thank you so much sir! The second inequality indeed works. But using the same technique in the first one, it seems we can only get a weaker result. Can you elaborate on how to prove the first inequality please?
– Zichong Li
Sep 3 at 13:09
You are right. Using the same argument, at the denominator you obtain $a(b+1)$.
– Rigel
Sep 3 at 19:04
You are right. Using the same argument, at the denominator you obtain $a(b+1)$.
– Rigel
Sep 3 at 19:04
Right, I'm just curious how to prove the stronger ($ab$ on denominator) inequality.
– Zichong Li
Sep 3 at 21:38
Right, I'm just curious how to prove the stronger ($ab$ on denominator) inequality.
– Zichong Li
Sep 3 at 21:38
I have added a proof of the first inequality.
– Rigel
Sep 4 at 10:06
I have added a proof of the first inequality.
– Rigel
Sep 4 at 10:06
Wow that works. Thank you!
– Zichong Li
Sep 5 at 14:12
Wow that works. Thank you!
– Zichong Li
Sep 5 at 14:12
add a comment |Â
Welcome to MSE. Your tags are in conflict. How can "real analysis" and "algebra precalculus" both apply, for example? What are you studying? What techniques do you know about? You have to give some context so we know hat kinds of answers are likely to be helpful to you. Also, you should show us what you have done so far, and where you are getting stuck, so we know where you need help.
– saulspatz
Sep 3 at 12:52
As you can see, this simple-looking inequality fits both tags. Also I made many failed attempts only, I have no idea if I made any "progress"
– Zichong Li
Sep 3 at 12:59