$(frac{n}{e})^{n} < n! < (frac{n}{e} + nvarepsilon)^{n}$ doesn't comply with the limit definition?












2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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.
    $endgroup$
    – 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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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.
    $endgroup$
    – 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














2












2








2





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 20:49







dpd

















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














  • 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










1 Answer
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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 $epsilon$.






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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 $epsilon$.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      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 $epsilon$.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        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 $epsilon$.






        share|cite|improve this answer









        $endgroup$



        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 $epsilon$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 10 at 20:42









        JoeJoe

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