If there is a positive sequence $left(x_nright)_ngeq1$ , s.t. $limsup_nrightarrowinfty(fracx_1+x_n+kx_n)^n=e^k$? The 2019 Stack Overflow Developer Survey Results Are InShow that $limlimits_n to infty suplimits_k geq n left(frac1+a_k+1a_kright)^k ge e$ for any positive sequence $a_n$Show that $limlimits_n to infty suplimits_k geq n left(frac1+a_k+1a_kright)^k ge e$ for any positive sequence $a_n$Definition of $limsup_$ and $lim_trightarrow0$Sequence $left x_nright _ngeq1$ s.t $left|x_n+1-x_nright|<2^-n$ for all $ngeq N$ , does this imply convergence?Proof $liminf_nrightarrowinftya_n=limsup_nrightarrowinftya_n=lim_nrightarrowinftya_n$True or false? $limsup_nrightarrowinftyleft((n mod 4)+frac1n+1 right )=frac92$An example of a sequence $left a_n right $ such that $lim_nrightarrow infty a_n=infty$Show $limsup_nrightarrow infty (-a_n) = -liminf_nrightarrow infty (a_n)$Prove $limsup_n rightarrow infty a_ngeq1$ if $lim_n rightarrow infty a_na_n+1=1$Show that $limsup_nrightarrowinfty a_n^frac1n leq limsup_nrightarrowinfty (a_n+1 / a_n).$Is $limsup_nrightarrowinftysqrt[n]a_n=frac1R=limsup_nrightarrowinfty|fraca_n+1a_n|$?
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If there is a positive sequence $left(x_nright)_ngeq1$ , s.t. $limsup_nrightarrowinfty(fracx_1+x_n+kx_n)^n=e^k$?
The 2019 Stack Overflow Developer Survey Results Are InShow that $limlimits_n to infty suplimits_k geq n left(frac1+a_k+1a_kright)^k ge e$ for any positive sequence $a_n$Show that $limlimits_n to infty suplimits_k geq n left(frac1+a_k+1a_kright)^k ge e$ for any positive sequence $a_n$Definition of $limsup_t$ and $lim_trightarrow0$Sequence $left x_nright _ngeq1$ s.t $left|x_n+1-x_nright|<2^-n$ for all $ngeq N$ , does this imply convergence?Proof $liminf_nrightarrowinftya_n=limsup_nrightarrowinftya_n=lim_nrightarrowinftya_n$True or false? $limsup_nrightarrowinftyleft((n mod 4)+frac1n+1 right )=frac92$An example of a sequence $left a_n right $ such that $lim_nrightarrow infty a_n=infty$Show $limsup_nrightarrow infty (-a_n) = -liminf_nrightarrow infty (a_n)$Prove $limsup_n rightarrow infty a_ngeq1$ if $lim_n rightarrow infty a_na_n+1=1$Show that $limsup_nrightarrowinfty a_n^frac1n leq limsup_nrightarrowinfty (a_n+1 / a_n).$Is $limsup_nrightarrowinftysqrt[n]=frac1R=limsup_nrightarrowinfty|fraca_n+1a_n|$?
$begingroup$
Let $x_1=varepsilon(>0),x_n=n(ngeq2),$ then $$limsup_nrightarrowinfty(fracx_1+x_n+kx_n)^n=limsup_nrightarrowinfty(1+frack+varepsilonn)^n=e^k+varepsilon>e^k.$$
So we have a sequence of positive numbers$left(x_nright)_ngeq1$ , such that $$limsup_nrightarrowinfty(fracx_1+x_n+kx_n)^n>e^k.$$ In fact,for a given $kinmathbbN,$ any sequence of positive numbers$left(x_nright)_ngeq1$ can satisfy$$limsup_nrightarrowinfty(fracx_1+x_n+kx_n)^ngeq e^k.$$ see here
For any given positive integer $k$,Do we have a sequence of positive numbers$left(x_nright)_ngeq1$ , such that
$$limsup_nrightarrowinfty(fracx_1+x_n+kx_n)^n=e^kquad?$$
real-analysis sequences-and-series limits
asked Mar 20 '17 at 14:08
james givensjames givens
1189
$endgroup$
add a comment |
$begingroup$
Let $x_1=varepsilon(>0),x_n=n(ngeq2),$ then $$limsup_nrightarrowinfty(fracx_1+x_n+kx_n)^n=limsup_nrightarrowinfty(1+frack+varepsilonn)^n=e^k+varepsilon>e^k.$$
So we have a sequence of positive numbers$left(x_nright)_ngeq1$ , such that $$limsup_nrightarrowinfty(fracx_1+x_n+kx_n)^n>e^k.$$ In fact,for a given $kinmathbbN,$ any sequence of positive numbers$left(x_nright)_ngeq1$ can satisfy$$limsup_nrightarrowinfty(fracx_1+x_n+kx_n)^ngeq e^k.$$ see here
For any given positive integer $k$,Do we have a sequence of positive numbers$left(x_nright)_ngeq1$ , such that
$$limsup_nrightarrowinfty(fracx_1+x_n+kx_n)^n=e^kquad?$$
real-analysis sequences-and-series limits
asked Mar 20 '17 at 14:08
james givensjames givens
1189
$endgroup$
add a comment |
$begingroup$
Let $x_1=varepsilon(>0),x_n=n(ngeq2),$ then $$limsup_nrightarrowinfty(fracx_1+x_n+kx_n)^n=limsup_nrightarrowinfty(1+frack+varepsilonn)^n=e^k+varepsilon>e^k.$$
So we have a sequence of positive numbers$left(x_nright)_ngeq1$ , such that $$limsup_nrightarrowinfty(fracx_1+x_n+kx_n)^n>e^k.$$ In fact,for a given $kinmathbbN,$ any sequence of positive numbers$left(x_nright)_ngeq1$ can satisfy$$limsup_nrightarrowinfty(fracx_1+x_n+kx_n)^ngeq e^k.$$ see here
For any given positive integer $k$,Do we have a sequence of positive numbers$left(x_nright)_ngeq1$ , such that
$$limsup_nrightarrowinfty(fracx_1+x_n+kx_n)^n=e^kquad?$$
real-analysis sequences-and-series limits
asked Mar 20 '17 at 14:08
james givensjames givens
1189
$endgroup$
Let $x_1=varepsilon(>0),x_n=n(ngeq2),$ then $$limsup_nrightarrowinfty(fracx_1+x_n+kx_n)^n=limsup_nrightarrowinfty(1+frack+varepsilonn)^n=e^k+varepsilon>e^k.$$
So we have a sequence of positive numbers$left(x_nright)_ngeq1$ , such that $$limsup_nrightarrowinfty(fracx_1+x_n+kx_n)^n>e^k.$$ In fact,for a given $kinmathbbN,$ any sequence of positive numbers$left(x_nright)_ngeq1$ can satisfy$$limsup_nrightarrowinfty(fracx_1+x_n+kx_n)^ngeq e^k.$$ see here
For any given positive integer $k$,Do we have a sequence of positive numbers$left(x_nright)_ngeq1$ , such that
$$limsup_nrightarrowinfty(fracx_1+x_n+kx_n)^n=e^kquad?$$
real-analysis sequences-and-series limits
real-analysis sequences-and-series limits
asked Mar 20 '17 at 14:08
james givensjames givens
1189
asked Mar 20 '17 at 14:08
james givensjames givens
1189
edited Mar 30 at 3:38
james givens
asked Mar 20 '17 at 14:08
james givensjames givens
1189
asked Mar 20 '17 at 14:08
james givensjames givens
1189
asked Mar 20 '17 at 14:08
james givensjames givens
1189
1189
add a comment |
add a comment |
1 Answer
1
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$begingroup$
$$x_n:= n-1implies left(fracx_1+x_n+kx_nright)^n=left(fracn+k-1n-1right)^n=left(1+frac kn-1right)^n=$$
$$=left(1+frac kn-1right)^n-1left(1+frac kn-1right)xrightarrow[ntoinfty]e^kcdot1=e^k$$
If it is very important for you the sequence is positive and not merely non-negative, you can take the above beginning at $;n=2;$ .
answered Mar 20 '17 at 14:21
DonAntonioDonAntonio
180k1494233
$endgroup$
$begingroup$
@K.Kwai I don't understand what you don't understand...perhaps because I didn't understand correctly. Yet I think in my example (1) we have a positive sequence, and (2) I showed that $;limleft(fracx_1+x_n+kx_nright)^n=e^k;$ , which means also the limit supremum and infimum are the same as the sequence converges...DId I miss anything?
$endgroup$
– DonAntonio
Mar 21 '17 at 10:54
$begingroup$
I really appreciate your reply.But in your example,$varliminf(fracx_1+x_n+kx_n)^n=varlimsup(fracx_1+x_n+kx_n)^n=e^1+kne e^k$
$endgroup$
– james givens
Mar 21 '17 at 11:08
$begingroup$
@K.Kwai I can't see how you see such a thing: the limit of a sequence exists iff it is equal to the common value of the suplim and inflim. If my sequence converges then the $;limsup;$ is exactly the same ...!
$endgroup$
– DonAntonio
Mar 21 '17 at 11:12
$begingroup$
I don't know what to say.Indeed,your sequence converges,but converges to $e^k+1.$So the $limsup$ also is $e^k+1,$ not $e^k.$Maybe both of us miss something from each other.
$endgroup$
– james givens
Mar 21 '17 at 11:20
$begingroup$
@K.Kwai Unless I am making a rather stupid mistake (wouldn't be the first time,. though...:)), I am seeing in my answer that the limit is exactly $;e^k;$ ...why do you think it is otherwise?! Can you point in my answer where is there a mistake, or where you can deduce the limit is $;e^k+1;$ ??
$endgroup$
– DonAntonio
Mar 21 '17 at 11:26
|
show 3 more comments
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$begingroup$
$$x_n:= n-1implies left(fracx_1+x_n+kx_nright)^n=left(fracn+k-1n-1right)^n=left(1+frac kn-1right)^n=$$
$$=left(1+frac kn-1right)^n-1left(1+frac kn-1right)xrightarrow[ntoinfty]e^kcdot1=e^k$$
If it is very important for you the sequence is positive and not merely non-negative, you can take the above beginning at $;n=2;$ .
answered Mar 20 '17 at 14:21
DonAntonioDonAntonio
180k1494233
$endgroup$
$begingroup$
@K.Kwai I don't understand what you don't understand...perhaps because I didn't understand correctly. Yet I think in my example (1) we have a positive sequence, and (2) I showed that $;limleft(fracx_1+x_n+kx_nright)^n=e^k;$ , which means also the limit supremum and infimum are the same as the sequence converges...DId I miss anything?
$endgroup$
– DonAntonio
Mar 21 '17 at 10:54
$begingroup$
I really appreciate your reply.But in your example,$varliminf(fracx_1+x_n+kx_n)^n=varlimsup(fracx_1+x_n+kx_n)^n=e^1+kne e^k$
$endgroup$
– james givens
Mar 21 '17 at 11:08
$begingroup$
@K.Kwai I can't see how you see such a thing: the limit of a sequence exists iff it is equal to the common value of the suplim and inflim. If my sequence converges then the $;limsup;$ is exactly the same ...!
$endgroup$
– DonAntonio
Mar 21 '17 at 11:12
$begingroup$
I don't know what to say.Indeed,your sequence converges,but converges to $e^k+1.$So the $limsup$ also is $e^k+1,$ not $e^k.$Maybe both of us miss something from each other.
$endgroup$
– james givens
Mar 21 '17 at 11:20
$begingroup$
@K.Kwai Unless I am making a rather stupid mistake (wouldn't be the first time,. though...:)), I am seeing in my answer that the limit is exactly $;e^k;$ ...why do you think it is otherwise?! Can you point in my answer where is there a mistake, or where you can deduce the limit is $;e^k+1;$ ??
$endgroup$
– DonAntonio
Mar 21 '17 at 11:26
|
show 3 more comments
$begingroup$
$$x_n:= n-1implies left(fracx_1+x_n+kx_nright)^n=left(fracn+k-1n-1right)^n=left(1+frac kn-1right)^n=$$
$$=left(1+frac kn-1right)^n-1left(1+frac kn-1right)xrightarrow[ntoinfty]e^kcdot1=e^k$$
If it is very important for you the sequence is positive and not merely non-negative, you can take the above beginning at $;n=2;$ .
answered Mar 20 '17 at 14:21
DonAntonioDonAntonio
180k1494233
$endgroup$
$begingroup$
@K.Kwai I don't understand what you don't understand...perhaps because I didn't understand correctly. Yet I think in my example (1) we have a positive sequence, and (2) I showed that $;limleft(fracx_1+x_n+kx_nright)^n=e^k;$ , which means also the limit supremum and infimum are the same as the sequence converges...DId I miss anything?
$endgroup$
– DonAntonio
Mar 21 '17 at 10:54
$begingroup$
I really appreciate your reply.But in your example,$varliminf(fracx_1+x_n+kx_n)^n=varlimsup(fracx_1+x_n+kx_n)^n=e^1+kne e^k$
$endgroup$
– james givens
Mar 21 '17 at 11:08
$begingroup$
@K.Kwai I can't see how you see such a thing: the limit of a sequence exists iff it is equal to the common value of the suplim and inflim. If my sequence converges then the $;limsup;$ is exactly the same ...!
$endgroup$
– DonAntonio
Mar 21 '17 at 11:12
$begingroup$
I don't know what to say.Indeed,your sequence converges,but converges to $e^k+1.$So the $limsup$ also is $e^k+1,$ not $e^k.$Maybe both of us miss something from each other.
$endgroup$
– james givens
Mar 21 '17 at 11:20
$begingroup$
@K.Kwai Unless I am making a rather stupid mistake (wouldn't be the first time,. though...:)), I am seeing in my answer that the limit is exactly $;e^k;$ ...why do you think it is otherwise?! Can you point in my answer where is there a mistake, or where you can deduce the limit is $;e^k+1;$ ??
$endgroup$
– DonAntonio
Mar 21 '17 at 11:26
|
show 3 more comments
$begingroup$
$$x_n:= n-1implies left(fracx_1+x_n+kx_nright)^n=left(fracn+k-1n-1right)^n=left(1+frac kn-1right)^n=$$
$$=left(1+frac kn-1right)^n-1left(1+frac kn-1right)xrightarrow[ntoinfty]e^kcdot1=e^k$$
If it is very important for you the sequence is positive and not merely non-negative, you can take the above beginning at $;n=2;$ .
answered Mar 20 '17 at 14:21
DonAntonioDonAntonio
180k1494233
$endgroup$
$$x_n:= n-1implies left(fracx_1+x_n+kx_nright)^n=left(fracn+k-1n-1right)^n=left(1+frac kn-1right)^n=$$
$$=left(1+frac kn-1right)^n-1left(1+frac kn-1right)xrightarrow[ntoinfty]e^kcdot1=e^k$$
If it is very important for you the sequence is positive and not merely non-negative, you can take the above beginning at $;n=2;$ .
answered Mar 20 '17 at 14:21
DonAntonioDonAntonio
180k1494233
answered Mar 20 '17 at 14:21
DonAntonioDonAntonio
180k1494233
answered Mar 20 '17 at 14:21
DonAntonioDonAntonio
180k1494233
answered Mar 20 '17 at 14:21
DonAntonioDonAntonio
180k1494233
180k1494233
$begingroup$
@K.Kwai I don't understand what you don't understand...perhaps because I didn't understand correctly. Yet I think in my example (1) we have a positive sequence, and (2) I showed that $;limleft(fracx_1+x_n+kx_nright)^n=e^k;$ , which means also the limit supremum and infimum are the same as the sequence converges...DId I miss anything?
$endgroup$
– DonAntonio
Mar 21 '17 at 10:54
$begingroup$
I really appreciate your reply.But in your example,$varliminf(fracx_1+x_n+kx_n)^n=varlimsup(fracx_1+x_n+kx_n)^n=e^1+kne e^k$
$endgroup$
– james givens
Mar 21 '17 at 11:08
$begingroup$
@K.Kwai I can't see how you see such a thing: the limit of a sequence exists iff it is equal to the common value of the suplim and inflim. If my sequence converges then the $;limsup;$ is exactly the same ...!
$endgroup$
– DonAntonio
Mar 21 '17 at 11:12
$begingroup$
I don't know what to say.Indeed,your sequence converges,but converges to $e^k+1.$So the $limsup$ also is $e^k+1,$ not $e^k.$Maybe both of us miss something from each other.
$endgroup$
– james givens
Mar 21 '17 at 11:20
$begingroup$
@K.Kwai Unless I am making a rather stupid mistake (wouldn't be the first time,. though...:)), I am seeing in my answer that the limit is exactly $;e^k;$ ...why do you think it is otherwise?! Can you point in my answer where is there a mistake, or where you can deduce the limit is $;e^k+1;$ ??
$endgroup$
– DonAntonio
Mar 21 '17 at 11:26
|
show 3 more comments
$begingroup$
@K.Kwai I don't understand what you don't understand...perhaps because I didn't understand correctly. Yet I think in my example (1) we have a positive sequence, and (2) I showed that $;limleft(fracx_1+x_n+kx_nright)^n=e^k;$ , which means also the limit supremum and infimum are the same as the sequence converges...DId I miss anything?
$endgroup$
– DonAntonio
Mar 21 '17 at 10:54
$begingroup$
I really appreciate your reply.But in your example,$varliminf(fracx_1+x_n+kx_n)^n=varlimsup(fracx_1+x_n+kx_n)^n=e^1+kne e^k$
$endgroup$
– james givens
Mar 21 '17 at 11:08
$begingroup$
@K.Kwai I can't see how you see such a thing: the limit of a sequence exists iff it is equal to the common value of the suplim and inflim. If my sequence converges then the $;limsup;$ is exactly the same ...!
$endgroup$
– DonAntonio
Mar 21 '17 at 11:12
$begingroup$
I don't know what to say.Indeed,your sequence converges,but converges to $e^k+1.$So the $limsup$ also is $e^k+1,$ not $e^k.$Maybe both of us miss something from each other.
$endgroup$
– james givens
Mar 21 '17 at 11:20
$begingroup$
@K.Kwai Unless I am making a rather stupid mistake (wouldn't be the first time,. though...:)), I am seeing in my answer that the limit is exactly $;e^k;$ ...why do you think it is otherwise?! Can you point in my answer where is there a mistake, or where you can deduce the limit is $;e^k+1;$ ??
$endgroup$
– DonAntonio
Mar 21 '17 at 11:26
$begingroup$
@K.Kwai I don't understand what you don't understand...perhaps because I didn't understand correctly. Yet I think in my example (1) we have a positive sequence, and (2) I showed that $;limleft(fracx_1+x_n+kx_nright)^n=e^k;$ , which means also the limit supremum and infimum are the same as the sequence converges...DId I miss anything?
$endgroup$
– DonAntonio
Mar 21 '17 at 10:54
$begingroup$
@K.Kwai I don't understand what you don't understand...perhaps because I didn't understand correctly. Yet I think in my example (1) we have a positive sequence, and (2) I showed that $;limleft(fracx_1+x_n+kx_nright)^n=e^k;$ , which means also the limit supremum and infimum are the same as the sequence converges...DId I miss anything?
$endgroup$
– DonAntonio
Mar 21 '17 at 10:54
$begingroup$
I really appreciate your reply.But in your example,$varliminf(fracx_1+x_n+kx_n)^n=varlimsup(fracx_1+x_n+kx_n)^n=e^1+kne e^k$
$endgroup$
– james givens
Mar 21 '17 at 11:08
$begingroup$
I really appreciate your reply.But in your example,$varliminf(fracx_1+x_n+kx_n)^n=varlimsup(fracx_1+x_n+kx_n)^n=e^1+kne e^k$
$endgroup$
– james givens
Mar 21 '17 at 11:08
$begingroup$
@K.Kwai I can't see how you see such a thing: the limit of a sequence exists iff it is equal to the common value of the suplim and inflim. If my sequence converges then the $;limsup;$ is exactly the same ...!
$endgroup$
– DonAntonio
Mar 21 '17 at 11:12
$begingroup$
@K.Kwai I can't see how you see such a thing: the limit of a sequence exists iff it is equal to the common value of the suplim and inflim. If my sequence converges then the $;limsup;$ is exactly the same ...!
$endgroup$
– DonAntonio
Mar 21 '17 at 11:12
$begingroup$
I don't know what to say.Indeed,your sequence converges,but converges to $e^k+1.$So the $limsup$ also is $e^k+1,$ not $e^k.$Maybe both of us miss something from each other.
$endgroup$
– james givens
Mar 21 '17 at 11:20
$begingroup$
I don't know what to say.Indeed,your sequence converges,but converges to $e^k+1.$So the $limsup$ also is $e^k+1,$ not $e^k.$Maybe both of us miss something from each other.
$endgroup$
– james givens
Mar 21 '17 at 11:20
$begingroup$
@K.Kwai Unless I am making a rather stupid mistake (wouldn't be the first time,. though...:)), I am seeing in my answer that the limit is exactly $;e^k;$ ...why do you think it is otherwise?! Can you point in my answer where is there a mistake, or where you can deduce the limit is $;e^k+1;$ ??
$endgroup$
– DonAntonio
Mar 21 '17 at 11:26
$begingroup$
@K.Kwai Unless I am making a rather stupid mistake (wouldn't be the first time,. though...:)), I am seeing in my answer that the limit is exactly $;e^k;$ ...why do you think it is otherwise?! Can you point in my answer where is there a mistake, or where you can deduce the limit is $;e^k+1;$ ??
$endgroup$
– DonAntonio
Mar 21 '17 at 11:26
|
show 3 more comments
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