limit of $left(1+frac{1}{n!}right)^n$
$begingroup$
I'm having trouble resolving the following limit: $$ lim_{n to infty} left(1+frac{1}{n!}right)^n $$ Intuituvely the limit is equal to 1, but the exercises requires me to resolve via calculation and I have no idea how I can accomplish this.
Can someone please explain it to me?
limits factorial
asked Nov 29 '18 at 15:51
SceptwoSceptwo
236
$endgroup$
add a comment |
$begingroup$
I'm having trouble resolving the following limit: $$ lim_{n to infty} left(1+frac{1}{n!}right)^n $$ Intuituvely the limit is equal to 1, but the exercises requires me to resolve via calculation and I have no idea how I can accomplish this.
Can someone please explain it to me?
limits factorial
asked Nov 29 '18 at 15:51
SceptwoSceptwo
236
$endgroup$
add a comment |
$begingroup$
I'm having trouble resolving the following limit: $$ lim_{n to infty} left(1+frac{1}{n!}right)^n $$ Intuituvely the limit is equal to 1, but the exercises requires me to resolve via calculation and I have no idea how I can accomplish this.
Can someone please explain it to me?
limits factorial
asked Nov 29 '18 at 15:51
SceptwoSceptwo
236
$endgroup$
I'm having trouble resolving the following limit: $$ lim_{n to infty} left(1+frac{1}{n!}right)^n $$ Intuituvely the limit is equal to 1, but the exercises requires me to resolve via calculation and I have no idea how I can accomplish this.
Can someone please explain it to me?
limits factorial
limits factorial
asked Nov 29 '18 at 15:51
SceptwoSceptwo
236
asked Nov 29 '18 at 15:51
SceptwoSceptwo
236
edited Nov 29 '18 at 16:14
Sceptwo
asked Nov 29 '18 at 15:51
SceptwoSceptwo
236
asked Nov 29 '18 at 15:51
SceptwoSceptwo
236
asked Nov 29 '18 at 15:51
SceptwoSceptwo
236
236
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
HINT
We have
$$lim_{n to infty} left(1+frac{1}{n!}right)^n=lim_{n to infty} left[left(1+frac{1}{n!}right)^{n!}right]^{frac1{(n-1)!}}$$
then refer to standard limit for $e$. How can we conclude form here?
answered Nov 29 '18 at 15:53
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
I thought it might be instructive to present a solution that relies on elementary inequalities and the squeeze theorem only. To that end, we now proceed.
Using the inequality $1+xle frac1{1-x}$, for $x<1$, along with Bernoulli's Inequality, we have for $nge 2$
$$1le left(1+frac1{n!}right)^nle frac1{left(1-frac1{n!}right)^n}le frac1{1-frac1{(n-1)!}}$$
whereby application of the squeeze theorem yields the coveted limit.
answered Nov 29 '18 at 16:00
Mark ViolaMark Viola
131k1275171
$endgroup$
1
$begingroup$
It is a very elegant way which only make use of simple results, the only issue is to have great the idea to start with the inequality $1+xle frac1{1-x}$ as $(x<1)$ and take is as a given.
$endgroup$
– gimusi
Nov 29 '18 at 17:00
1
$begingroup$
Well, for all $x$, $x^2ge 0$. Hence, $(1+x)(1-x)=1-x^2le 1$. For $x<1$, $1+xle frac1{1-x}$. So, that is a fairly standard "trick."
$endgroup$
– Mark Viola
Nov 29 '18 at 17:04
$begingroup$
Yes indeed the proof is simple maybe it is less obviuos have the idea to strat with it. Of course it is a good trick to keep in mind :)
$endgroup$
– gimusi
Nov 29 '18 at 17:07
1
$begingroup$
Yes, this "trick" can be used, for example, to show that $exp(x)exp(y)=exp(x+y)$ where we use the limit characterization $exp(x)=lim_{ntoinfty}left(1+frac xnright)^n$ of the exponential function.
$endgroup$
– Mark Viola
Nov 29 '18 at 17:09
add a comment |
$begingroup$
It suffices to compute the limit of the logarithm and show that it equals zero. Namely note that
$$
nlogleft(1+frac{1}{n!}right)=ntimes frac{1}{n!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}
$$
Now let $nto infty$. Using the fact that
$$
lim_{uto 0}frac{log(1+u)-log1}{u-0}=1
$$
by definition of the derivative we get that
$$
frac{1}{(n-1)!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}to0times 1=0
$$
as $nto infty$.
answered Nov 29 '18 at 16:01
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
add a comment |
$begingroup$
Note that
$$
frac{1}{n!}<frac{1}{n(n-1)}<frac{1}{(n-1)^2}
$$
Now prove that
$$
lim_{tto0}frac{log(1+t^2)}{t}=0
$$
and observe that
begin{align}
left(1+frac{1}{(n-1)^2}right)^n&=
left(1+frac{1}{(n-1)^2}right)left(1+frac{1}{(n-1)^2}right)^{n-1}
\[6px]
&=left(1+frac{1}{(n-1)^2}right)expleft(fleft(frac{1}{n-1}right)right)
end{align}
Finally note that
$$
1leleft(1+frac{1}{n!}right)^n
$$
answered Nov 29 '18 at 16:05
egregegreg
180k1485202
$endgroup$
add a comment |
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Since Mark Viola has already used the elementary approach of Bernoulli inequality here is one more approach via binomial theorem.
We use the following lemma of Thomas Andrews (proved using binomial theorem):
Lemma: If ${a_n} $ is a sequence of real or complex terms such that $n(a_n-1)to 0$ then $a_n^nto 1$.
Now use $a_n=1+(1/n!)$ and check that $n(a_n-1)=1/(n-1)!to 0$. The desired limit is $1$ by the above lemma.
answered Nov 30 '18 at 2:52
Paramanand SinghParamanand Singh
49.4k555162
$endgroup$
$begingroup$
(+1) Nice use of that "lemma."
$endgroup$
– Mark Viola
Nov 30 '18 at 15:23
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT
We have
$$lim_{n to infty} left(1+frac{1}{n!}right)^n=lim_{n to infty} left[left(1+frac{1}{n!}right)^{n!}right]^{frac1{(n-1)!}}$$
then refer to standard limit for $e$. How can we conclude form here?
answered Nov 29 '18 at 15:53
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
HINT
We have
$$lim_{n to infty} left(1+frac{1}{n!}right)^n=lim_{n to infty} left[left(1+frac{1}{n!}right)^{n!}right]^{frac1{(n-1)!}}$$
then refer to standard limit for $e$. How can we conclude form here?
answered Nov 29 '18 at 15:53
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
HINT
We have
$$lim_{n to infty} left(1+frac{1}{n!}right)^n=lim_{n to infty} left[left(1+frac{1}{n!}right)^{n!}right]^{frac1{(n-1)!}}$$
then refer to standard limit for $e$. How can we conclude form here?
answered Nov 29 '18 at 15:53
gimusigimusi
1
$endgroup$
HINT
We have
$$lim_{n to infty} left(1+frac{1}{n!}right)^n=lim_{n to infty} left[left(1+frac{1}{n!}right)^{n!}right]^{frac1{(n-1)!}}$$
then refer to standard limit for $e$. How can we conclude form here?
answered Nov 29 '18 at 15:53
gimusigimusi
1
edited Nov 29 '18 at 15:54
answered Nov 29 '18 at 15:53
gimusigimusi
1
answered Nov 29 '18 at 15:53
gimusigimusi
1
answered Nov 29 '18 at 15:53
gimusigimusi
1
1
add a comment |
add a comment |
$begingroup$
I thought it might be instructive to present a solution that relies on elementary inequalities and the squeeze theorem only. To that end, we now proceed.
Using the inequality $1+xle frac1{1-x}$, for $x<1$, along with Bernoulli's Inequality, we have for $nge 2$
$$1le left(1+frac1{n!}right)^nle frac1{left(1-frac1{n!}right)^n}le frac1{1-frac1{(n-1)!}}$$
whereby application of the squeeze theorem yields the coveted limit.
answered Nov 29 '18 at 16:00
Mark ViolaMark Viola
131k1275171
$endgroup$
1
$begingroup$
It is a very elegant way which only make use of simple results, the only issue is to have great the idea to start with the inequality $1+xle frac1{1-x}$ as $(x<1)$ and take is as a given.
$endgroup$
– gimusi
Nov 29 '18 at 17:00
1
$begingroup$
Well, for all $x$, $x^2ge 0$. Hence, $(1+x)(1-x)=1-x^2le 1$. For $x<1$, $1+xle frac1{1-x}$. So, that is a fairly standard "trick."
$endgroup$
– Mark Viola
Nov 29 '18 at 17:04
$begingroup$
Yes indeed the proof is simple maybe it is less obviuos have the idea to strat with it. Of course it is a good trick to keep in mind :)
$endgroup$
– gimusi
Nov 29 '18 at 17:07
1
$begingroup$
Yes, this "trick" can be used, for example, to show that $exp(x)exp(y)=exp(x+y)$ where we use the limit characterization $exp(x)=lim_{ntoinfty}left(1+frac xnright)^n$ of the exponential function.
$endgroup$
– Mark Viola
Nov 29 '18 at 17:09
add a comment |
$begingroup$
I thought it might be instructive to present a solution that relies on elementary inequalities and the squeeze theorem only. To that end, we now proceed.
Using the inequality $1+xle frac1{1-x}$, for $x<1$, along with Bernoulli's Inequality, we have for $nge 2$
$$1le left(1+frac1{n!}right)^nle frac1{left(1-frac1{n!}right)^n}le frac1{1-frac1{(n-1)!}}$$
whereby application of the squeeze theorem yields the coveted limit.
answered Nov 29 '18 at 16:00
Mark ViolaMark Viola
131k1275171
$endgroup$
1
$begingroup$
It is a very elegant way which only make use of simple results, the only issue is to have great the idea to start with the inequality $1+xle frac1{1-x}$ as $(x<1)$ and take is as a given.
$endgroup$
– gimusi
Nov 29 '18 at 17:00
1
$begingroup$
Well, for all $x$, $x^2ge 0$. Hence, $(1+x)(1-x)=1-x^2le 1$. For $x<1$, $1+xle frac1{1-x}$. So, that is a fairly standard "trick."
$endgroup$
– Mark Viola
Nov 29 '18 at 17:04
$begingroup$
Yes indeed the proof is simple maybe it is less obviuos have the idea to strat with it. Of course it is a good trick to keep in mind :)
$endgroup$
– gimusi
Nov 29 '18 at 17:07
1
$begingroup$
Yes, this "trick" can be used, for example, to show that $exp(x)exp(y)=exp(x+y)$ where we use the limit characterization $exp(x)=lim_{ntoinfty}left(1+frac xnright)^n$ of the exponential function.
$endgroup$
– Mark Viola
Nov 29 '18 at 17:09
add a comment |
$begingroup$
I thought it might be instructive to present a solution that relies on elementary inequalities and the squeeze theorem only. To that end, we now proceed.
Using the inequality $1+xle frac1{1-x}$, for $x<1$, along with Bernoulli's Inequality, we have for $nge 2$
$$1le left(1+frac1{n!}right)^nle frac1{left(1-frac1{n!}right)^n}le frac1{1-frac1{(n-1)!}}$$
whereby application of the squeeze theorem yields the coveted limit.
answered Nov 29 '18 at 16:00
Mark ViolaMark Viola
131k1275171
$endgroup$
I thought it might be instructive to present a solution that relies on elementary inequalities and the squeeze theorem only. To that end, we now proceed.
Using the inequality $1+xle frac1{1-x}$, for $x<1$, along with Bernoulli's Inequality, we have for $nge 2$
$$1le left(1+frac1{n!}right)^nle frac1{left(1-frac1{n!}right)^n}le frac1{1-frac1{(n-1)!}}$$
whereby application of the squeeze theorem yields the coveted limit.
answered Nov 29 '18 at 16:00
Mark ViolaMark Viola
131k1275171
edited Nov 29 '18 at 17:07
answered Nov 29 '18 at 16:00
Mark ViolaMark Viola
131k1275171
answered Nov 29 '18 at 16:00
Mark ViolaMark Viola
131k1275171
answered Nov 29 '18 at 16:00
Mark ViolaMark Viola
131k1275171
131k1275171
1
$begingroup$
It is a very elegant way which only make use of simple results, the only issue is to have great the idea to start with the inequality $1+xle frac1{1-x}$ as $(x<1)$ and take is as a given.
$endgroup$
– gimusi
Nov 29 '18 at 17:00
1
$begingroup$
Well, for all $x$, $x^2ge 0$. Hence, $(1+x)(1-x)=1-x^2le 1$. For $x<1$, $1+xle frac1{1-x}$. So, that is a fairly standard "trick."
$endgroup$
– Mark Viola
Nov 29 '18 at 17:04
$begingroup$
Yes indeed the proof is simple maybe it is less obviuos have the idea to strat with it. Of course it is a good trick to keep in mind :)
$endgroup$
– gimusi
Nov 29 '18 at 17:07
1
$begingroup$
Yes, this "trick" can be used, for example, to show that $exp(x)exp(y)=exp(x+y)$ where we use the limit characterization $exp(x)=lim_{ntoinfty}left(1+frac xnright)^n$ of the exponential function.
$endgroup$
– Mark Viola
Nov 29 '18 at 17:09
add a comment |
1
$begingroup$
It is a very elegant way which only make use of simple results, the only issue is to have great the idea to start with the inequality $1+xle frac1{1-x}$ as $(x<1)$ and take is as a given.
$endgroup$
– gimusi
Nov 29 '18 at 17:00
1
$begingroup$
Well, for all $x$, $x^2ge 0$. Hence, $(1+x)(1-x)=1-x^2le 1$. For $x<1$, $1+xle frac1{1-x}$. So, that is a fairly standard "trick."
$endgroup$
– Mark Viola
Nov 29 '18 at 17:04
$begingroup$
Yes indeed the proof is simple maybe it is less obviuos have the idea to strat with it. Of course it is a good trick to keep in mind :)
$endgroup$
– gimusi
Nov 29 '18 at 17:07
1
$begingroup$
Yes, this "trick" can be used, for example, to show that $exp(x)exp(y)=exp(x+y)$ where we use the limit characterization $exp(x)=lim_{ntoinfty}left(1+frac xnright)^n$ of the exponential function.
$endgroup$
– Mark Viola
Nov 29 '18 at 17:09
1
1
$begingroup$
It is a very elegant way which only make use of simple results, the only issue is to have great the idea to start with the inequality $1+xle frac1{1-x}$ as $(x<1)$ and take is as a given.
$endgroup$
– gimusi
Nov 29 '18 at 17:00
$begingroup$
It is a very elegant way which only make use of simple results, the only issue is to have great the idea to start with the inequality $1+xle frac1{1-x}$ as $(x<1)$ and take is as a given.
$endgroup$
– gimusi
Nov 29 '18 at 17:00
1
1
$begingroup$
Well, for all $x$, $x^2ge 0$. Hence, $(1+x)(1-x)=1-x^2le 1$. For $x<1$, $1+xle frac1{1-x}$. So, that is a fairly standard "trick."
$endgroup$
– Mark Viola
Nov 29 '18 at 17:04
$begingroup$
Well, for all $x$, $x^2ge 0$. Hence, $(1+x)(1-x)=1-x^2le 1$. For $x<1$, $1+xle frac1{1-x}$. So, that is a fairly standard "trick."
$endgroup$
– Mark Viola
Nov 29 '18 at 17:04
$begingroup$
Yes indeed the proof is simple maybe it is less obviuos have the idea to strat with it. Of course it is a good trick to keep in mind :)
$endgroup$
– gimusi
Nov 29 '18 at 17:07
$begingroup$
Yes indeed the proof is simple maybe it is less obviuos have the idea to strat with it. Of course it is a good trick to keep in mind :)
$endgroup$
– gimusi
Nov 29 '18 at 17:07
1
1
$begingroup$
Yes, this "trick" can be used, for example, to show that $exp(x)exp(y)=exp(x+y)$ where we use the limit characterization $exp(x)=lim_{ntoinfty}left(1+frac xnright)^n$ of the exponential function.
$endgroup$
– Mark Viola
Nov 29 '18 at 17:09
$begingroup$
Yes, this "trick" can be used, for example, to show that $exp(x)exp(y)=exp(x+y)$ where we use the limit characterization $exp(x)=lim_{ntoinfty}left(1+frac xnright)^n$ of the exponential function.
$endgroup$
– Mark Viola
Nov 29 '18 at 17:09
add a comment |
$begingroup$
It suffices to compute the limit of the logarithm and show that it equals zero. Namely note that
$$
nlogleft(1+frac{1}{n!}right)=ntimes frac{1}{n!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}
$$
Now let $nto infty$. Using the fact that
$$
lim_{uto 0}frac{log(1+u)-log1}{u-0}=1
$$
by definition of the derivative we get that
$$
frac{1}{(n-1)!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}to0times 1=0
$$
as $nto infty$.
answered Nov 29 '18 at 16:01
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
add a comment |
$begingroup$
It suffices to compute the limit of the logarithm and show that it equals zero. Namely note that
$$
nlogleft(1+frac{1}{n!}right)=ntimes frac{1}{n!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}
$$
Now let $nto infty$. Using the fact that
$$
lim_{uto 0}frac{log(1+u)-log1}{u-0}=1
$$
by definition of the derivative we get that
$$
frac{1}{(n-1)!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}to0times 1=0
$$
as $nto infty$.
answered Nov 29 '18 at 16:01
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
add a comment |
$begingroup$
It suffices to compute the limit of the logarithm and show that it equals zero. Namely note that
$$
nlogleft(1+frac{1}{n!}right)=ntimes frac{1}{n!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}
$$
Now let $nto infty$. Using the fact that
$$
lim_{uto 0}frac{log(1+u)-log1}{u-0}=1
$$
by definition of the derivative we get that
$$
frac{1}{(n-1)!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}to0times 1=0
$$
as $nto infty$.
answered Nov 29 '18 at 16:01
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
It suffices to compute the limit of the logarithm and show that it equals zero. Namely note that
$$
nlogleft(1+frac{1}{n!}right)=ntimes frac{1}{n!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}
$$
Now let $nto infty$. Using the fact that
$$
lim_{uto 0}frac{log(1+u)-log1}{u-0}=1
$$
by definition of the derivative we get that
$$
frac{1}{(n-1)!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}to0times 1=0
$$
as $nto infty$.
answered Nov 29 '18 at 16:01
Foobaz JohnFoobaz John
21.6k41352
answered Nov 29 '18 at 16:01
Foobaz JohnFoobaz John
21.6k41352
answered Nov 29 '18 at 16:01
Foobaz JohnFoobaz John
21.6k41352
answered Nov 29 '18 at 16:01
Foobaz JohnFoobaz John
21.6k41352
21.6k41352
add a comment |
add a comment |
$begingroup$
Note that
$$
frac{1}{n!}<frac{1}{n(n-1)}<frac{1}{(n-1)^2}
$$
Now prove that
$$
lim_{tto0}frac{log(1+t^2)}{t}=0
$$
and observe that
begin{align}
left(1+frac{1}{(n-1)^2}right)^n&=
left(1+frac{1}{(n-1)^2}right)left(1+frac{1}{(n-1)^2}right)^{n-1}
\[6px]
&=left(1+frac{1}{(n-1)^2}right)expleft(fleft(frac{1}{n-1}right)right)
end{align}
Finally note that
$$
1leleft(1+frac{1}{n!}right)^n
$$
answered Nov 29 '18 at 16:05
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
Note that
$$
frac{1}{n!}<frac{1}{n(n-1)}<frac{1}{(n-1)^2}
$$
Now prove that
$$
lim_{tto0}frac{log(1+t^2)}{t}=0
$$
and observe that
begin{align}
left(1+frac{1}{(n-1)^2}right)^n&=
left(1+frac{1}{(n-1)^2}right)left(1+frac{1}{(n-1)^2}right)^{n-1}
\[6px]
&=left(1+frac{1}{(n-1)^2}right)expleft(fleft(frac{1}{n-1}right)right)
end{align}
Finally note that
$$
1leleft(1+frac{1}{n!}right)^n
$$
answered Nov 29 '18 at 16:05
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
Note that
$$
frac{1}{n!}<frac{1}{n(n-1)}<frac{1}{(n-1)^2}
$$
Now prove that
$$
lim_{tto0}frac{log(1+t^2)}{t}=0
$$
and observe that
begin{align}
left(1+frac{1}{(n-1)^2}right)^n&=
left(1+frac{1}{(n-1)^2}right)left(1+frac{1}{(n-1)^2}right)^{n-1}
\[6px]
&=left(1+frac{1}{(n-1)^2}right)expleft(fleft(frac{1}{n-1}right)right)
end{align}
Finally note that
$$
1leleft(1+frac{1}{n!}right)^n
$$
answered Nov 29 '18 at 16:05
egregegreg
180k1485202
$endgroup$
Note that
$$
frac{1}{n!}<frac{1}{n(n-1)}<frac{1}{(n-1)^2}
$$
Now prove that
$$
lim_{tto0}frac{log(1+t^2)}{t}=0
$$
and observe that
begin{align}
left(1+frac{1}{(n-1)^2}right)^n&=
left(1+frac{1}{(n-1)^2}right)left(1+frac{1}{(n-1)^2}right)^{n-1}
\[6px]
&=left(1+frac{1}{(n-1)^2}right)expleft(fleft(frac{1}{n-1}right)right)
end{align}
Finally note that
$$
1leleft(1+frac{1}{n!}right)^n
$$
answered Nov 29 '18 at 16:05
egregegreg
180k1485202
edited Nov 29 '18 at 16:45
answered Nov 29 '18 at 16:05
egregegreg
180k1485202
answered Nov 29 '18 at 16:05
egregegreg
180k1485202
answered Nov 29 '18 at 16:05
egregegreg
180k1485202
180k1485202
add a comment |
add a comment |
$begingroup$
Since Mark Viola has already used the elementary approach of Bernoulli inequality here is one more approach via binomial theorem.
We use the following lemma of Thomas Andrews (proved using binomial theorem):
Lemma: If ${a_n} $ is a sequence of real or complex terms such that $n(a_n-1)to 0$ then $a_n^nto 1$.
Now use $a_n=1+(1/n!)$ and check that $n(a_n-1)=1/(n-1)!to 0$. The desired limit is $1$ by the above lemma.
answered Nov 30 '18 at 2:52
Paramanand SinghParamanand Singh
49.4k555162
$endgroup$
$begingroup$
(+1) Nice use of that "lemma."
$endgroup$
– Mark Viola
Nov 30 '18 at 15:23
add a comment |
$begingroup$
Since Mark Viola has already used the elementary approach of Bernoulli inequality here is one more approach via binomial theorem.
We use the following lemma of Thomas Andrews (proved using binomial theorem):
Lemma: If ${a_n} $ is a sequence of real or complex terms such that $n(a_n-1)to 0$ then $a_n^nto 1$.
Now use $a_n=1+(1/n!)$ and check that $n(a_n-1)=1/(n-1)!to 0$. The desired limit is $1$ by the above lemma.
answered Nov 30 '18 at 2:52
Paramanand SinghParamanand Singh
49.4k555162
$endgroup$
$begingroup$
(+1) Nice use of that "lemma."
$endgroup$
– Mark Viola
Nov 30 '18 at 15:23
add a comment |
$begingroup$
Since Mark Viola has already used the elementary approach of Bernoulli inequality here is one more approach via binomial theorem.
We use the following lemma of Thomas Andrews (proved using binomial theorem):
Lemma: If ${a_n} $ is a sequence of real or complex terms such that $n(a_n-1)to 0$ then $a_n^nto 1$.
Now use $a_n=1+(1/n!)$ and check that $n(a_n-1)=1/(n-1)!to 0$. The desired limit is $1$ by the above lemma.
answered Nov 30 '18 at 2:52
Paramanand SinghParamanand Singh
49.4k555162
$endgroup$
Since Mark Viola has already used the elementary approach of Bernoulli inequality here is one more approach via binomial theorem.
We use the following lemma of Thomas Andrews (proved using binomial theorem):
Lemma: If ${a_n} $ is a sequence of real or complex terms such that $n(a_n-1)to 0$ then $a_n^nto 1$.
Now use $a_n=1+(1/n!)$ and check that $n(a_n-1)=1/(n-1)!to 0$. The desired limit is $1$ by the above lemma.
answered Nov 30 '18 at 2:52
Paramanand SinghParamanand Singh
49.4k555162
answered Nov 30 '18 at 2:52
Paramanand SinghParamanand Singh
49.4k555162
answered Nov 30 '18 at 2:52
Paramanand SinghParamanand Singh
49.4k555162
answered Nov 30 '18 at 2:52
Paramanand SinghParamanand Singh
49.4k555162
49.4k555162
$begingroup$
(+1) Nice use of that "lemma."
$endgroup$
– Mark Viola
Nov 30 '18 at 15:23
add a comment |
$begingroup$
(+1) Nice use of that "lemma."
$endgroup$
– Mark Viola
Nov 30 '18 at 15:23
$begingroup$
(+1) Nice use of that "lemma."
$endgroup$
– Mark Viola
Nov 30 '18 at 15:23
$begingroup$
(+1) Nice use of that "lemma."
$endgroup$
– Mark Viola
Nov 30 '18 at 15:23
add a comment |
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