If $a_{1},a_{2},a_{3},…a_{n}$ are positive real numbers,then The Next CEO of Stack...
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If $a_{1},a_{2},a_{3},…a_{n}$ are positive real numbers,then
The Next CEO of Stack OverflowIf $a_n =int^{pi}_0 frac{sin(2n-1)x}{sin x}dx$ ,${a_n}_{ngeq 1}$ be a sequence of strictly decreasing positive numbers then…Can anyone prove D'Alembert Criterion (Dalambert) criterion for converging positive sequences?About a largest integral value of this sum of reciprocal numbers.Ratios between index and sequence elementProof of this inequalityHow to show that $lim_{ntoinfty}[a_1,cdots,a_n]$ exists if $a_kgeq 2$ for all $k$?find the formula of $ a_n $ in terms of $ a_1, a_2 and F_n $.For positive double sequence, limit of sum = sum of limit ?Find maximum of succession $a_{n+1}=frac{n^2+n+4}{2n^2+1}a_n$ knowing $a_1=1$
$begingroup$
Q) If $a_{1},a_{2},a_{3},....a_{n}$ are positive real numbers,then
$frac{a_{1}}{a_{2}} + frac{a_{2}}{a_{3}}+......+frac{a_{n-1}}{a_{n}}+frac{a_{n}}{a_{1}}$
is always
(A) $geq n ;$ (B) $leq n;$ (C) $leq n^{1/n};$ (D) none of the above
I have taken some small values of $n$ like $n=2,3$ and sample inputs as $(a_1,a_2,a_3)=(2,3,5)$,$(a_1,a_2,a_3)=(1/2,1/3,1/5)$ , $(a_1,a_2,a_3)=(0.1,0.2,0.3)$, $(a_1,a_2)=(2,3)$ for which I am getting answer as $(A)$ but I am not sure that it is always true.If it is not true always then answer will be $(D).$ I don't know how to prove it formally. Please help.
sequences-and-series
asked Mar 17 at 7:06
ankitankit
987
$endgroup$
|
show 3 more comments
$begingroup$
Q) If $a_{1},a_{2},a_{3},....a_{n}$ are positive real numbers,then
$frac{a_{1}}{a_{2}} + frac{a_{2}}{a_{3}}+......+frac{a_{n-1}}{a_{n}}+frac{a_{n}}{a_{1}}$
is always
(A) $geq n ;$ (B) $leq n;$ (C) $leq n^{1/n};$ (D) none of the above
I have taken some small values of $n$ like $n=2,3$ and sample inputs as $(a_1,a_2,a_3)=(2,3,5)$,$(a_1,a_2,a_3)=(1/2,1/3,1/5)$ , $(a_1,a_2,a_3)=(0.1,0.2,0.3)$, $(a_1,a_2)=(2,3)$ for which I am getting answer as $(A)$ but I am not sure that it is always true.If it is not true always then answer will be $(D).$ I don't know how to prove it formally. Please help.
sequences-and-series
asked Mar 17 at 7:06
ankitankit
987
$endgroup$
2
$begingroup$
Use AM-GM. First will be right answer.
$endgroup$
– Love Invariants
Mar 17 at 7:09
$begingroup$
Thank you @Love Invariants, got it..Now I can continue.
$endgroup$
– ankit
Mar 17 at 7:13
$begingroup$
@LoveInvariants Why are you answering in a comment?
$endgroup$
– Arthur
Mar 17 at 7:15
$begingroup$
@Arthur-Because a hint would satisfy. OP should do the rest of the work. It helps askers very much.
$endgroup$
– Love Invariants
Mar 17 at 7:16
$begingroup$
@LoveInvariants Hints are still answers more than they are clarification requests. So they belong in answer posts, not in comments. Also, you have a direct answer to the question by saying which alternative was right, so calling it just a hint isn't quite right.
$endgroup$
– Arthur
Mar 17 at 7:18
|
show 3 more comments
$begingroup$
Q) If $a_{1},a_{2},a_{3},....a_{n}$ are positive real numbers,then
$frac{a_{1}}{a_{2}} + frac{a_{2}}{a_{3}}+......+frac{a_{n-1}}{a_{n}}+frac{a_{n}}{a_{1}}$
is always
(A) $geq n ;$ (B) $leq n;$ (C) $leq n^{1/n};$ (D) none of the above
I have taken some small values of $n$ like $n=2,3$ and sample inputs as $(a_1,a_2,a_3)=(2,3,5)$,$(a_1,a_2,a_3)=(1/2,1/3,1/5)$ , $(a_1,a_2,a_3)=(0.1,0.2,0.3)$, $(a_1,a_2)=(2,3)$ for which I am getting answer as $(A)$ but I am not sure that it is always true.If it is not true always then answer will be $(D).$ I don't know how to prove it formally. Please help.
sequences-and-series
asked Mar 17 at 7:06
ankitankit
987
$endgroup$
Q) If $a_{1},a_{2},a_{3},....a_{n}$ are positive real numbers,then
$frac{a_{1}}{a_{2}} + frac{a_{2}}{a_{3}}+......+frac{a_{n-1}}{a_{n}}+frac{a_{n}}{a_{1}}$
is always
(A) $geq n ;$ (B) $leq n;$ (C) $leq n^{1/n};$ (D) none of the above
I have taken some small values of $n$ like $n=2,3$ and sample inputs as $(a_1,a_2,a_3)=(2,3,5)$,$(a_1,a_2,a_3)=(1/2,1/3,1/5)$ , $(a_1,a_2,a_3)=(0.1,0.2,0.3)$, $(a_1,a_2)=(2,3)$ for which I am getting answer as $(A)$ but I am not sure that it is always true.If it is not true always then answer will be $(D).$ I don't know how to prove it formally. Please help.
sequences-and-series
sequences-and-series
asked Mar 17 at 7:06
ankitankit
987
asked Mar 17 at 7:06
ankitankit
987
asked Mar 17 at 7:06
ankitankit
987
asked Mar 17 at 7:06
ankitankit
987
asked Mar 17 at 7:06
ankitankit
987
987
2
$begingroup$
Use AM-GM. First will be right answer.
$endgroup$
– Love Invariants
Mar 17 at 7:09
$begingroup$
Thank you @Love Invariants, got it..Now I can continue.
$endgroup$
– ankit
Mar 17 at 7:13
$begingroup$
@LoveInvariants Why are you answering in a comment?
$endgroup$
– Arthur
Mar 17 at 7:15
$begingroup$
@Arthur-Because a hint would satisfy. OP should do the rest of the work. It helps askers very much.
$endgroup$
– Love Invariants
Mar 17 at 7:16
$begingroup$
@LoveInvariants Hints are still answers more than they are clarification requests. So they belong in answer posts, not in comments. Also, you have a direct answer to the question by saying which alternative was right, so calling it just a hint isn't quite right.
$endgroup$
– Arthur
Mar 17 at 7:18
|
show 3 more comments
2
$begingroup$
Use AM-GM. First will be right answer.
$endgroup$
– Love Invariants
Mar 17 at 7:09
$begingroup$
Thank you @Love Invariants, got it..Now I can continue.
$endgroup$
– ankit
Mar 17 at 7:13
$begingroup$
@LoveInvariants Why are you answering in a comment?
$endgroup$
– Arthur
Mar 17 at 7:15
$begingroup$
@Arthur-Because a hint would satisfy. OP should do the rest of the work. It helps askers very much.
$endgroup$
– Love Invariants
Mar 17 at 7:16
$begingroup$
@LoveInvariants Hints are still answers more than they are clarification requests. So they belong in answer posts, not in comments. Also, you have a direct answer to the question by saying which alternative was right, so calling it just a hint isn't quite right.
$endgroup$
– Arthur
Mar 17 at 7:18
2
2
$begingroup$
Use AM-GM. First will be right answer.
$endgroup$
– Love Invariants
Mar 17 at 7:09
$begingroup$
Use AM-GM. First will be right answer.
$endgroup$
– Love Invariants
Mar 17 at 7:09
$begingroup$
Thank you @Love Invariants, got it..Now I can continue.
$endgroup$
– ankit
Mar 17 at 7:13
$begingroup$
Thank you @Love Invariants, got it..Now I can continue.
$endgroup$
– ankit
Mar 17 at 7:13
$begingroup$
@LoveInvariants Why are you answering in a comment?
$endgroup$
– Arthur
Mar 17 at 7:15
$begingroup$
@LoveInvariants Why are you answering in a comment?
$endgroup$
– Arthur
Mar 17 at 7:15
$begingroup$
@Arthur-Because a hint would satisfy. OP should do the rest of the work. It helps askers very much.
$endgroup$
– Love Invariants
Mar 17 at 7:16
$begingroup$
@Arthur-Because a hint would satisfy. OP should do the rest of the work. It helps askers very much.
$endgroup$
– Love Invariants
Mar 17 at 7:16
$begingroup$
@LoveInvariants Hints are still answers more than they are clarification requests. So they belong in answer posts, not in comments. Also, you have a direct answer to the question by saying which alternative was right, so calling it just a hint isn't quite right.
$endgroup$
– Arthur
Mar 17 at 7:18
$begingroup$
@LoveInvariants Hints are still answers more than they are clarification requests. So they belong in answer posts, not in comments. Also, you have a direct answer to the question by saying which alternative was right, so calling it just a hint isn't quite right.
$endgroup$
– Arthur
Mar 17 at 7:18
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
This may be solved using the rearrangement inequality. We have the two sequences
$$
a_1, a_2, a_3,ldots,a_n\
frac1{a_1}, frac1{a_2},frac1{a_3}, ldots,frac1{a_n}
$$
If we multiply each $a_i$ with $frac1{a_i}$, then the sum is $n$. By the rearrangement inequality, this is the smallest possible value we can get, so rearranging must give us a result $geq n$.
Or one may use the AM-GM inequality to get the same result.
answered Mar 17 at 7:13
ArthurArthur
121k7121207
$endgroup$
add a comment |
$begingroup$
Hint: Use AM-GM. First will be right answer.
AM-GM inequality
answered Mar 17 at 7:23
Love InvariantsLove Invariants
89015
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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oldest
votes
$begingroup$
This may be solved using the rearrangement inequality. We have the two sequences
$$
a_1, a_2, a_3,ldots,a_n\
frac1{a_1}, frac1{a_2},frac1{a_3}, ldots,frac1{a_n}
$$
If we multiply each $a_i$ with $frac1{a_i}$, then the sum is $n$. By the rearrangement inequality, this is the smallest possible value we can get, so rearranging must give us a result $geq n$.
Or one may use the AM-GM inequality to get the same result.
answered Mar 17 at 7:13
ArthurArthur
121k7121207
$endgroup$
add a comment |
$begingroup$
This may be solved using the rearrangement inequality. We have the two sequences
$$
a_1, a_2, a_3,ldots,a_n\
frac1{a_1}, frac1{a_2},frac1{a_3}, ldots,frac1{a_n}
$$
If we multiply each $a_i$ with $frac1{a_i}$, then the sum is $n$. By the rearrangement inequality, this is the smallest possible value we can get, so rearranging must give us a result $geq n$.
Or one may use the AM-GM inequality to get the same result.
answered Mar 17 at 7:13
ArthurArthur
121k7121207
$endgroup$
add a comment |
$begingroup$
This may be solved using the rearrangement inequality. We have the two sequences
$$
a_1, a_2, a_3,ldots,a_n\
frac1{a_1}, frac1{a_2},frac1{a_3}, ldots,frac1{a_n}
$$
If we multiply each $a_i$ with $frac1{a_i}$, then the sum is $n$. By the rearrangement inequality, this is the smallest possible value we can get, so rearranging must give us a result $geq n$.
Or one may use the AM-GM inequality to get the same result.
answered Mar 17 at 7:13
ArthurArthur
121k7121207
$endgroup$
This may be solved using the rearrangement inequality. We have the two sequences
$$
a_1, a_2, a_3,ldots,a_n\
frac1{a_1}, frac1{a_2},frac1{a_3}, ldots,frac1{a_n}
$$
If we multiply each $a_i$ with $frac1{a_i}$, then the sum is $n$. By the rearrangement inequality, this is the smallest possible value we can get, so rearranging must give us a result $geq n$.
Or one may use the AM-GM inequality to get the same result.
answered Mar 17 at 7:13
ArthurArthur
121k7121207
answered Mar 17 at 7:13
ArthurArthur
121k7121207
answered Mar 17 at 7:13
ArthurArthur
121k7121207
answered Mar 17 at 7:13
ArthurArthur
121k7121207
121k7121207
add a comment |
add a comment |
$begingroup$
Hint: Use AM-GM. First will be right answer.
AM-GM inequality
answered Mar 17 at 7:23
Love InvariantsLove Invariants
89015
$endgroup$
add a comment |
$begingroup$
Hint: Use AM-GM. First will be right answer.
AM-GM inequality
answered Mar 17 at 7:23
Love InvariantsLove Invariants
89015
$endgroup$
add a comment |
$begingroup$
Hint: Use AM-GM. First will be right answer.
AM-GM inequality
answered Mar 17 at 7:23
Love InvariantsLove Invariants
89015
$endgroup$
Hint: Use AM-GM. First will be right answer.
AM-GM inequality
answered Mar 17 at 7:23
Love InvariantsLove Invariants
89015
answered Mar 17 at 7:23
Love InvariantsLove Invariants
89015
answered Mar 17 at 7:23
Love InvariantsLove Invariants
89015
answered Mar 17 at 7:23
Love InvariantsLove Invariants
89015
89015
add a comment |
add a comment |
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2
$begingroup$
Use AM-GM. First will be right answer.
$endgroup$
– Love Invariants
Mar 17 at 7:09
$begingroup$
Thank you @Love Invariants, got it..Now I can continue.
$endgroup$
– ankit
Mar 17 at 7:13
$begingroup$
@LoveInvariants Why are you answering in a comment?
$endgroup$
– Arthur
Mar 17 at 7:15
$begingroup$
@Arthur-Because a hint would satisfy. OP should do the rest of the work. It helps askers very much.
$endgroup$
– Love Invariants
Mar 17 at 7:16
$begingroup$
@LoveInvariants Hints are still answers more than they are clarification requests. So they belong in answer posts, not in comments. Also, you have a direct answer to the question by saying which alternative was right, so calling it just a hint isn't quite right.
$endgroup$
– Arthur
Mar 17 at 7:18