Mixing
$frac{1}{15}<(frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot cdot cdot cdotfrac{99}{100})<frac{1}{10}$.
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Show that
$$frac{1}{15}<(frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot cdot cdotcdotfrac{99}{100})<frac{1}{10}$$
My attempt:
This problem form text book were are introduce https://en.wikipedia.org/wiki/Generalized_mean
This wouldn't be a problem if I know the sum or a product of the given numbers. Then I will use AG inequality, but I don't know.
inequality
edited 1 hour ago
idea
2,0112923
asked 1 hour ago
Lovro Sindičić
234216
add a comment |
up vote
3
down vote
favorite
Show that
$$frac{1}{15}<(frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot cdot cdotcdotfrac{99}{100})<frac{1}{10}$$
My attempt:
This problem form text book were are introduce https://en.wikipedia.org/wiki/Generalized_mean
This wouldn't be a problem if I know the sum or a product of the given numbers. Then I will use AG inequality, but I don't know.
inequality
edited 1 hour ago
idea
2,0112923
asked 1 hour ago
Lovro Sindičić
234216
1
Related
– Kemono Chen
41 mins ago
In fact the link above has tighter bounds, so answers this.
– Macavity
32 mins ago
Note that the middle term is the coefficient of $x^{50}$ in $(x+tfrac14)^{100}$.
– Servaes
28 mins ago
1
But this question is still worth answering because there is a method showing $frac2{3sqrt n}<frac12cdotfrac34cdotsfrac{(2n-1)}{2n}<frac1{sqrt n}$ and it's not suitable for that question (unfortunately I forgot that) :(
– Kemono Chen
26 mins ago
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Show that
$$frac{1}{15}<(frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot cdot cdotcdotfrac{99}{100})<frac{1}{10}$$
My attempt:
This problem form text book were are introduce https://en.wikipedia.org/wiki/Generalized_mean
This wouldn't be a problem if I know the sum or a product of the given numbers. Then I will use AG inequality, but I don't know.
inequality
edited 1 hour ago
idea
2,0112923
asked 1 hour ago
Lovro Sindičić
234216
Show that
$$frac{1}{15}<(frac{1}{2}cdotfrac{3}{4}cdotfrac{5}{6}cdot cdot cdotcdotfrac{99}{100})<frac{1}{10}$$
My attempt:
This problem form text book were are introduce https://en.wikipedia.org/wiki/Generalized_mean
This wouldn't be a problem if I know the sum or a product of the given numbers. Then I will use AG inequality, but I don't know.
inequality
inequality
edited 1 hour ago
idea
2,0112923
asked 1 hour ago
Lovro Sindičić
234216
edited 1 hour ago
idea
2,0112923
asked 1 hour ago
Lovro Sindičić
234216
edited 1 hour ago
idea
2,0112923
edited 1 hour ago
idea
2,0112923
edited 1 hour ago
idea
2,0112923
2,0112923
asked 1 hour ago
Lovro Sindičić
234216
asked 1 hour ago
Lovro Sindičić
234216
asked 1 hour ago
Lovro Sindičić
234216
234216
1
Related
– Kemono Chen
41 mins ago
In fact the link above has tighter bounds, so answers this.
– Macavity
32 mins ago
Note that the middle term is the coefficient of $x^{50}$ in $(x+tfrac14)^{100}$.
– Servaes
28 mins ago
1
But this question is still worth answering because there is a method showing $frac2{3sqrt n}<frac12cdotfrac34cdotsfrac{(2n-1)}{2n}<frac1{sqrt n}$ and it's not suitable for that question (unfortunately I forgot that) :(
– Kemono Chen
26 mins ago
add a comment |
1
Related
– Kemono Chen
41 mins ago
In fact the link above has tighter bounds, so answers this.
– Macavity
32 mins ago
Note that the middle term is the coefficient of $x^{50}$ in $(x+tfrac14)^{100}$.
– Servaes
28 mins ago
1
But this question is still worth answering because there is a method showing $frac2{3sqrt n}<frac12cdotfrac34cdotsfrac{(2n-1)}{2n}<frac1{sqrt n}$ and it's not suitable for that question (unfortunately I forgot that) :(
– Kemono Chen
26 mins ago
1
1
Related
– Kemono Chen
41 mins ago
Related
– Kemono Chen
41 mins ago
In fact the link above has tighter bounds, so answers this.
– Macavity
32 mins ago
In fact the link above has tighter bounds, so answers this.
– Macavity
32 mins ago
Note that the middle term is the coefficient of $x^{50}$ in $(x+tfrac14)^{100}$.
– Servaes
28 mins ago
Note that the middle term is the coefficient of $x^{50}$ in $(x+tfrac14)^{100}$.
– Servaes
28 mins ago
1
1
But this question is still worth answering because there is a method showing $frac2{3sqrt n}<frac12cdotfrac34cdotsfrac{(2n-1)}{2n}<frac1{sqrt n}$ and it's not suitable for that question (unfortunately I forgot that) :(
– Kemono Chen
26 mins ago
But this question is still worth answering because there is a method showing $frac2{3sqrt n}<frac12cdotfrac34cdotsfrac{(2n-1)}{2n}<frac1{sqrt n}$ and it's not suitable for that question (unfortunately I forgot that) :(
– Kemono Chen
26 mins ago
add a comment |
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1
Related
– Kemono Chen
41 mins ago
In fact the link above has tighter bounds, so answers this.
– Macavity
32 mins ago
Note that the middle term is the coefficient of $x^{50}$ in $(x+tfrac14)^{100}$.
– Servaes
28 mins ago
1
But this question is still worth answering because there is a method showing $frac2{3sqrt n}<frac12cdotfrac34cdotsfrac{(2n-1)}{2n}<frac1{sqrt n}$ and it's not suitable for that question (unfortunately I forgot that) :(
– Kemono Chen
26 mins ago