$(frac{n}{e})^{n} < n! < (frac{n}{e} + nvarepsilon)^{n}$ doesn't comply with the limit definition?
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I try to understand what I've overlooked, when I came up with this inequality:
First, we have this limit:
$$limlimits_{n to infty} sqrt[n]{frac{n!}{n^n}} = frac{1}{e}$$
Which gives, by the definition of limit and some simple transformations:
$frac{1}{e} - varepsilon < sqrt[n]{frac{n!}{n^n}} < frac{1}{e} + varepsilon$
$(frac{n}{e} - nvarepsilon)^{n} < n! < (frac{n}{e} + nvarepsilon)^{n}quadforall varepsilon > 0$
Then, we have this well-known inequality (multiple proofs can be found on math.stackexchange):
$$(frac{n}{e})^{n} < n!$$
So we have:
$$(frac{n}{e})^{n} < n! < (frac{n}{e} + nvarepsilon)^{n}$$
According to this inequality, we cannot make $varepsilon$ arbitrary small, which contradicts the definition of limit.
What am I missing here?
sequences-and-series limits factorial
asked Jan 10 at 20:36
dpddpd
134
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add a comment |
$begingroup$
I try to understand what I've overlooked, when I came up with this inequality:
First, we have this limit:
$$limlimits_{n to infty} sqrt[n]{frac{n!}{n^n}} = frac{1}{e}$$
Which gives, by the definition of limit and some simple transformations:
$frac{1}{e} - varepsilon < sqrt[n]{frac{n!}{n^n}} < frac{1}{e} + varepsilon$
$(frac{n}{e} - nvarepsilon)^{n} < n! < (frac{n}{e} + nvarepsilon)^{n}quadforall varepsilon > 0$
Then, we have this well-known inequality (multiple proofs can be found on math.stackexchange):
$$(frac{n}{e})^{n} < n!$$
So we have:
$$(frac{n}{e})^{n} < n! < (frac{n}{e} + nvarepsilon)^{n}$$
According to this inequality, we cannot make $varepsilon$ arbitrary small, which contradicts the definition of limit.
What am I missing here?
sequences-and-series limits factorial
asked Jan 10 at 20:36
dpddpd
134
$endgroup$
1
$begingroup$
What do you mean by $exp$? As far as I know, it's meant to be a function, but you're not giving it an input.
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– Calvin Godfrey
Jan 10 at 20:39
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@CalvinGodfrey, they just mean $e$.
$endgroup$
– Joe
Jan 10 at 20:40
3
$begingroup$
@Joe In that case $exp(1)$ would be more accurate.
$endgroup$
– cansomeonehelpmeout
Jan 10 at 20:41
$begingroup$
@cansomeonehelpmeout, yes that would have been more appropriate. I'm just explaining what OP meant. I absolutely do not condone their notation.
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– Joe
Jan 10 at 20:43
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@mfl, thanks, I was stuck because I didn't consider that all 3 of expressions in this inequality are sequences and thought in terms of fixed n
$endgroup$
– dpd
Jan 10 at 21:02
add a comment |
$begingroup$
I try to understand what I've overlooked, when I came up with this inequality:
First, we have this limit:
$$limlimits_{n to infty} sqrt[n]{frac{n!}{n^n}} = frac{1}{e}$$
Which gives, by the definition of limit and some simple transformations:
$frac{1}{e} - varepsilon < sqrt[n]{frac{n!}{n^n}} < frac{1}{e} + varepsilon$
$(frac{n}{e} - nvarepsilon)^{n} < n! < (frac{n}{e} + nvarepsilon)^{n}quadforall varepsilon > 0$
Then, we have this well-known inequality (multiple proofs can be found on math.stackexchange):
$$(frac{n}{e})^{n} < n!$$
So we have:
$$(frac{n}{e})^{n} < n! < (frac{n}{e} + nvarepsilon)^{n}$$
According to this inequality, we cannot make $varepsilon$ arbitrary small, which contradicts the definition of limit.
What am I missing here?
sequences-and-series limits factorial
asked Jan 10 at 20:36
dpddpd
134
$endgroup$
I try to understand what I've overlooked, when I came up with this inequality:
First, we have this limit:
$$limlimits_{n to infty} sqrt[n]{frac{n!}{n^n}} = frac{1}{e}$$
Which gives, by the definition of limit and some simple transformations:
$frac{1}{e} - varepsilon < sqrt[n]{frac{n!}{n^n}} < frac{1}{e} + varepsilon$
$(frac{n}{e} - nvarepsilon)^{n} < n! < (frac{n}{e} + nvarepsilon)^{n}quadforall varepsilon > 0$
Then, we have this well-known inequality (multiple proofs can be found on math.stackexchange):
$$(frac{n}{e})^{n} < n!$$
So we have:
$$(frac{n}{e})^{n} < n! < (frac{n}{e} + nvarepsilon)^{n}$$
According to this inequality, we cannot make $varepsilon$ arbitrary small, which contradicts the definition of limit.
What am I missing here?
sequences-and-series limits factorial
sequences-and-series limits factorial
asked Jan 10 at 20:36
dpddpd
134
asked Jan 10 at 20:36
dpddpd
134
edited Jan 10 at 20:49
dpd
asked Jan 10 at 20:36
dpddpd
134
asked Jan 10 at 20:36
dpddpd
134
asked Jan 10 at 20:36
dpddpd
134
134
1
$begingroup$
What do you mean by $exp$? As far as I know, it's meant to be a function, but you're not giving it an input.
$endgroup$
– Calvin Godfrey
Jan 10 at 20:39
$begingroup$
@CalvinGodfrey, they just mean $e$.
$endgroup$
– Joe
Jan 10 at 20:40
3
$begingroup$
@Joe In that case $exp(1)$ would be more accurate.
$endgroup$
– cansomeonehelpmeout
Jan 10 at 20:41
$begingroup$
@cansomeonehelpmeout, yes that would have been more appropriate. I'm just explaining what OP meant. I absolutely do not condone their notation.
$endgroup$
– Joe
Jan 10 at 20:43
$begingroup$
@mfl, thanks, I was stuck because I didn't consider that all 3 of expressions in this inequality are sequences and thought in terms of fixed n
$endgroup$
– dpd
Jan 10 at 21:02
add a comment |
1
$begingroup$
What do you mean by $exp$? As far as I know, it's meant to be a function, but you're not giving it an input.
$endgroup$
– Calvin Godfrey
Jan 10 at 20:39
$begingroup$
@CalvinGodfrey, they just mean $e$.
$endgroup$
– Joe
Jan 10 at 20:40
3
$begingroup$
@Joe In that case $exp(1)$ would be more accurate.
$endgroup$
– cansomeonehelpmeout
Jan 10 at 20:41
$begingroup$
@cansomeonehelpmeout, yes that would have been more appropriate. I'm just explaining what OP meant. I absolutely do not condone their notation.
$endgroup$
– Joe
Jan 10 at 20:43
$begingroup$
@mfl, thanks, I was stuck because I didn't consider that all 3 of expressions in this inequality are sequences and thought in terms of fixed n
$endgroup$
– dpd
Jan 10 at 21:02
1
1
$begingroup$
What do you mean by $exp$? As far as I know, it's meant to be a function, but you're not giving it an input.
$endgroup$
– Calvin Godfrey
Jan 10 at 20:39
$begingroup$
What do you mean by $exp$? As far as I know, it's meant to be a function, but you're not giving it an input.
$endgroup$
– Calvin Godfrey
Jan 10 at 20:39
$begingroup$
@CalvinGodfrey, they just mean $e$.
$endgroup$
– Joe
Jan 10 at 20:40
$begingroup$
@CalvinGodfrey, they just mean $e$.
$endgroup$
– Joe
Jan 10 at 20:40
3
3
$begingroup$
@Joe In that case $exp(1)$ would be more accurate.
$endgroup$
– cansomeonehelpmeout
Jan 10 at 20:41
$begingroup$
@Joe In that case $exp(1)$ would be more accurate.
$endgroup$
– cansomeonehelpmeout
Jan 10 at 20:41
$begingroup$
@cansomeonehelpmeout, yes that would have been more appropriate. I'm just explaining what OP meant. I absolutely do not condone their notation.
$endgroup$
– Joe
Jan 10 at 20:43
$begingroup$
@cansomeonehelpmeout, yes that would have been more appropriate. I'm just explaining what OP meant. I absolutely do not condone their notation.
$endgroup$
– Joe
Jan 10 at 20:43
$begingroup$
@mfl, thanks, I was stuck because I didn't consider that all 3 of expressions in this inequality are sequences and thought in terms of fixed n
$endgroup$
– dpd
Jan 10 at 21:02
$begingroup$
@mfl, thanks, I was stuck because I didn't consider that all 3 of expressions in this inequality are sequences and thought in terms of fixed n
$endgroup$
– dpd
Jan 10 at 21:02
add a comment |
1 Answer
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Actually yes, we can make $epsilon$ arbitrarily small. Note that your argument is a limit argument, meaning it doesn't hold for every $n$. It only holds for all $n geq N$, where $N$ depends on $epsilon$.
answered Jan 10 at 20:42
JoeJoe
75129
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$begingroup$
Actually yes, we can make $epsilon$ arbitrarily small. Note that your argument is a limit argument, meaning it doesn't hold for every $n$. It only holds for all $n geq N$, where $N$ depends on $epsilon$.
answered Jan 10 at 20:42
JoeJoe
75129
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add a comment |
$begingroup$
Actually yes, we can make $epsilon$ arbitrarily small. Note that your argument is a limit argument, meaning it doesn't hold for every $n$. It only holds for all $n geq N$, where $N$ depends on $epsilon$.
answered Jan 10 at 20:42
JoeJoe
75129
$endgroup$
add a comment |
$begingroup$
Actually yes, we can make $epsilon$ arbitrarily small. Note that your argument is a limit argument, meaning it doesn't hold for every $n$. It only holds for all $n geq N$, where $N$ depends on $epsilon$.
answered Jan 10 at 20:42
JoeJoe
75129
$endgroup$
Actually yes, we can make $epsilon$ arbitrarily small. Note that your argument is a limit argument, meaning it doesn't hold for every $n$. It only holds for all $n geq N$, where $N$ depends on $epsilon$.
answered Jan 10 at 20:42
JoeJoe
75129
answered Jan 10 at 20:42
JoeJoe
75129
answered Jan 10 at 20:42
JoeJoe
75129
answered Jan 10 at 20:42
JoeJoe
75129
75129
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1
$begingroup$
What do you mean by $exp$? As far as I know, it's meant to be a function, but you're not giving it an input.
$endgroup$
– Calvin Godfrey
Jan 10 at 20:39
$begingroup$
@CalvinGodfrey, they just mean $e$.
$endgroup$
– Joe
Jan 10 at 20:40
3
$begingroup$
@Joe In that case $exp(1)$ would be more accurate.
$endgroup$
– cansomeonehelpmeout
Jan 10 at 20:41
$begingroup$
@cansomeonehelpmeout, yes that would have been more appropriate. I'm just explaining what OP meant. I absolutely do not condone their notation.
$endgroup$
– Joe
Jan 10 at 20:43
$begingroup$
@mfl, thanks, I was stuck because I didn't consider that all 3 of expressions in this inequality are sequences and thought in terms of fixed n
$endgroup$
– dpd
Jan 10 at 21:02