Given $lim_n rightarrow infty a_n a_n+1 = L$, how to show: $lim_n rightarrow infty a_n a_n+3 = L$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Prove: If $lim_nrightarrowinfty|a_n| = 0$, then $lim_nrightarrowinftya_n = 0$Proof the limit of sequence $cos(a_n)_n=0^infty $ where $a_n in mathbb R$ and $lim_n rightarrow infty a_n = 0$ is equal to $1$Show, using the definitions, that $lim_n rightarrow infty inf a_n = infty$ implies that $lim_n rightarrow infty a_n = infty$How to prove $lim_n to inftya_n=1 rightarrow lim_n to inftysqrt[n] a_n=1$Proof $liminf_nrightarrowinftya_n=limsup_nrightarrowinftya_n=lim_nrightarrowinftya_n$Prove $limsup_n rightarrow infty a_ngeq1$ if $lim_n rightarrow infty a_na_n+1=1$Prove $lim_ntoinfty inf (a_n) leq lim_ntoinfty sup (a_n)$$lim_n rightarrow infty a_n = +infty, lim_n rightarrow infty b_n = +infty$ and $lim_n rightarrow infty(a_n + b_n ) = -infty$.$lim_nrightarrow infty (a_n+b_n)=0$ and $lim_nrightarrow infty c_n=L$ imply $lim_nrightarrow infty exp(a_n)*c_n-L*exp(-b_n)=0$Finding $lim_n rightarrow inftya_n$ given $lim_n rightarrow inftyfraca_n -1a_n + 1$
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Given $lim_n rightarrow infty a_n a_n+1 = L$, how to show: $lim_n rightarrow infty a_n a_n+3 = L$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Prove: If $lim_nrightarrowinfty|a_n| = 0$, then $lim_nrightarrowinftya_n = 0$Proof the limit of sequence $cos(a_n)_n=0^infty $ where $a_n in mathbb R$ and $lim_n rightarrow infty a_n = 0$ is equal to $1$Show, using the definitions, that $lim_n rightarrow infty inf a_n = infty$ implies that $lim_n rightarrow infty a_n = infty$How to prove $lim_n to inftya_n=1 rightarrow lim_n to inftysqrt[n] a_n=1$Proof $liminf_nrightarrowinftya_n=limsup_nrightarrowinftya_n=lim_nrightarrowinftya_n$Prove $limsup_n rightarrow infty a_ngeq1$ if $lim_n rightarrow infty a_na_n+1=1$Prove $lim_ntoinfty inf (a_n) leq lim_ntoinfty sup (a_n)$$lim_n rightarrow infty a_n = +infty, lim_n rightarrow infty b_n = +infty$ and $lim_n rightarrow infty(a_n + b_n ) = -infty$.$lim_nrightarrow infty (a_n+b_n)=0$ and $lim_nrightarrow infty c_n=L$ imply $lim_nrightarrow infty exp(a_n)*c_n-L*exp(-b_n)=0$Finding $lim_n rightarrow inftya_n$ given $lim_n rightarrow inftyfraca_n -1a_n + 1$
$begingroup$
let $ a_n $ be a sequence where for each $n in mathbb N$ $ a_n neq 0 $ and where $lim_n rightarrow infty a_n a_n+1 = L$ with $L neq 0$
I want to prove that
$lim_n rightarrow infty a_n a_n+3 = L$
and that
$lim_n rightarrow infty a_n a_n+2 neq -1$
Any ideas?
Thanks!
Edit: Intuitively it's clear but I am looking for a real regorous proof..
sequences-and-series limits limsup-and-liminf
edited Apr 2 at 10:12
Bernard
124k742117
asked Apr 2 at 10:05
user135172user135172
449210
$endgroup$
add a comment |
$begingroup$
let $ a_n $ be a sequence where for each $n in mathbb N$ $ a_n neq 0 $ and where $lim_n rightarrow infty a_n a_n+1 = L$ with $L neq 0$
I want to prove that
$lim_n rightarrow infty a_n a_n+3 = L$
and that
$lim_n rightarrow infty a_n a_n+2 neq -1$
Any ideas?
Thanks!
Edit: Intuitively it's clear but I am looking for a real regorous proof..
sequences-and-series limits limsup-and-liminf
edited Apr 2 at 10:12
Bernard
124k742117
asked Apr 2 at 10:05
user135172user135172
449210
$endgroup$
3
$begingroup$
For the first part, note that $$a_na_n+3=frac(a_na_n+1)(a_n+2a_n+3)a_n+1a_n+2.$$Similarly, for the second part, note that $$a_na_n+2=frac(a_na_n+1)(a_n+1a_n+2)a_n+1^2.$$ Since the numerator converges to the positive number $L^2$ and the denominator is positive, the right-hand side is positive for all large $n$, and so, it cannot converge to a negative number.
$endgroup$
– Sangchul Lee
Apr 2 at 10:12
$begingroup$
Interesting question! It does not follow that $lim a_n a_n+2 = L$, or even that it exists.
$endgroup$
– GEdgar
Apr 2 at 13:03
add a comment |
$begingroup$
let $ a_n $ be a sequence where for each $n in mathbb N$ $ a_n neq 0 $ and where $lim_n rightarrow infty a_n a_n+1 = L$ with $L neq 0$
I want to prove that
$lim_n rightarrow infty a_n a_n+3 = L$
and that
$lim_n rightarrow infty a_n a_n+2 neq -1$
Any ideas?
Thanks!
Edit: Intuitively it's clear but I am looking for a real regorous proof..
sequences-and-series limits limsup-and-liminf
edited Apr 2 at 10:12
Bernard
124k742117
asked Apr 2 at 10:05
user135172user135172
449210
$endgroup$
let $ a_n $ be a sequence where for each $n in mathbb N$ $ a_n neq 0 $ and where $lim_n rightarrow infty a_n a_n+1 = L$ with $L neq 0$
I want to prove that
$lim_n rightarrow infty a_n a_n+3 = L$
and that
$lim_n rightarrow infty a_n a_n+2 neq -1$
Any ideas?
Thanks!
Edit: Intuitively it's clear but I am looking for a real regorous proof..
sequences-and-series limits limsup-and-liminf
sequences-and-series limits limsup-and-liminf
edited Apr 2 at 10:12
Bernard
124k742117
asked Apr 2 at 10:05
user135172user135172
449210
edited Apr 2 at 10:12
Bernard
124k742117
asked Apr 2 at 10:05
user135172user135172
449210
edited Apr 2 at 10:12
Bernard
124k742117
edited Apr 2 at 10:12
Bernard
124k742117
edited Apr 2 at 10:12
Bernard
124k742117
124k742117
asked Apr 2 at 10:05
user135172user135172
449210
asked Apr 2 at 10:05
user135172user135172
449210
asked Apr 2 at 10:05
user135172user135172
449210
449210
3
$begingroup$
For the first part, note that $$a_na_n+3=frac(a_na_n+1)(a_n+2a_n+3)a_n+1a_n+2.$$Similarly, for the second part, note that $$a_na_n+2=frac(a_na_n+1)(a_n+1a_n+2)a_n+1^2.$$ Since the numerator converges to the positive number $L^2$ and the denominator is positive, the right-hand side is positive for all large $n$, and so, it cannot converge to a negative number.
$endgroup$
– Sangchul Lee
Apr 2 at 10:12
$begingroup$
Interesting question! It does not follow that $lim a_n a_n+2 = L$, or even that it exists.
$endgroup$
– GEdgar
Apr 2 at 13:03
add a comment |
3
$begingroup$
For the first part, note that $$a_na_n+3=frac(a_na_n+1)(a_n+2a_n+3)a_n+1a_n+2.$$Similarly, for the second part, note that $$a_na_n+2=frac(a_na_n+1)(a_n+1a_n+2)a_n+1^2.$$ Since the numerator converges to the positive number $L^2$ and the denominator is positive, the right-hand side is positive for all large $n$, and so, it cannot converge to a negative number.
$endgroup$
– Sangchul Lee
Apr 2 at 10:12
$begingroup$
Interesting question! It does not follow that $lim a_n a_n+2 = L$, or even that it exists.
$endgroup$
– GEdgar
Apr 2 at 13:03
3
3
$begingroup$
For the first part, note that $$a_na_n+3=frac(a_na_n+1)(a_n+2a_n+3)a_n+1a_n+2.$$Similarly, for the second part, note that $$a_na_n+2=frac(a_na_n+1)(a_n+1a_n+2)a_n+1^2.$$ Since the numerator converges to the positive number $L^2$ and the denominator is positive, the right-hand side is positive for all large $n$, and so, it cannot converge to a negative number.
$endgroup$
– Sangchul Lee
Apr 2 at 10:12
$begingroup$
For the first part, note that $$a_na_n+3=frac(a_na_n+1)(a_n+2a_n+3)a_n+1a_n+2.$$Similarly, for the second part, note that $$a_na_n+2=frac(a_na_n+1)(a_n+1a_n+2)a_n+1^2.$$ Since the numerator converges to the positive number $L^2$ and the denominator is positive, the right-hand side is positive for all large $n$, and so, it cannot converge to a negative number.
$endgroup$
– Sangchul Lee
Apr 2 at 10:12
$begingroup$
Interesting question! It does not follow that $lim a_n a_n+2 = L$, or even that it exists.
$endgroup$
– GEdgar
Apr 2 at 13:03
$begingroup$
Interesting question! It does not follow that $lim a_n a_n+2 = L$, or even that it exists.
$endgroup$
– GEdgar
Apr 2 at 13:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First part : Because $a_n neq 0$ for all $n$, you can write $$a_na_n+3 = fracleft( a_n a_n+1 right) left( a_n+2 a_n+3right) a_n+1 a_n+2$$
Now it is easy to see that it tends to
$$ fracL times LL = L$$
Second part : Similarly you can write
$$a_n+1^2= fracleft( a_n a_n+1 right) left( a_n+1 a_n+2right)a_n a_n+2 $$
Suppose that $a_n a_n+2$ tends to $-1$ ; then you would deduce that $a_n+1^2$ tends to $-L^2$ which is impossible because $-L^2 < 0$ and $a_n+1^2 > 0$ for all $n$.
answered Apr 2 at 12:51
TheSilverDoeTheSilverDoe
5,593316
$endgroup$
add a comment |
$begingroup$
$a_na_n+1 to L$, $a_n+1a_n+2 to L$,$a_n+2a_n+3 to L$. Multiply the first and the third and divide by the second to get $a_na_n+3 to L$. For the second part note that $a_na_n+2 a_n+1^2 to L^2$. Can you see why $a_na_n+2$ cannot be negative for large $n$?.
answered Apr 2 at 10:12
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
$endgroup$
add a comment |
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2 Answers
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2 Answers
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votes
$begingroup$
First part : Because $a_n neq 0$ for all $n$, you can write $$a_na_n+3 = fracleft( a_n a_n+1 right) left( a_n+2 a_n+3right) a_n+1 a_n+2$$
Now it is easy to see that it tends to
$$ fracL times LL = L$$
Second part : Similarly you can write
$$a_n+1^2= fracleft( a_n a_n+1 right) left( a_n+1 a_n+2right)a_n a_n+2 $$
Suppose that $a_n a_n+2$ tends to $-1$ ; then you would deduce that $a_n+1^2$ tends to $-L^2$ which is impossible because $-L^2 < 0$ and $a_n+1^2 > 0$ for all $n$.
answered Apr 2 at 12:51
TheSilverDoeTheSilverDoe
5,593316
$endgroup$
add a comment |
$begingroup$
First part : Because $a_n neq 0$ for all $n$, you can write $$a_na_n+3 = fracleft( a_n a_n+1 right) left( a_n+2 a_n+3right) a_n+1 a_n+2$$
Now it is easy to see that it tends to
$$ fracL times LL = L$$
Second part : Similarly you can write
$$a_n+1^2= fracleft( a_n a_n+1 right) left( a_n+1 a_n+2right)a_n a_n+2 $$
Suppose that $a_n a_n+2$ tends to $-1$ ; then you would deduce that $a_n+1^2$ tends to $-L^2$ which is impossible because $-L^2 < 0$ and $a_n+1^2 > 0$ for all $n$.
answered Apr 2 at 12:51
TheSilverDoeTheSilverDoe
5,593316
$endgroup$
add a comment |
$begingroup$
First part : Because $a_n neq 0$ for all $n$, you can write $$a_na_n+3 = fracleft( a_n a_n+1 right) left( a_n+2 a_n+3right) a_n+1 a_n+2$$
Now it is easy to see that it tends to
$$ fracL times LL = L$$
Second part : Similarly you can write
$$a_n+1^2= fracleft( a_n a_n+1 right) left( a_n+1 a_n+2right)a_n a_n+2 $$
Suppose that $a_n a_n+2$ tends to $-1$ ; then you would deduce that $a_n+1^2$ tends to $-L^2$ which is impossible because $-L^2 < 0$ and $a_n+1^2 > 0$ for all $n$.
answered Apr 2 at 12:51
TheSilverDoeTheSilverDoe
5,593316
$endgroup$
First part : Because $a_n neq 0$ for all $n$, you can write $$a_na_n+3 = fracleft( a_n a_n+1 right) left( a_n+2 a_n+3right) a_n+1 a_n+2$$
Now it is easy to see that it tends to
$$ fracL times LL = L$$
Second part : Similarly you can write
$$a_n+1^2= fracleft( a_n a_n+1 right) left( a_n+1 a_n+2right)a_n a_n+2 $$
Suppose that $a_n a_n+2$ tends to $-1$ ; then you would deduce that $a_n+1^2$ tends to $-L^2$ which is impossible because $-L^2 < 0$ and $a_n+1^2 > 0$ for all $n$.
answered Apr 2 at 12:51
TheSilverDoeTheSilverDoe
5,593316
answered Apr 2 at 12:51
TheSilverDoeTheSilverDoe
5,593316
answered Apr 2 at 12:51
TheSilverDoeTheSilverDoe
5,593316
answered Apr 2 at 12:51
TheSilverDoeTheSilverDoe
5,593316
5,593316
add a comment |
add a comment |
$begingroup$
$a_na_n+1 to L$, $a_n+1a_n+2 to L$,$a_n+2a_n+3 to L$. Multiply the first and the third and divide by the second to get $a_na_n+3 to L$. For the second part note that $a_na_n+2 a_n+1^2 to L^2$. Can you see why $a_na_n+2$ cannot be negative for large $n$?.
answered Apr 2 at 10:12
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
$endgroup$
add a comment |
$begingroup$
$a_na_n+1 to L$, $a_n+1a_n+2 to L$,$a_n+2a_n+3 to L$. Multiply the first and the third and divide by the second to get $a_na_n+3 to L$. For the second part note that $a_na_n+2 a_n+1^2 to L^2$. Can you see why $a_na_n+2$ cannot be negative for large $n$?.
answered Apr 2 at 10:12
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
$endgroup$
add a comment |
$begingroup$
$a_na_n+1 to L$, $a_n+1a_n+2 to L$,$a_n+2a_n+3 to L$. Multiply the first and the third and divide by the second to get $a_na_n+3 to L$. For the second part note that $a_na_n+2 a_n+1^2 to L^2$. Can you see why $a_na_n+2$ cannot be negative for large $n$?.
answered Apr 2 at 10:12
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
$endgroup$
$a_na_n+1 to L$, $a_n+1a_n+2 to L$,$a_n+2a_n+3 to L$. Multiply the first and the third and divide by the second to get $a_na_n+3 to L$. For the second part note that $a_na_n+2 a_n+1^2 to L^2$. Can you see why $a_na_n+2$ cannot be negative for large $n$?.
answered Apr 2 at 10:12
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
answered Apr 2 at 10:12
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
answered Apr 2 at 10:12
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
answered Apr 2 at 10:12
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
76.4k53370
add a comment |
add a comment |
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3
$begingroup$
For the first part, note that $$a_na_n+3=frac(a_na_n+1)(a_n+2a_n+3)a_n+1a_n+2.$$Similarly, for the second part, note that $$a_na_n+2=frac(a_na_n+1)(a_n+1a_n+2)a_n+1^2.$$ Since the numerator converges to the positive number $L^2$ and the denominator is positive, the right-hand side is positive for all large $n$, and so, it cannot converge to a negative number.
$endgroup$
– Sangchul Lee
Apr 2 at 10:12
$begingroup$
Interesting question! It does not follow that $lim a_n a_n+2 = L$, or even that it exists.
$endgroup$
– GEdgar
Apr 2 at 13:03