Clarification on step in proof for Ratio test
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My thoughts: So apart from this step, the rest of the proof is fairly simple. Now in terms of lim sup, I know that it is just equal to the regular limit for convergent sequences and for bounded, divergent sequences it is what the "peaks" of the sequence converges towards. Of course there is a formal definition as well but I would like to have an intiutive understanding as well.
However I am not sure how to deduce the the inequality between the fraction and t from the definition of limsup... Can someone help me? Am I missing some key property of lim sup?
sequences-and-series convergence limsup-and-liminf
asked Dec 20 '18 at 11:06
CruZCruZ
638
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add a comment |
$begingroup$
My thoughts: So apart from this step, the rest of the proof is fairly simple. Now in terms of lim sup, I know that it is just equal to the regular limit for convergent sequences and for bounded, divergent sequences it is what the "peaks" of the sequence converges towards. Of course there is a formal definition as well but I would like to have an intiutive understanding as well.
However I am not sure how to deduce the the inequality between the fraction and t from the definition of limsup... Can someone help me? Am I missing some key property of lim sup?
sequences-and-series convergence limsup-and-liminf
asked Dec 20 '18 at 11:06
CruZCruZ
638
$endgroup$
add a comment |
$begingroup$
My thoughts: So apart from this step, the rest of the proof is fairly simple. Now in terms of lim sup, I know that it is just equal to the regular limit for convergent sequences and for bounded, divergent sequences it is what the "peaks" of the sequence converges towards. Of course there is a formal definition as well but I would like to have an intiutive understanding as well.
However I am not sure how to deduce the the inequality between the fraction and t from the definition of limsup... Can someone help me? Am I missing some key property of lim sup?
sequences-and-series convergence limsup-and-liminf
asked Dec 20 '18 at 11:06
CruZCruZ
638
$endgroup$
My thoughts: So apart from this step, the rest of the proof is fairly simple. Now in terms of lim sup, I know that it is just equal to the regular limit for convergent sequences and for bounded, divergent sequences it is what the "peaks" of the sequence converges towards. Of course there is a formal definition as well but I would like to have an intiutive understanding as well.
However I am not sure how to deduce the the inequality between the fraction and t from the definition of limsup... Can someone help me? Am I missing some key property of lim sup?
sequences-and-series convergence limsup-and-liminf
sequences-and-series convergence limsup-and-liminf
asked Dec 20 '18 at 11:06
CruZCruZ
638
asked Dec 20 '18 at 11:06
CruZCruZ
638
asked Dec 20 '18 at 11:06
CruZCruZ
638
asked Dec 20 '18 at 11:06
CruZCruZ
638
asked Dec 20 '18 at 11:06
CruZCruZ
638
638
add a comment |
add a comment |
2 Answers
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Okay, If you consider simply the definition of lim sup The limit superior of ${x_{n}} $ is the smallest real number $b$ such that, for any positive real number $epsilon$ there exists a natural number $N$ such that
$x_{n}<b+epsilon forall n>N$
So, in your case if we simply consider in particular $epsilon=t-b$ there exists a natural number $N_epsilon$ such that $frac{|a_{k+1}|}{|a_{k}|} leq b+t-b=b forall k>N_epsilon$
Even intuitively if you consider any number $t$ greater than the $limsup$ of your sequence. If there was no such natural number N such that $ forall kgeq N $ the sequence $frac{|a_{k+1}|}{|a_{k}|}$ was bounded above by $t$ this implies that the limsup is nessecarily greater than or equal to the number t as the sequence "peaks" beyond the value t for larger and larger values of k.
answered Dec 20 '18 at 12:33
Uday KhannaUday Khanna
21718
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Let $x_k = frac{|a_{k+1}|}{|a_k|}$.
Assume that $limsup$ $x_k leq Q$ and let $t>Q$.
By the definition of limsup, limsup $x_k =$ $inf_{n geq 1} sup_{m geq n}x_m$ $leq Q$. By the definition of the infimum this implies that there exists an N such that $sup_{m geq N}x_m leq t$ (note that $y_n = sup_{m geq n}x_m$ is a decreasing sequence). But this implies that $x_m leq t$ $forall m geq N$, as needed.
answered Dec 20 '18 at 12:32
Sorin TircSorin Tirc
1,845213
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2 Answers
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2 Answers
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$begingroup$
Okay, If you consider simply the definition of lim sup The limit superior of ${x_{n}} $ is the smallest real number $b$ such that, for any positive real number $epsilon$ there exists a natural number $N$ such that
$x_{n}<b+epsilon forall n>N$
So, in your case if we simply consider in particular $epsilon=t-b$ there exists a natural number $N_epsilon$ such that $frac{|a_{k+1}|}{|a_{k}|} leq b+t-b=b forall k>N_epsilon$
Even intuitively if you consider any number $t$ greater than the $limsup$ of your sequence. If there was no such natural number N such that $ forall kgeq N $ the sequence $frac{|a_{k+1}|}{|a_{k}|}$ was bounded above by $t$ this implies that the limsup is nessecarily greater than or equal to the number t as the sequence "peaks" beyond the value t for larger and larger values of k.
answered Dec 20 '18 at 12:33
Uday KhannaUday Khanna
21718
$endgroup$
add a comment |
$begingroup$
Okay, If you consider simply the definition of lim sup The limit superior of ${x_{n}} $ is the smallest real number $b$ such that, for any positive real number $epsilon$ there exists a natural number $N$ such that
$x_{n}<b+epsilon forall n>N$
So, in your case if we simply consider in particular $epsilon=t-b$ there exists a natural number $N_epsilon$ such that $frac{|a_{k+1}|}{|a_{k}|} leq b+t-b=b forall k>N_epsilon$
Even intuitively if you consider any number $t$ greater than the $limsup$ of your sequence. If there was no such natural number N such that $ forall kgeq N $ the sequence $frac{|a_{k+1}|}{|a_{k}|}$ was bounded above by $t$ this implies that the limsup is nessecarily greater than or equal to the number t as the sequence "peaks" beyond the value t for larger and larger values of k.
answered Dec 20 '18 at 12:33
Uday KhannaUday Khanna
21718
$endgroup$
add a comment |
$begingroup$
Okay, If you consider simply the definition of lim sup The limit superior of ${x_{n}} $ is the smallest real number $b$ such that, for any positive real number $epsilon$ there exists a natural number $N$ such that
$x_{n}<b+epsilon forall n>N$
So, in your case if we simply consider in particular $epsilon=t-b$ there exists a natural number $N_epsilon$ such that $frac{|a_{k+1}|}{|a_{k}|} leq b+t-b=b forall k>N_epsilon$
Even intuitively if you consider any number $t$ greater than the $limsup$ of your sequence. If there was no such natural number N such that $ forall kgeq N $ the sequence $frac{|a_{k+1}|}{|a_{k}|}$ was bounded above by $t$ this implies that the limsup is nessecarily greater than or equal to the number t as the sequence "peaks" beyond the value t for larger and larger values of k.
answered Dec 20 '18 at 12:33
Uday KhannaUday Khanna
21718
$endgroup$
Okay, If you consider simply the definition of lim sup The limit superior of ${x_{n}} $ is the smallest real number $b$ such that, for any positive real number $epsilon$ there exists a natural number $N$ such that
$x_{n}<b+epsilon forall n>N$
So, in your case if we simply consider in particular $epsilon=t-b$ there exists a natural number $N_epsilon$ such that $frac{|a_{k+1}|}{|a_{k}|} leq b+t-b=b forall k>N_epsilon$
Even intuitively if you consider any number $t$ greater than the $limsup$ of your sequence. If there was no such natural number N such that $ forall kgeq N $ the sequence $frac{|a_{k+1}|}{|a_{k}|}$ was bounded above by $t$ this implies that the limsup is nessecarily greater than or equal to the number t as the sequence "peaks" beyond the value t for larger and larger values of k.
answered Dec 20 '18 at 12:33
Uday KhannaUday Khanna
21718
answered Dec 20 '18 at 12:33
Uday KhannaUday Khanna
21718
answered Dec 20 '18 at 12:33
Uday KhannaUday Khanna
21718
answered Dec 20 '18 at 12:33
Uday KhannaUday Khanna
21718
21718
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add a comment |
$begingroup$
Let $x_k = frac{|a_{k+1}|}{|a_k|}$.
Assume that $limsup$ $x_k leq Q$ and let $t>Q$.
By the definition of limsup, limsup $x_k =$ $inf_{n geq 1} sup_{m geq n}x_m$ $leq Q$. By the definition of the infimum this implies that there exists an N such that $sup_{m geq N}x_m leq t$ (note that $y_n = sup_{m geq n}x_m$ is a decreasing sequence). But this implies that $x_m leq t$ $forall m geq N$, as needed.
answered Dec 20 '18 at 12:32
Sorin TircSorin Tirc
1,845213
$endgroup$
add a comment |
$begingroup$
Let $x_k = frac{|a_{k+1}|}{|a_k|}$.
Assume that $limsup$ $x_k leq Q$ and let $t>Q$.
By the definition of limsup, limsup $x_k =$ $inf_{n geq 1} sup_{m geq n}x_m$ $leq Q$. By the definition of the infimum this implies that there exists an N such that $sup_{m geq N}x_m leq t$ (note that $y_n = sup_{m geq n}x_m$ is a decreasing sequence). But this implies that $x_m leq t$ $forall m geq N$, as needed.
answered Dec 20 '18 at 12:32
Sorin TircSorin Tirc
1,845213
$endgroup$
add a comment |
$begingroup$
Let $x_k = frac{|a_{k+1}|}{|a_k|}$.
Assume that $limsup$ $x_k leq Q$ and let $t>Q$.
By the definition of limsup, limsup $x_k =$ $inf_{n geq 1} sup_{m geq n}x_m$ $leq Q$. By the definition of the infimum this implies that there exists an N such that $sup_{m geq N}x_m leq t$ (note that $y_n = sup_{m geq n}x_m$ is a decreasing sequence). But this implies that $x_m leq t$ $forall m geq N$, as needed.
answered Dec 20 '18 at 12:32
Sorin TircSorin Tirc
1,845213
$endgroup$
Let $x_k = frac{|a_{k+1}|}{|a_k|}$.
Assume that $limsup$ $x_k leq Q$ and let $t>Q$.
By the definition of limsup, limsup $x_k =$ $inf_{n geq 1} sup_{m geq n}x_m$ $leq Q$. By the definition of the infimum this implies that there exists an N such that $sup_{m geq N}x_m leq t$ (note that $y_n = sup_{m geq n}x_m$ is a decreasing sequence). But this implies that $x_m leq t$ $forall m geq N$, as needed.
answered Dec 20 '18 at 12:32
Sorin TircSorin Tirc
1,845213
answered Dec 20 '18 at 12:32
Sorin TircSorin Tirc
1,845213
answered Dec 20 '18 at 12:32
Sorin TircSorin Tirc
1,845213
answered Dec 20 '18 at 12:32
Sorin TircSorin Tirc
1,845213
1,845213
add a comment |
add a comment |
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