Limit of $n left(frac12-e^-nleft(1+n+fracn^22!+…+fracn^nn!right)right)$
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It is known around here that $$lim_ntoinfty e^-nleft(1+n+fracn^22!+...+fracn^nn!right)=frac12$$ So what about the following limit:
$$L=lim_n to inftyn left(frac12-frac1e^nleft(1+n+fracn^22!+...+fracn^nn!right)right)$$ My thought was to apply Stolz-Cesaro theorem, after rewritting as $$L=lim_n to inftyfrac left(frac12-frac1e^nleft(1+n+fracn^22!+...+fracn^nn!right)right)frac1n$$ would produce: $$L=lim_n to inftyfrace^-(n+1)left(1+(n+1)+frac(n+1)^22!+...+frac(n+1)^n+1(n+1)!right)-e^-nleft(1+n+fracn^22!+...+fracn^nn!right)frac1n(n+1)$$
$$=lim_n to inftyfrace^-nleft(frac1-ee +frac1efrac(n+1)^n+1(n+1)! +nleft(fracleft(1+frac1nright)e-1right) +fracn^22!left(fracleft(1+frac1nright)^2e-1right)+cdots +fracn^nn!left(fracleft(1+frac1nright)^ne-1right)right)frac1n(n+1)$$ But I dont see how can I go further, can you help me evaluate it? Also, are limits of this form already known?
limits exponential-function
asked Aug 15 at 10:36
Zacky
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up vote
5
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It is known around here that $$lim_ntoinfty e^-nleft(1+n+fracn^22!+...+fracn^nn!right)=frac12$$ So what about the following limit:
$$L=lim_n to inftyn left(frac12-frac1e^nleft(1+n+fracn^22!+...+fracn^nn!right)right)$$ My thought was to apply Stolz-Cesaro theorem, after rewritting as $$L=lim_n to inftyfrac left(frac12-frac1e^nleft(1+n+fracn^22!+...+fracn^nn!right)right)frac1n$$ would produce: $$L=lim_n to inftyfrace^-(n+1)left(1+(n+1)+frac(n+1)^22!+...+frac(n+1)^n+1(n+1)!right)-e^-nleft(1+n+fracn^22!+...+fracn^nn!right)frac1n(n+1)$$
$$=lim_n to inftyfrace^-nleft(frac1-ee +frac1efrac(n+1)^n+1(n+1)! +nleft(fracleft(1+frac1nright)e-1right) +fracn^22!left(fracleft(1+frac1nright)^2e-1right)+cdots +fracn^nn!left(fracleft(1+frac1nright)^ne-1right)right)frac1n(n+1)$$ But I dont see how can I go further, can you help me evaluate it? Also, are limits of this form already known?
limits exponential-function
asked Aug 15 at 10:36
Zacky
1
Where is that first equation from? Surely $e^-1cdotleft(1+1right)neq frac12$. Or did you mean this as an asymptotic relation?
– Jam
Aug 15 at 10:41
You mean this one: math.stackexchange.com/questions/160248/… ?
– Zacky
Aug 15 at 10:44
1
Right, you should probably put the $lim_nto+infty$ in there, otherwise it isn't true.
– Jam
Aug 15 at 10:45
1
From plotting $y=lfloor xrfloorleft(frac12-e^-lfloor x rfloor cdot sum_i=0^lfloor xrfloorfraclfloor x rfloor ^ii! right)$ in Desmos, I think the expression is divergent. imgur.com/a/t7ENPKv
– Jam
Aug 15 at 10:52
1
Probably not, but you could potentially still get a nice asymptotic relationship out of it since the purple graph appears to approach a certain curve as $xto+infty$. It's often good to graph limits to get a feel for whether they are convergent (although this method isn't necessarily foolproof).
– Jam
Aug 15 at 10:58
 |Â
show 2 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
It is known around here that $$lim_ntoinfty e^-nleft(1+n+fracn^22!+...+fracn^nn!right)=frac12$$ So what about the following limit:
$$L=lim_n to inftyn left(frac12-frac1e^nleft(1+n+fracn^22!+...+fracn^nn!right)right)$$ My thought was to apply Stolz-Cesaro theorem, after rewritting as $$L=lim_n to inftyfrac left(frac12-frac1e^nleft(1+n+fracn^22!+...+fracn^nn!right)right)frac1n$$ would produce: $$L=lim_n to inftyfrace^-(n+1)left(1+(n+1)+frac(n+1)^22!+...+frac(n+1)^n+1(n+1)!right)-e^-nleft(1+n+fracn^22!+...+fracn^nn!right)frac1n(n+1)$$
$$=lim_n to inftyfrace^-nleft(frac1-ee +frac1efrac(n+1)^n+1(n+1)! +nleft(fracleft(1+frac1nright)e-1right) +fracn^22!left(fracleft(1+frac1nright)^2e-1right)+cdots +fracn^nn!left(fracleft(1+frac1nright)^ne-1right)right)frac1n(n+1)$$ But I dont see how can I go further, can you help me evaluate it? Also, are limits of this form already known?
limits exponential-function
asked Aug 15 at 10:36
Zacky
1
It is known around here that $$lim_ntoinfty e^-nleft(1+n+fracn^22!+...+fracn^nn!right)=frac12$$ So what about the following limit:
$$L=lim_n to inftyn left(frac12-frac1e^nleft(1+n+fracn^22!+...+fracn^nn!right)right)$$ My thought was to apply Stolz-Cesaro theorem, after rewritting as $$L=lim_n to inftyfrac left(frac12-frac1e^nleft(1+n+fracn^22!+...+fracn^nn!right)right)frac1n$$ would produce: $$L=lim_n to inftyfrace^-(n+1)left(1+(n+1)+frac(n+1)^22!+...+frac(n+1)^n+1(n+1)!right)-e^-nleft(1+n+fracn^22!+...+fracn^nn!right)frac1n(n+1)$$
$$=lim_n to inftyfrace^-nleft(frac1-ee +frac1efrac(n+1)^n+1(n+1)! +nleft(fracleft(1+frac1nright)e-1right) +fracn^22!left(fracleft(1+frac1nright)^2e-1right)+cdots +fracn^nn!left(fracleft(1+frac1nright)^ne-1right)right)frac1n(n+1)$$ But I dont see how can I go further, can you help me evaluate it? Also, are limits of this form already known?
limits exponential-function
asked Aug 15 at 10:36
Zacky
1
edited Aug 15 at 10:46
asked Aug 15 at 10:36
Zacky
1
asked Aug 15 at 10:36
Zacky
1
asked Aug 15 at 10:36
Zacky
1
1
Where is that first equation from? Surely $e^-1cdotleft(1+1right)neq frac12$. Or did you mean this as an asymptotic relation?
– Jam
Aug 15 at 10:41
You mean this one: math.stackexchange.com/questions/160248/… ?
– Zacky
Aug 15 at 10:44
1
Right, you should probably put the $lim_nto+infty$ in there, otherwise it isn't true.
– Jam
Aug 15 at 10:45
1
From plotting $y=lfloor xrfloorleft(frac12-e^-lfloor x rfloor cdot sum_i=0^lfloor xrfloorfraclfloor x rfloor ^ii! right)$ in Desmos, I think the expression is divergent. imgur.com/a/t7ENPKv
– Jam
Aug 15 at 10:52
1
Probably not, but you could potentially still get a nice asymptotic relationship out of it since the purple graph appears to approach a certain curve as $xto+infty$. It's often good to graph limits to get a feel for whether they are convergent (although this method isn't necessarily foolproof).
– Jam
Aug 15 at 10:58
 |Â
show 2 more comments
Where is that first equation from? Surely $e^-1cdotleft(1+1right)neq frac12$. Or did you mean this as an asymptotic relation?
– Jam
Aug 15 at 10:41
You mean this one: math.stackexchange.com/questions/160248/… ?
– Zacky
Aug 15 at 10:44
1
Right, you should probably put the $lim_nto+infty$ in there, otherwise it isn't true.
– Jam
Aug 15 at 10:45
1
From plotting $y=lfloor xrfloorleft(frac12-e^-lfloor x rfloor cdot sum_i=0^lfloor xrfloorfraclfloor x rfloor ^ii! right)$ in Desmos, I think the expression is divergent. imgur.com/a/t7ENPKv
– Jam
Aug 15 at 10:52
1
Probably not, but you could potentially still get a nice asymptotic relationship out of it since the purple graph appears to approach a certain curve as $xto+infty$. It's often good to graph limits to get a feel for whether they are convergent (although this method isn't necessarily foolproof).
– Jam
Aug 15 at 10:58
Where is that first equation from? Surely $e^-1cdotleft(1+1right)neq frac12$. Or did you mean this as an asymptotic relation?
– Jam
Aug 15 at 10:41
Where is that first equation from? Surely $e^-1cdotleft(1+1right)neq frac12$. Or did you mean this as an asymptotic relation?
– Jam
Aug 15 at 10:41
You mean this one: math.stackexchange.com/questions/160248/… ?
– Zacky
Aug 15 at 10:44
You mean this one: math.stackexchange.com/questions/160248/… ?
– Zacky
Aug 15 at 10:44
1
1
Right, you should probably put the $lim_nto+infty$ in there, otherwise it isn't true.
– Jam
Aug 15 at 10:45
Right, you should probably put the $lim_nto+infty$ in there, otherwise it isn't true.
– Jam
Aug 15 at 10:45
1
1
From plotting $y=lfloor xrfloorleft(frac12-e^-lfloor x rfloor cdot sum_i=0^lfloor xrfloorfraclfloor x rfloor ^ii! right)$ in Desmos, I think the expression is divergent. imgur.com/a/t7ENPKv
– Jam
Aug 15 at 10:52
From plotting $y=lfloor xrfloorleft(frac12-e^-lfloor x rfloor cdot sum_i=0^lfloor xrfloorfraclfloor x rfloor ^ii! right)$ in Desmos, I think the expression is divergent. imgur.com/a/t7ENPKv
– Jam
Aug 15 at 10:52
1
1
Probably not, but you could potentially still get a nice asymptotic relationship out of it since the purple graph appears to approach a certain curve as $xto+infty$. It's often good to graph limits to get a feel for whether they are convergent (although this method isn't necessarily foolproof).
– Jam
Aug 15 at 10:58
Probably not, but you could potentially still get a nice asymptotic relationship out of it since the purple graph appears to approach a certain curve as $xto+infty$. It's often good to graph limits to get a feel for whether they are convergent (although this method isn't necessarily foolproof).
– Jam
Aug 15 at 10:58
 |Â
show 2 more comments
1 Answer
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The follow equation is from
Evaluating $limlimits_ntoinfty e^-n sumlimits_k=0^n fracn^kk!$.
beginalign
e^-nsum_k=0^nfracn^kk!
&=frac1n!int_n^infty e^-t,t^n,mathrmdt\
&=frac12+frac2/3sqrt2pi n+O(n^-1)tag11
endalign
Use this equation we know:
$$lim_ntoinftysqrtn left(e^-n sum_k=0^n fracn^kk!-frac12right)=frac23sqrt2pi .$$
So your limit is $infty$.
answered Aug 15 at 11:31
Riemann
2,4811219
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
The follow equation is from
Evaluating $limlimits_ntoinfty e^-n sumlimits_k=0^n fracn^kk!$.
beginalign
e^-nsum_k=0^nfracn^kk!
&=frac1n!int_n^infty e^-t,t^n,mathrmdt\
&=frac12+frac2/3sqrt2pi n+O(n^-1)tag11
endalign
Use this equation we know:
$$lim_ntoinftysqrtn left(e^-n sum_k=0^n fracn^kk!-frac12right)=frac23sqrt2pi .$$
So your limit is $infty$.
answered Aug 15 at 11:31
Riemann
2,4811219
add a comment |Â
up vote
6
down vote
The follow equation is from
Evaluating $limlimits_ntoinfty e^-n sumlimits_k=0^n fracn^kk!$.
beginalign
e^-nsum_k=0^nfracn^kk!
&=frac1n!int_n^infty e^-t,t^n,mathrmdt\
&=frac12+frac2/3sqrt2pi n+O(n^-1)tag11
endalign
Use this equation we know:
$$lim_ntoinftysqrtn left(e^-n sum_k=0^n fracn^kk!-frac12right)=frac23sqrt2pi .$$
So your limit is $infty$.
answered Aug 15 at 11:31
Riemann
2,4811219
add a comment |Â
up vote
6
down vote
up vote
6
down vote
The follow equation is from
Evaluating $limlimits_ntoinfty e^-n sumlimits_k=0^n fracn^kk!$.
beginalign
e^-nsum_k=0^nfracn^kk!
&=frac1n!int_n^infty e^-t,t^n,mathrmdt\
&=frac12+frac2/3sqrt2pi n+O(n^-1)tag11
endalign
Use this equation we know:
$$lim_ntoinftysqrtn left(e^-n sum_k=0^n fracn^kk!-frac12right)=frac23sqrt2pi .$$
So your limit is $infty$.
answered Aug 15 at 11:31
Riemann
2,4811219
The follow equation is from
Evaluating $limlimits_ntoinfty e^-n sumlimits_k=0^n fracn^kk!$.
beginalign
e^-nsum_k=0^nfracn^kk!
&=frac1n!int_n^infty e^-t,t^n,mathrmdt\
&=frac12+frac2/3sqrt2pi n+O(n^-1)tag11
endalign
Use this equation we know:
$$lim_ntoinftysqrtn left(e^-n sum_k=0^n fracn^kk!-frac12right)=frac23sqrt2pi .$$
So your limit is $infty$.
answered Aug 15 at 11:31
Riemann
2,4811219
answered Aug 15 at 11:31
Riemann
2,4811219
answered Aug 15 at 11:31
Riemann
2,4811219
answered Aug 15 at 11:31
Riemann
2,4811219
2,4811219
add a comment |Â
add a comment |Â
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Where is that first equation from? Surely $e^-1cdotleft(1+1right)neq frac12$. Or did you mean this as an asymptotic relation?
– Jam
Aug 15 at 10:41
You mean this one: math.stackexchange.com/questions/160248/… ?
– Zacky
Aug 15 at 10:44
1
Right, you should probably put the $lim_nto+infty$ in there, otherwise it isn't true.
– Jam
Aug 15 at 10:45
1
From plotting $y=lfloor xrfloorleft(frac12-e^-lfloor x rfloor cdot sum_i=0^lfloor xrfloorfraclfloor x rfloor ^ii! right)$ in Desmos, I think the expression is divergent. imgur.com/a/t7ENPKv
– Jam
Aug 15 at 10:52
1
Probably not, but you could potentially still get a nice asymptotic relationship out of it since the purple graph appears to approach a certain curve as $xto+infty$. It's often good to graph limits to get a feel for whether they are convergent (although this method isn't necessarily foolproof).
– Jam
Aug 15 at 10:58