Bound for $(1-frac{1}{n})^t$
I'm having trouble proving that:
For any constant $epsilon > 0$ and $n > 1$:
$$
left(1-frac{1}{n}right)^{n lgleft(n^{epsilon}right)} leq frac{1}{n^{epsilon}}$$
I'm using $lg(n)$ as $log_2(n)$. Any help is appreciated, hints for critical points of the proof are welcome as well.
calculus inequality upper-lower-bounds
asked Nov 25 '18 at 9:47
figure09
11
add a comment |
I'm having trouble proving that:
For any constant $epsilon > 0$ and $n > 1$:
$$
left(1-frac{1}{n}right)^{n lgleft(n^{epsilon}right)} leq frac{1}{n^{epsilon}}$$
I'm using $lg(n)$ as $log_2(n)$. Any help is appreciated, hints for critical points of the proof are welcome as well.
calculus inequality upper-lower-bounds
asked Nov 25 '18 at 9:47
figure09
11
Take $log_2$ on both sides. $implies n:lg(n^{varepsilon}):lgleft(1+dfrac1nright)leq-lg(n^{varepsilon})$.
– Yadati Kiran
Nov 25 '18 at 9:56
Observe $left(1+dfrac1nright)<2:forall:n>1implies lgleft(1+dfrac1nright)<0.$
– Yadati Kiran
Nov 25 '18 at 10:01
Thank you, sir!
– figure09
Nov 25 '18 at 10:21
$left(1-dfrac1nright)$ and not $left(1+dfrac1nright)$. For being sure check $displaystyle lim_{ntoinfty}n:lgleft(1-dfrac1nright)$.
– Yadati Kiran
Nov 25 '18 at 10:49
add a comment |
I'm having trouble proving that:
For any constant $epsilon > 0$ and $n > 1$:
$$
left(1-frac{1}{n}right)^{n lgleft(n^{epsilon}right)} leq frac{1}{n^{epsilon}}$$
I'm using $lg(n)$ as $log_2(n)$. Any help is appreciated, hints for critical points of the proof are welcome as well.
calculus inequality upper-lower-bounds
asked Nov 25 '18 at 9:47
figure09
11
I'm having trouble proving that:
For any constant $epsilon > 0$ and $n > 1$:
$$
left(1-frac{1}{n}right)^{n lgleft(n^{epsilon}right)} leq frac{1}{n^{epsilon}}$$
I'm using $lg(n)$ as $log_2(n)$. Any help is appreciated, hints for critical points of the proof are welcome as well.
calculus inequality upper-lower-bounds
calculus inequality upper-lower-bounds
asked Nov 25 '18 at 9:47
figure09
11
asked Nov 25 '18 at 9:47
figure09
11
edited Nov 25 '18 at 9:57
asked Nov 25 '18 at 9:47
figure09
11
asked Nov 25 '18 at 9:47
figure09
11
asked Nov 25 '18 at 9:47
figure09
11
11
Take $log_2$ on both sides. $implies n:lg(n^{varepsilon}):lgleft(1+dfrac1nright)leq-lg(n^{varepsilon})$.
– Yadati Kiran
Nov 25 '18 at 9:56
Observe $left(1+dfrac1nright)<2:forall:n>1implies lgleft(1+dfrac1nright)<0.$
– Yadati Kiran
Nov 25 '18 at 10:01
Thank you, sir!
– figure09
Nov 25 '18 at 10:21
$left(1-dfrac1nright)$ and not $left(1+dfrac1nright)$. For being sure check $displaystyle lim_{ntoinfty}n:lgleft(1-dfrac1nright)$.
– Yadati Kiran
Nov 25 '18 at 10:49
add a comment |
Take $log_2$ on both sides. $implies n:lg(n^{varepsilon}):lgleft(1+dfrac1nright)leq-lg(n^{varepsilon})$.
– Yadati Kiran
Nov 25 '18 at 9:56
Observe $left(1+dfrac1nright)<2:forall:n>1implies lgleft(1+dfrac1nright)<0.$
– Yadati Kiran
Nov 25 '18 at 10:01
Thank you, sir!
– figure09
Nov 25 '18 at 10:21
$left(1-dfrac1nright)$ and not $left(1+dfrac1nright)$. For being sure check $displaystyle lim_{ntoinfty}n:lgleft(1-dfrac1nright)$.
– Yadati Kiran
Nov 25 '18 at 10:49
Take $log_2$ on both sides. $implies n:lg(n^{varepsilon}):lgleft(1+dfrac1nright)leq-lg(n^{varepsilon})$.
– Yadati Kiran
Nov 25 '18 at 9:56
Take $log_2$ on both sides. $implies n:lg(n^{varepsilon}):lgleft(1+dfrac1nright)leq-lg(n^{varepsilon})$.
– Yadati Kiran
Nov 25 '18 at 9:56
Observe $left(1+dfrac1nright)<2:forall:n>1implies lgleft(1+dfrac1nright)<0.$
– Yadati Kiran
Nov 25 '18 at 10:01
Observe $left(1+dfrac1nright)<2:forall:n>1implies lgleft(1+dfrac1nright)<0.$
– Yadati Kiran
Nov 25 '18 at 10:01
Thank you, sir!
– figure09
Nov 25 '18 at 10:21
Thank you, sir!
– figure09
Nov 25 '18 at 10:21
$left(1-dfrac1nright)$ and not $left(1+dfrac1nright)$. For being sure check $displaystyle lim_{ntoinfty}n:lgleft(1-dfrac1nright)$.
– Yadati Kiran
Nov 25 '18 at 10:49
$left(1-dfrac1nright)$ and not $left(1+dfrac1nright)$. For being sure check $displaystyle lim_{ntoinfty}n:lgleft(1-dfrac1nright)$.
– Yadati Kiran
Nov 25 '18 at 10:49
add a comment |
1 Answer
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Hint :
The $ln$ function is strictly increasing and defined for numbers $>0$. Since these hold in your expression, it is :
$$bigg(1-frac{1}{n}bigg)^{nln{n^epsilon}} leq frac{1}{n^epsilon} Rightarrow lnbigg(1-frac{1}{n}bigg)^{nln{n^epsilon}} leq lnbigg(frac{1}{n^epsilon}bigg)$$
$$Leftrightarrow$$
$$ncdot epsilon ln n cdot lnbigg(1-frac{1}{n}bigg)leq -epsilonln n$$
$$Leftrightarrow$$
$$bigg[nlnbigg(1-frac{1}{n}bigg)-1bigg]epsilonln n leq 0$$
answered Nov 25 '18 at 10:42
Rebellos
14.4k31245
add a comment |
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1 Answer
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1 Answer
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active
oldest
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Hint :
The $ln$ function is strictly increasing and defined for numbers $>0$. Since these hold in your expression, it is :
$$bigg(1-frac{1}{n}bigg)^{nln{n^epsilon}} leq frac{1}{n^epsilon} Rightarrow lnbigg(1-frac{1}{n}bigg)^{nln{n^epsilon}} leq lnbigg(frac{1}{n^epsilon}bigg)$$
$$Leftrightarrow$$
$$ncdot epsilon ln n cdot lnbigg(1-frac{1}{n}bigg)leq -epsilonln n$$
$$Leftrightarrow$$
$$bigg[nlnbigg(1-frac{1}{n}bigg)-1bigg]epsilonln n leq 0$$
answered Nov 25 '18 at 10:42
Rebellos
14.4k31245
add a comment |
Hint :
The $ln$ function is strictly increasing and defined for numbers $>0$. Since these hold in your expression, it is :
$$bigg(1-frac{1}{n}bigg)^{nln{n^epsilon}} leq frac{1}{n^epsilon} Rightarrow lnbigg(1-frac{1}{n}bigg)^{nln{n^epsilon}} leq lnbigg(frac{1}{n^epsilon}bigg)$$
$$Leftrightarrow$$
$$ncdot epsilon ln n cdot lnbigg(1-frac{1}{n}bigg)leq -epsilonln n$$
$$Leftrightarrow$$
$$bigg[nlnbigg(1-frac{1}{n}bigg)-1bigg]epsilonln n leq 0$$
answered Nov 25 '18 at 10:42
Rebellos
14.4k31245
add a comment |
Hint :
The $ln$ function is strictly increasing and defined for numbers $>0$. Since these hold in your expression, it is :
$$bigg(1-frac{1}{n}bigg)^{nln{n^epsilon}} leq frac{1}{n^epsilon} Rightarrow lnbigg(1-frac{1}{n}bigg)^{nln{n^epsilon}} leq lnbigg(frac{1}{n^epsilon}bigg)$$
$$Leftrightarrow$$
$$ncdot epsilon ln n cdot lnbigg(1-frac{1}{n}bigg)leq -epsilonln n$$
$$Leftrightarrow$$
$$bigg[nlnbigg(1-frac{1}{n}bigg)-1bigg]epsilonln n leq 0$$
answered Nov 25 '18 at 10:42
Rebellos
14.4k31245
Hint :
The $ln$ function is strictly increasing and defined for numbers $>0$. Since these hold in your expression, it is :
$$bigg(1-frac{1}{n}bigg)^{nln{n^epsilon}} leq frac{1}{n^epsilon} Rightarrow lnbigg(1-frac{1}{n}bigg)^{nln{n^epsilon}} leq lnbigg(frac{1}{n^epsilon}bigg)$$
$$Leftrightarrow$$
$$ncdot epsilon ln n cdot lnbigg(1-frac{1}{n}bigg)leq -epsilonln n$$
$$Leftrightarrow$$
$$bigg[nlnbigg(1-frac{1}{n}bigg)-1bigg]epsilonln n leq 0$$
answered Nov 25 '18 at 10:42
Rebellos
14.4k31245
answered Nov 25 '18 at 10:42
Rebellos
14.4k31245
answered Nov 25 '18 at 10:42
Rebellos
14.4k31245
answered Nov 25 '18 at 10:42
Rebellos
14.4k31245
14.4k31245
add a comment |
add a comment |
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Take $log_2$ on both sides. $implies n:lg(n^{varepsilon}):lgleft(1+dfrac1nright)leq-lg(n^{varepsilon})$.
– Yadati Kiran
Nov 25 '18 at 9:56
Observe $left(1+dfrac1nright)<2:forall:n>1implies lgleft(1+dfrac1nright)<0.$
– Yadati Kiran
Nov 25 '18 at 10:01
Thank you, sir!
– figure09
Nov 25 '18 at 10:21
$left(1-dfrac1nright)$ and not $left(1+dfrac1nright)$. For being sure check $displaystyle lim_{ntoinfty}n:lgleft(1-dfrac1nright)$.
– Yadati Kiran
Nov 25 '18 at 10:49