Showing that $a_n to 0$ if $a_n/a_n+1 to l > 1$.
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
How can we prove that $lim_nrightarrowinftya_n=0$
if $a_n>0$ and $lim_ntoinfty a_n / a_n+1 = l >1$?
calculus
edited Sep 7 at 10:44
Jendrik Stelzner
7,69121137
asked May 24 '12 at 13:35
Francis King
367
add a comment |Â
up vote
1
down vote
favorite
How can we prove that $lim_nrightarrowinftya_n=0$
if $a_n>0$ and $lim_ntoinfty a_n / a_n+1 = l >1$?
calculus
edited Sep 7 at 10:44
Jendrik Stelzner
7,69121137
asked May 24 '12 at 13:35
Francis King
367
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How can we prove that $lim_nrightarrowinftya_n=0$
if $a_n>0$ and $lim_ntoinfty a_n / a_n+1 = l >1$?
calculus
edited Sep 7 at 10:44
Jendrik Stelzner
7,69121137
asked May 24 '12 at 13:35
Francis King
367
How can we prove that $lim_nrightarrowinftya_n=0$
if $a_n>0$ and $lim_ntoinfty a_n / a_n+1 = l >1$?
calculus
calculus
edited Sep 7 at 10:44
Jendrik Stelzner
7,69121137
asked May 24 '12 at 13:35
Francis King
367
edited Sep 7 at 10:44
Jendrik Stelzner
7,69121137
asked May 24 '12 at 13:35
Francis King
367
edited Sep 7 at 10:44
Jendrik Stelzner
7,69121137
edited Sep 7 at 10:44
Jendrik Stelzner
7,69121137
edited Sep 7 at 10:44
Jendrik Stelzner
7,69121137
7,69121137
asked May 24 '12 at 13:35
Francis King
367
asked May 24 '12 at 13:35
Francis King
367
asked May 24 '12 at 13:35
Francis King
367
367
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
There is some $N$ sufficiently large so that, for all $n > N$, we have $a_n+1/a_n leq c < 1$.
For for $n > 0$ we have $0 < a_n+N leq a_N c^n$. By the squeeze theorem, $a_n+N rightarrow 0$.
answered May 24 '12 at 13:46
David Harris
706411
The key observation in applying the squeeze theorem is $a_N$ is fixed.
– toypajme
May 24 '12 at 14:18
I don't certainly get the point that how we reach the inequality $a_n+N<a_Nc^n$ thanks
– Francis King
May 24 '12 at 15:09
I have get the point.thanks
– Francis King
May 24 '12 at 15:19
add a comment |Â
up vote
1
down vote
Hint:
If $limlimits_nrightarrowinftya_nover a_n+1=l>1$, then $limlimits_nrightarrowinftya_n+1over a_n =1over l<1$. Choose $0<c<1$, and choose $N$ so that if $nge N$, then $a_n+1over a_n<c$.
Then note:
$ a_N+1<c, a_N$,
$ a_N+2<c,a_N+1<c^2a_N$,
$ a_N+3<c,a_N+2<c^3a_N$,
$ vdots$
answered May 24 '12 at 13:46
David Mitra
62k695159
thanks. your step-by-step induction helps me a lot.
– Francis King
May 24 '12 at 15:20
add a comment |Â
up vote
1
down vote
Another proof not-by-the-definition: the series $,displaystylesum_n=1^infty a_n,$ is a positive one and it converges by D'Alembert's test (or the quotient test), since $,displaystylefraca_n+1a_nto frac1l<1,$ , thus it must be $,a_nto 0$
answered May 24 '12 at 14:04
DonAntonio
173k1487220
Right. But I guess that D'Alembert's test is proved the same way as the fact asked here. Your suggestion might be a circular reasoning.
– Siminore
May 24 '12 at 16:27
No, or at least not the proof I know, which proves the sequence of partial sums of the series is bounded (using a geometric convergente series) and thus the series converges.
– DonAntonio
May 24 '12 at 18:01
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
There is some $N$ sufficiently large so that, for all $n > N$, we have $a_n+1/a_n leq c < 1$.
For for $n > 0$ we have $0 < a_n+N leq a_N c^n$. By the squeeze theorem, $a_n+N rightarrow 0$.
answered May 24 '12 at 13:46
David Harris
706411
The key observation in applying the squeeze theorem is $a_N$ is fixed.
– toypajme
May 24 '12 at 14:18
I don't certainly get the point that how we reach the inequality $a_n+N<a_Nc^n$ thanks
– Francis King
May 24 '12 at 15:09
I have get the point.thanks
– Francis King
May 24 '12 at 15:19
add a comment |Â
up vote
3
down vote
accepted
There is some $N$ sufficiently large so that, for all $n > N$, we have $a_n+1/a_n leq c < 1$.
For for $n > 0$ we have $0 < a_n+N leq a_N c^n$. By the squeeze theorem, $a_n+N rightarrow 0$.
answered May 24 '12 at 13:46
David Harris
706411
The key observation in applying the squeeze theorem is $a_N$ is fixed.
– toypajme
May 24 '12 at 14:18
I don't certainly get the point that how we reach the inequality $a_n+N<a_Nc^n$ thanks
– Francis King
May 24 '12 at 15:09
I have get the point.thanks
– Francis King
May 24 '12 at 15:19
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
There is some $N$ sufficiently large so that, for all $n > N$, we have $a_n+1/a_n leq c < 1$.
For for $n > 0$ we have $0 < a_n+N leq a_N c^n$. By the squeeze theorem, $a_n+N rightarrow 0$.
answered May 24 '12 at 13:46
David Harris
706411
There is some $N$ sufficiently large so that, for all $n > N$, we have $a_n+1/a_n leq c < 1$.
For for $n > 0$ we have $0 < a_n+N leq a_N c^n$. By the squeeze theorem, $a_n+N rightarrow 0$.
answered May 24 '12 at 13:46
David Harris
706411
answered May 24 '12 at 13:46
David Harris
706411
answered May 24 '12 at 13:46
David Harris
706411
answered May 24 '12 at 13:46
David Harris
706411
706411
The key observation in applying the squeeze theorem is $a_N$ is fixed.
– toypajme
May 24 '12 at 14:18
I don't certainly get the point that how we reach the inequality $a_n+N<a_Nc^n$ thanks
– Francis King
May 24 '12 at 15:09
I have get the point.thanks
– Francis King
May 24 '12 at 15:19
add a comment |Â
The key observation in applying the squeeze theorem is $a_N$ is fixed.
– toypajme
May 24 '12 at 14:18
I don't certainly get the point that how we reach the inequality $a_n+N<a_Nc^n$ thanks
– Francis King
May 24 '12 at 15:09
I have get the point.thanks
– Francis King
May 24 '12 at 15:19
The key observation in applying the squeeze theorem is $a_N$ is fixed.
– toypajme
May 24 '12 at 14:18
The key observation in applying the squeeze theorem is $a_N$ is fixed.
– toypajme
May 24 '12 at 14:18
I don't certainly get the point that how we reach the inequality $a_n+N<a_Nc^n$ thanks
– Francis King
May 24 '12 at 15:09
I don't certainly get the point that how we reach the inequality $a_n+N<a_Nc^n$ thanks
– Francis King
May 24 '12 at 15:09
I have get the point.thanks
– Francis King
May 24 '12 at 15:19
I have get the point.thanks
– Francis King
May 24 '12 at 15:19
add a comment |Â
up vote
1
down vote
Hint:
If $limlimits_nrightarrowinftya_nover a_n+1=l>1$, then $limlimits_nrightarrowinftya_n+1over a_n =1over l<1$. Choose $0<c<1$, and choose $N$ so that if $nge N$, then $a_n+1over a_n<c$.
Then note:
$ a_N+1<c, a_N$,
$ a_N+2<c,a_N+1<c^2a_N$,
$ a_N+3<c,a_N+2<c^3a_N$,
$ vdots$
answered May 24 '12 at 13:46
David Mitra
62k695159
thanks. your step-by-step induction helps me a lot.
– Francis King
May 24 '12 at 15:20
add a comment |Â
up vote
1
down vote
Hint:
If $limlimits_nrightarrowinftya_nover a_n+1=l>1$, then $limlimits_nrightarrowinftya_n+1over a_n =1over l<1$. Choose $0<c<1$, and choose $N$ so that if $nge N$, then $a_n+1over a_n<c$.
Then note:
$ a_N+1<c, a_N$,
$ a_N+2<c,a_N+1<c^2a_N$,
$ a_N+3<c,a_N+2<c^3a_N$,
$ vdots$
answered May 24 '12 at 13:46
David Mitra
62k695159
thanks. your step-by-step induction helps me a lot.
– Francis King
May 24 '12 at 15:20
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint:
If $limlimits_nrightarrowinftya_nover a_n+1=l>1$, then $limlimits_nrightarrowinftya_n+1over a_n =1over l<1$. Choose $0<c<1$, and choose $N$ so that if $nge N$, then $a_n+1over a_n<c$.
Then note:
$ a_N+1<c, a_N$,
$ a_N+2<c,a_N+1<c^2a_N$,
$ a_N+3<c,a_N+2<c^3a_N$,
$ vdots$
answered May 24 '12 at 13:46
David Mitra
62k695159
Hint:
If $limlimits_nrightarrowinftya_nover a_n+1=l>1$, then $limlimits_nrightarrowinftya_n+1over a_n =1over l<1$. Choose $0<c<1$, and choose $N$ so that if $nge N$, then $a_n+1over a_n<c$.
Then note:
$ a_N+1<c, a_N$,
$ a_N+2<c,a_N+1<c^2a_N$,
$ a_N+3<c,a_N+2<c^3a_N$,
$ vdots$
answered May 24 '12 at 13:46
David Mitra
62k695159
answered May 24 '12 at 13:46
David Mitra
62k695159
answered May 24 '12 at 13:46
David Mitra
62k695159
answered May 24 '12 at 13:46
David Mitra
62k695159
62k695159
thanks. your step-by-step induction helps me a lot.
– Francis King
May 24 '12 at 15:20
add a comment |Â
thanks. your step-by-step induction helps me a lot.
– Francis King
May 24 '12 at 15:20
thanks. your step-by-step induction helps me a lot.
– Francis King
May 24 '12 at 15:20
thanks. your step-by-step induction helps me a lot.
– Francis King
May 24 '12 at 15:20
add a comment |Â
up vote
1
down vote
Another proof not-by-the-definition: the series $,displaystylesum_n=1^infty a_n,$ is a positive one and it converges by D'Alembert's test (or the quotient test), since $,displaystylefraca_n+1a_nto frac1l<1,$ , thus it must be $,a_nto 0$
answered May 24 '12 at 14:04
DonAntonio
173k1487220
Right. But I guess that D'Alembert's test is proved the same way as the fact asked here. Your suggestion might be a circular reasoning.
– Siminore
May 24 '12 at 16:27
No, or at least not the proof I know, which proves the sequence of partial sums of the series is bounded (using a geometric convergente series) and thus the series converges.
– DonAntonio
May 24 '12 at 18:01
add a comment |Â
up vote
1
down vote
Another proof not-by-the-definition: the series $,displaystylesum_n=1^infty a_n,$ is a positive one and it converges by D'Alembert's test (or the quotient test), since $,displaystylefraca_n+1a_nto frac1l<1,$ , thus it must be $,a_nto 0$
answered May 24 '12 at 14:04
DonAntonio
173k1487220
Right. But I guess that D'Alembert's test is proved the same way as the fact asked here. Your suggestion might be a circular reasoning.
– Siminore
May 24 '12 at 16:27
No, or at least not the proof I know, which proves the sequence of partial sums of the series is bounded (using a geometric convergente series) and thus the series converges.
– DonAntonio
May 24 '12 at 18:01
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Another proof not-by-the-definition: the series $,displaystylesum_n=1^infty a_n,$ is a positive one and it converges by D'Alembert's test (or the quotient test), since $,displaystylefraca_n+1a_nto frac1l<1,$ , thus it must be $,a_nto 0$
answered May 24 '12 at 14:04
DonAntonio
173k1487220
Another proof not-by-the-definition: the series $,displaystylesum_n=1^infty a_n,$ is a positive one and it converges by D'Alembert's test (or the quotient test), since $,displaystylefraca_n+1a_nto frac1l<1,$ , thus it must be $,a_nto 0$
answered May 24 '12 at 14:04
DonAntonio
173k1487220
answered May 24 '12 at 14:04
DonAntonio
173k1487220
answered May 24 '12 at 14:04
DonAntonio
173k1487220
answered May 24 '12 at 14:04
DonAntonio
173k1487220
173k1487220
Right. But I guess that D'Alembert's test is proved the same way as the fact asked here. Your suggestion might be a circular reasoning.
– Siminore
May 24 '12 at 16:27
No, or at least not the proof I know, which proves the sequence of partial sums of the series is bounded (using a geometric convergente series) and thus the series converges.
– DonAntonio
May 24 '12 at 18:01
add a comment |Â
Right. But I guess that D'Alembert's test is proved the same way as the fact asked here. Your suggestion might be a circular reasoning.
– Siminore
May 24 '12 at 16:27
No, or at least not the proof I know, which proves the sequence of partial sums of the series is bounded (using a geometric convergente series) and thus the series converges.
– DonAntonio
May 24 '12 at 18:01
Right. But I guess that D'Alembert's test is proved the same way as the fact asked here. Your suggestion might be a circular reasoning.
– Siminore
May 24 '12 at 16:27
Right. But I guess that D'Alembert's test is proved the same way as the fact asked here. Your suggestion might be a circular reasoning.
– Siminore
May 24 '12 at 16:27
No, or at least not the proof I know, which proves the sequence of partial sums of the series is bounded (using a geometric convergente series) and thus the series converges.
– DonAntonio
May 24 '12 at 18:01
No, or at least not the proof I know, which proves the sequence of partial sums of the series is bounded (using a geometric convergente series) and thus the series converges.
– DonAntonio
May 24 '12 at 18:01
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f149216%2fshowing-that-a-n-to-0-if-a-n-a-n1-to-l-1%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
