Finding the sup and inf of the following sequence
$begingroup$
I am trying to find the $underline{s}=liminf s_n$ and $overline{s}=limsup s_n$ where $s_n=r^{2n}/(1+r^n)$.
$a)$ If $r=1:overline{s}=underline{s}=1/2$
$b)$ If $|r|<1:overline{s}=underline{s}=0$
$c)$ If $r>1:overline{s}=infty,quadunderline{s}=1/2$
$d)$ If $r<-1:overline{s}=infty,quadunderline{s}=-infty$
However, apparently the solution for $(c)$ is $,overline{s}=underline{s}=infty$, but I do not understand why.
Another example is for $s_n=n^2-n$ which I stated that $overline{s}=infty,quadunderline{s}=0$, but again the book states that $,overline{s}=underline{s}=infty$.
real-analysis sequences-and-series
asked Dec 18 '18 at 17:52
DMH16DMH16
592317
$endgroup$
add a comment |
$begingroup$
I am trying to find the $underline{s}=liminf s_n$ and $overline{s}=limsup s_n$ where $s_n=r^{2n}/(1+r^n)$.
$a)$ If $r=1:overline{s}=underline{s}=1/2$
$b)$ If $|r|<1:overline{s}=underline{s}=0$
$c)$ If $r>1:overline{s}=infty,quadunderline{s}=1/2$
$d)$ If $r<-1:overline{s}=infty,quadunderline{s}=-infty$
However, apparently the solution for $(c)$ is $,overline{s}=underline{s}=infty$, but I do not understand why.
Another example is for $s_n=n^2-n$ which I stated that $overline{s}=infty,quadunderline{s}=0$, but again the book states that $,overline{s}=underline{s}=infty$.
real-analysis sequences-and-series
asked Dec 18 '18 at 17:52
DMH16DMH16
592317
$endgroup$
add a comment |
$begingroup$
I am trying to find the $underline{s}=liminf s_n$ and $overline{s}=limsup s_n$ where $s_n=r^{2n}/(1+r^n)$.
$a)$ If $r=1:overline{s}=underline{s}=1/2$
$b)$ If $|r|<1:overline{s}=underline{s}=0$
$c)$ If $r>1:overline{s}=infty,quadunderline{s}=1/2$
$d)$ If $r<-1:overline{s}=infty,quadunderline{s}=-infty$
However, apparently the solution for $(c)$ is $,overline{s}=underline{s}=infty$, but I do not understand why.
Another example is for $s_n=n^2-n$ which I stated that $overline{s}=infty,quadunderline{s}=0$, but again the book states that $,overline{s}=underline{s}=infty$.
real-analysis sequences-and-series
asked Dec 18 '18 at 17:52
DMH16DMH16
592317
$endgroup$
I am trying to find the $underline{s}=liminf s_n$ and $overline{s}=limsup s_n$ where $s_n=r^{2n}/(1+r^n)$.
$a)$ If $r=1:overline{s}=underline{s}=1/2$
$b)$ If $|r|<1:overline{s}=underline{s}=0$
$c)$ If $r>1:overline{s}=infty,quadunderline{s}=1/2$
$d)$ If $r<-1:overline{s}=infty,quadunderline{s}=-infty$
However, apparently the solution for $(c)$ is $,overline{s}=underline{s}=infty$, but I do not understand why.
Another example is for $s_n=n^2-n$ which I stated that $overline{s}=infty,quadunderline{s}=0$, but again the book states that $,overline{s}=underline{s}=infty$.
real-analysis sequences-and-series
real-analysis sequences-and-series
asked Dec 18 '18 at 17:52
DMH16DMH16
592317
asked Dec 18 '18 at 17:52
DMH16DMH16
592317
asked Dec 18 '18 at 17:52
DMH16DMH16
592317
asked Dec 18 '18 at 17:52
DMH16DMH16
592317
asked Dec 18 '18 at 17:52
DMH16DMH16
592317
592317
add a comment |
add a comment |
1 Answer
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$begingroup$
Note that if $r>1$, then
$$
frac{r^{2n}}{1+r^n}=frac{r^{n}}{1+r^{-n}}to infty
$$
as $nto infty$. In particular every subsequence $(s_{n_k})$ is such that $s_{n_k}to infty$. So the infimum of the set of subsequential limits (which is $liminf s_n$) and supremum of the set of subsequential limits(which is $limsup s_n$) are both equal to infinity.
answered Dec 18 '18 at 17:57
Foobaz JohnFoobaz John
22.8k41452
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1 Answer
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1 Answer
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$begingroup$
Note that if $r>1$, then
$$
frac{r^{2n}}{1+r^n}=frac{r^{n}}{1+r^{-n}}to infty
$$
as $nto infty$. In particular every subsequence $(s_{n_k})$ is such that $s_{n_k}to infty$. So the infimum of the set of subsequential limits (which is $liminf s_n$) and supremum of the set of subsequential limits(which is $limsup s_n$) are both equal to infinity.
answered Dec 18 '18 at 17:57
Foobaz JohnFoobaz John
22.8k41452
$endgroup$
add a comment |
$begingroup$
Note that if $r>1$, then
$$
frac{r^{2n}}{1+r^n}=frac{r^{n}}{1+r^{-n}}to infty
$$
as $nto infty$. In particular every subsequence $(s_{n_k})$ is such that $s_{n_k}to infty$. So the infimum of the set of subsequential limits (which is $liminf s_n$) and supremum of the set of subsequential limits(which is $limsup s_n$) are both equal to infinity.
answered Dec 18 '18 at 17:57
Foobaz JohnFoobaz John
22.8k41452
$endgroup$
add a comment |
$begingroup$
Note that if $r>1$, then
$$
frac{r^{2n}}{1+r^n}=frac{r^{n}}{1+r^{-n}}to infty
$$
as $nto infty$. In particular every subsequence $(s_{n_k})$ is such that $s_{n_k}to infty$. So the infimum of the set of subsequential limits (which is $liminf s_n$) and supremum of the set of subsequential limits(which is $limsup s_n$) are both equal to infinity.
answered Dec 18 '18 at 17:57
Foobaz JohnFoobaz John
22.8k41452
$endgroup$
Note that if $r>1$, then
$$
frac{r^{2n}}{1+r^n}=frac{r^{n}}{1+r^{-n}}to infty
$$
as $nto infty$. In particular every subsequence $(s_{n_k})$ is such that $s_{n_k}to infty$. So the infimum of the set of subsequential limits (which is $liminf s_n$) and supremum of the set of subsequential limits(which is $limsup s_n$) are both equal to infinity.
answered Dec 18 '18 at 17:57
Foobaz JohnFoobaz John
22.8k41452
answered Dec 18 '18 at 17:57
Foobaz JohnFoobaz John
22.8k41452
answered Dec 18 '18 at 17:57
Foobaz JohnFoobaz John
22.8k41452
answered Dec 18 '18 at 17:57
Foobaz JohnFoobaz John
22.8k41452
22.8k41452
add a comment |
add a comment |
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