On a method to solve certain recursive sequences - looking for counterexamples? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Proving the convergence of $a_n = fracnn+sqrt n$A conjecture concerning convergence of a kind of recursive sequenceMonotone Convergence Theorem (for real sequences) equivalent to the Least Upper Bound Property?Can I use Dominated Convergence Theorem for this sum?Uniform Limit vs. Integral MeanInequalities with Cauchy sequences.Second order non linear recurrence relation$a_n$ does not have an upper bound$iff a_n$ has a subsequence which diverges to infinityBounds on a limit related to the Euclid numbersOn a conjecture that $sumlimits_n=1^kfrac1pi^1/np_nstackrelktoinftylongrightarrow 2$.
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On a method to solve certain recursive sequences - looking for counterexamples?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Proving the convergence of $a_n = fracnn+sqrt n$A conjecture concerning convergence of a kind of recursive sequenceMonotone Convergence Theorem (for real sequences) equivalent to the Least Upper Bound Property?Can I use Dominated Convergence Theorem for this sum?Uniform Limit vs. Integral MeanInequalities with Cauchy sequences.Second order non linear recurrence relation$a_n$ does not have an upper bound$iff a_n$ has a subsequence which diverges to infinityBounds on a limit related to the Euclid numbersOn a conjecture that $sumlimits_n=1^kfrac1pi^1/np_nstackrelktoinftylongrightarrow 2$.
$begingroup$
When I started with this question, I wanted to know why my reasoning was wrong. Nevertheless, after checking some examples, I've noticed that my conjecture was actually - or at least seems to be - right! Therefore I'm looking for counterexamples or rigorous proofs.
Suppose that a recursive sequence with $L_0=5$ is defined recursively as follows
$$L_n+1=fracL_n+1L_n+2$$
Sidenote: Please notice that this is only an example.
When it comes to evaluating the limit of a recursive sequence, you're trying to find $$lim_ntoinfty bigg(fracL_n+1L_n+2bigg)=lim_nto inftyL_n+1colorredstackrel?=lim_nto inftyL_n$$
At least to me, it sounds intuitive that, if the sequence has an upper bound, and the difference between consecutive terms approaches 0 $$L_nsim L_n+1quadtextas $n$ approaches infinity$$
Therefore, evaluating an upper bound should be a simple as solving the equation $$L_n=L_n+1stackreltextin the exampleimpliesL_n=fracL_n+1L_n+2implies L_n=frac-1+sqrt52$$
Another example would be, for instance, the sequence defined recursively as follows $$M_n+1=fracM_n3+frac1M_nqquad textwherequad M_0=4$$ which converges to $sqrtfrac32$.
This leads to
Conjecture
Suppose there's a recursive sequence $L_n^infty_n=0$ that converges to $k$, where $L_n+1-L_nto 0$ as $n$ approaches infinity. Then, $k$ is one of the solutions to the equation $$L_n=L_n+1$$
sequences-and-series recurrence-relations conjectures
asked Apr 8 at 19:53
Dr. MathvaDr. Mathva
3,519630
$endgroup$
add a comment |
$begingroup$
When I started with this question, I wanted to know why my reasoning was wrong. Nevertheless, after checking some examples, I've noticed that my conjecture was actually - or at least seems to be - right! Therefore I'm looking for counterexamples or rigorous proofs.
Suppose that a recursive sequence with $L_0=5$ is defined recursively as follows
$$L_n+1=fracL_n+1L_n+2$$
Sidenote: Please notice that this is only an example.
When it comes to evaluating the limit of a recursive sequence, you're trying to find $$lim_ntoinfty bigg(fracL_n+1L_n+2bigg)=lim_nto inftyL_n+1colorredstackrel?=lim_nto inftyL_n$$
At least to me, it sounds intuitive that, if the sequence has an upper bound, and the difference between consecutive terms approaches 0 $$L_nsim L_n+1quadtextas $n$ approaches infinity$$
Therefore, evaluating an upper bound should be a simple as solving the equation $$L_n=L_n+1stackreltextin the exampleimpliesL_n=fracL_n+1L_n+2implies L_n=frac-1+sqrt52$$
Another example would be, for instance, the sequence defined recursively as follows $$M_n+1=fracM_n3+frac1M_nqquad textwherequad M_0=4$$ which converges to $sqrtfrac32$.
This leads to
Conjecture
Suppose there's a recursive sequence $L_n^infty_n=0$ that converges to $k$, where $L_n+1-L_nto 0$ as $n$ approaches infinity. Then, $k$ is one of the solutions to the equation $$L_n=L_n+1$$
sequences-and-series recurrence-relations conjectures
asked Apr 8 at 19:53
Dr. MathvaDr. Mathva
3,519630
$endgroup$
$begingroup$
As written, your conjecture makes no sense. If $k$ is an upper bound, then so is any number $> k$. Are these all solutions to $L_n=L_n+1$?
$endgroup$
– Robert Israel
Apr 8 at 20:09
$begingroup$
I meant a 'convergence value' (is there a word for this?). In fact 'upper bound' makes no sense... Thanks @RobertIsrael
$endgroup$
– Dr. Mathva
Apr 8 at 20:12
1
$begingroup$
@RobertIsrael Edited
$endgroup$
– Dr. Mathva
Apr 8 at 20:29
add a comment |
$begingroup$
When I started with this question, I wanted to know why my reasoning was wrong. Nevertheless, after checking some examples, I've noticed that my conjecture was actually - or at least seems to be - right! Therefore I'm looking for counterexamples or rigorous proofs.
Suppose that a recursive sequence with $L_0=5$ is defined recursively as follows
$$L_n+1=fracL_n+1L_n+2$$
Sidenote: Please notice that this is only an example.
When it comes to evaluating the limit of a recursive sequence, you're trying to find $$lim_ntoinfty bigg(fracL_n+1L_n+2bigg)=lim_nto inftyL_n+1colorredstackrel?=lim_nto inftyL_n$$
At least to me, it sounds intuitive that, if the sequence has an upper bound, and the difference between consecutive terms approaches 0 $$L_nsim L_n+1quadtextas $n$ approaches infinity$$
Therefore, evaluating an upper bound should be a simple as solving the equation $$L_n=L_n+1stackreltextin the exampleimpliesL_n=fracL_n+1L_n+2implies L_n=frac-1+sqrt52$$
Another example would be, for instance, the sequence defined recursively as follows $$M_n+1=fracM_n3+frac1M_nqquad textwherequad M_0=4$$ which converges to $sqrtfrac32$.
This leads to
Conjecture
Suppose there's a recursive sequence $L_n^infty_n=0$ that converges to $k$, where $L_n+1-L_nto 0$ as $n$ approaches infinity. Then, $k$ is one of the solutions to the equation $$L_n=L_n+1$$
sequences-and-series recurrence-relations conjectures
asked Apr 8 at 19:53
Dr. MathvaDr. Mathva
3,519630
$endgroup$
When I started with this question, I wanted to know why my reasoning was wrong. Nevertheless, after checking some examples, I've noticed that my conjecture was actually - or at least seems to be - right! Therefore I'm looking for counterexamples or rigorous proofs.
Suppose that a recursive sequence with $L_0=5$ is defined recursively as follows
$$L_n+1=fracL_n+1L_n+2$$
Sidenote: Please notice that this is only an example.
When it comes to evaluating the limit of a recursive sequence, you're trying to find $$lim_ntoinfty bigg(fracL_n+1L_n+2bigg)=lim_nto inftyL_n+1colorredstackrel?=lim_nto inftyL_n$$
At least to me, it sounds intuitive that, if the sequence has an upper bound, and the difference between consecutive terms approaches 0 $$L_nsim L_n+1quadtextas $n$ approaches infinity$$
Therefore, evaluating an upper bound should be a simple as solving the equation $$L_n=L_n+1stackreltextin the exampleimpliesL_n=fracL_n+1L_n+2implies L_n=frac-1+sqrt52$$
Another example would be, for instance, the sequence defined recursively as follows $$M_n+1=fracM_n3+frac1M_nqquad textwherequad M_0=4$$ which converges to $sqrtfrac32$.
This leads to
Conjecture
Suppose there's a recursive sequence $L_n^infty_n=0$ that converges to $k$, where $L_n+1-L_nto 0$ as $n$ approaches infinity. Then, $k$ is one of the solutions to the equation $$L_n=L_n+1$$
sequences-and-series recurrence-relations conjectures
sequences-and-series recurrence-relations conjectures
asked Apr 8 at 19:53
Dr. MathvaDr. Mathva
3,519630
asked Apr 8 at 19:53
Dr. MathvaDr. Mathva
3,519630
edited Apr 8 at 20:29
Dr. Mathva
asked Apr 8 at 19:53
Dr. MathvaDr. Mathva
3,519630
asked Apr 8 at 19:53
Dr. MathvaDr. Mathva
3,519630
asked Apr 8 at 19:53
Dr. MathvaDr. Mathva
3,519630
3,519630
$begingroup$
As written, your conjecture makes no sense. If $k$ is an upper bound, then so is any number $> k$. Are these all solutions to $L_n=L_n+1$?
$endgroup$
– Robert Israel
Apr 8 at 20:09
$begingroup$
I meant a 'convergence value' (is there a word for this?). In fact 'upper bound' makes no sense... Thanks @RobertIsrael
$endgroup$
– Dr. Mathva
Apr 8 at 20:12
1
$begingroup$
@RobertIsrael Edited
$endgroup$
– Dr. Mathva
Apr 8 at 20:29
add a comment |
$begingroup$
As written, your conjecture makes no sense. If $k$ is an upper bound, then so is any number $> k$. Are these all solutions to $L_n=L_n+1$?
$endgroup$
– Robert Israel
Apr 8 at 20:09
$begingroup$
I meant a 'convergence value' (is there a word for this?). In fact 'upper bound' makes no sense... Thanks @RobertIsrael
$endgroup$
– Dr. Mathva
Apr 8 at 20:12
1
$begingroup$
@RobertIsrael Edited
$endgroup$
– Dr. Mathva
Apr 8 at 20:29
$begingroup$
As written, your conjecture makes no sense. If $k$ is an upper bound, then so is any number $> k$. Are these all solutions to $L_n=L_n+1$?
$endgroup$
– Robert Israel
Apr 8 at 20:09
$begingroup$
As written, your conjecture makes no sense. If $k$ is an upper bound, then so is any number $> k$. Are these all solutions to $L_n=L_n+1$?
$endgroup$
– Robert Israel
Apr 8 at 20:09
$begingroup$
I meant a 'convergence value' (is there a word for this?). In fact 'upper bound' makes no sense... Thanks @RobertIsrael
$endgroup$
– Dr. Mathva
Apr 8 at 20:12
$begingroup$
I meant a 'convergence value' (is there a word for this?). In fact 'upper bound' makes no sense... Thanks @RobertIsrael
$endgroup$
– Dr. Mathva
Apr 8 at 20:12
1
1
$begingroup$
@RobertIsrael Edited
$endgroup$
– Dr. Mathva
Apr 8 at 20:29
$begingroup$
@RobertIsrael Edited
$endgroup$
– Dr. Mathva
Apr 8 at 20:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $lim_n to infty L_n = L$, then this is also $lim_n to infty L_n+1$ (or, for that matter, $L_n+k$ for any $k$). So if the sequence satisfies
some equation
$f(L_n, L_n+1) = 0$ where $f$ is a continuous function on some subset of $mathbb R^2$ containing $(L_n, L_n+1)$ for all $n$ and also containing $(L,L)$, we have
$$ f(L,L) = lim_n to infty f(L_n, L_n+1) = 0$$
answered Apr 8 at 19:59
Robert IsraelRobert Israel
331k23221478
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $lim_n to infty L_n = L$, then this is also $lim_n to infty L_n+1$ (or, for that matter, $L_n+k$ for any $k$). So if the sequence satisfies
some equation
$f(L_n, L_n+1) = 0$ where $f$ is a continuous function on some subset of $mathbb R^2$ containing $(L_n, L_n+1)$ for all $n$ and also containing $(L,L)$, we have
$$ f(L,L) = lim_n to infty f(L_n, L_n+1) = 0$$
answered Apr 8 at 19:59
Robert IsraelRobert Israel
331k23221478
$endgroup$
add a comment |
$begingroup$
If $lim_n to infty L_n = L$, then this is also $lim_n to infty L_n+1$ (or, for that matter, $L_n+k$ for any $k$). So if the sequence satisfies
some equation
$f(L_n, L_n+1) = 0$ where $f$ is a continuous function on some subset of $mathbb R^2$ containing $(L_n, L_n+1)$ for all $n$ and also containing $(L,L)$, we have
$$ f(L,L) = lim_n to infty f(L_n, L_n+1) = 0$$
answered Apr 8 at 19:59
Robert IsraelRobert Israel
331k23221478
$endgroup$
add a comment |
$begingroup$
If $lim_n to infty L_n = L$, then this is also $lim_n to infty L_n+1$ (or, for that matter, $L_n+k$ for any $k$). So if the sequence satisfies
some equation
$f(L_n, L_n+1) = 0$ where $f$ is a continuous function on some subset of $mathbb R^2$ containing $(L_n, L_n+1)$ for all $n$ and also containing $(L,L)$, we have
$$ f(L,L) = lim_n to infty f(L_n, L_n+1) = 0$$
answered Apr 8 at 19:59
Robert IsraelRobert Israel
331k23221478
$endgroup$
If $lim_n to infty L_n = L$, then this is also $lim_n to infty L_n+1$ (or, for that matter, $L_n+k$ for any $k$). So if the sequence satisfies
some equation
$f(L_n, L_n+1) = 0$ where $f$ is a continuous function on some subset of $mathbb R^2$ containing $(L_n, L_n+1)$ for all $n$ and also containing $(L,L)$, we have
$$ f(L,L) = lim_n to infty f(L_n, L_n+1) = 0$$
answered Apr 8 at 19:59
Robert IsraelRobert Israel
331k23221478
answered Apr 8 at 19:59
Robert IsraelRobert Israel
331k23221478
answered Apr 8 at 19:59
Robert IsraelRobert Israel
331k23221478
answered Apr 8 at 19:59
Robert IsraelRobert Israel
331k23221478
331k23221478
add a comment |
add a comment |
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$begingroup$
As written, your conjecture makes no sense. If $k$ is an upper bound, then so is any number $> k$. Are these all solutions to $L_n=L_n+1$?
$endgroup$
– Robert Israel
Apr 8 at 20:09
$begingroup$
I meant a 'convergence value' (is there a word for this?). In fact 'upper bound' makes no sense... Thanks @RobertIsrael
$endgroup$
– Dr. Mathva
Apr 8 at 20:12
1
$begingroup$
@RobertIsrael Edited
$endgroup$
– Dr. Mathva
Apr 8 at 20:29