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What is $lim_n to infty (1 - (1/n)^y)^n $?
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I'm trying to evaluate the following. Say $y in mathbbR$ is given,
$lim_n to infty (1 - (1/n)^y)^n $
I tried:
Only found some similarity with the formula for $e^-1$. That is,
$lim_n to infty (1 - (1/n))^n = e^-1$
limits
asked Jul 18 at 23:58
moreblue
1738
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up vote
0
down vote
favorite
1
I'm trying to evaluate the following. Say $y in mathbbR$ is given,
$lim_n to infty (1 - (1/n)^y)^n $
I tried:
Only found some similarity with the formula for $e^-1$. That is,
$lim_n to infty (1 - (1/n))^n = e^-1$
limits
asked Jul 18 at 23:58
moreblue
1738
What is $y$...?
– amsmath
Jul 19 at 0:00
@amsmath oh, I wrote $y$ as a given real number.
– moreblue
Jul 19 at 0:05
You will need to distinguish cases: $y>0$, $y=0$, and $y<0$. For the most interesting case, $y>0$, write it as $[(1-n^-y)^-n^y]^-n^1-y$, and take into account that $(1-n^-y)^-n^yto e$.
– user574889
Jul 19 at 0:08
add a comment |Â
up vote
0
down vote
favorite
1
up vote
0
down vote
favorite
1
1
I'm trying to evaluate the following. Say $y in mathbbR$ is given,
$lim_n to infty (1 - (1/n)^y)^n $
I tried:
Only found some similarity with the formula for $e^-1$. That is,
$lim_n to infty (1 - (1/n))^n = e^-1$
limits
asked Jul 18 at 23:58
moreblue
1738
I'm trying to evaluate the following. Say $y in mathbbR$ is given,
$lim_n to infty (1 - (1/n)^y)^n $
I tried:
Only found some similarity with the formula for $e^-1$. That is,
$lim_n to infty (1 - (1/n))^n = e^-1$
limits
asked Jul 18 at 23:58
moreblue
1738
edited Jul 19 at 0:05
asked Jul 18 at 23:58
moreblue
1738
asked Jul 18 at 23:58
moreblue
1738
asked Jul 18 at 23:58
moreblue
1738
1738
What is $y$...?
– amsmath
Jul 19 at 0:00
@amsmath oh, I wrote $y$ as a given real number.
– moreblue
Jul 19 at 0:05
You will need to distinguish cases: $y>0$, $y=0$, and $y<0$. For the most interesting case, $y>0$, write it as $[(1-n^-y)^-n^y]^-n^1-y$, and take into account that $(1-n^-y)^-n^yto e$.
– user574889
Jul 19 at 0:08
add a comment |Â
What is $y$...?
– amsmath
Jul 19 at 0:00
@amsmath oh, I wrote $y$ as a given real number.
– moreblue
Jul 19 at 0:05
You will need to distinguish cases: $y>0$, $y=0$, and $y<0$. For the most interesting case, $y>0$, write it as $[(1-n^-y)^-n^y]^-n^1-y$, and take into account that $(1-n^-y)^-n^yto e$.
– user574889
Jul 19 at 0:08
What is $y$...?
– amsmath
Jul 19 at 0:00
What is $y$...?
– amsmath
Jul 19 at 0:00
@amsmath oh, I wrote $y$ as a given real number.
– moreblue
Jul 19 at 0:05
@amsmath oh, I wrote $y$ as a given real number.
– moreblue
Jul 19 at 0:05
You will need to distinguish cases: $y>0$, $y=0$, and $y<0$. For the most interesting case, $y>0$, write it as $[(1-n^-y)^-n^y]^-n^1-y$, and take into account that $(1-n^-y)^-n^yto e$.
– user574889
Jul 19 at 0:08
You will need to distinguish cases: $y>0$, $y=0$, and $y<0$. For the most interesting case, $y>0$, write it as $[(1-n^-y)^-n^y]^-n^1-y$, and take into account that $(1-n^-y)^-n^yto e$.
– user574889
Jul 19 at 0:08
add a comment |Â
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What is $y$...?
– amsmath
Jul 19 at 0:00
@amsmath oh, I wrote $y$ as a given real number.
– moreblue
Jul 19 at 0:05
You will need to distinguish cases: $y>0$, $y=0$, and $y<0$. For the most interesting case, $y>0$, write it as $[(1-n^-y)^-n^y]^-n^1-y$, and take into account that $(1-n^-y)^-n^yto e$.
– user574889
Jul 19 at 0:08