Using induction to prove a sequence $a_1=1$ and $a_{n+1}=3-frac{1}{a_n}$ is increasing
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I have a sequence defined by $a_1=1$ and $a_{n+1}=3-frac{1}{a_n}$. I need to use induction to show that the sequence is increasing and $a_n<3$ for all $n$.
Also to deduce that $a_n$ is convergence and find its limit.
So far I have found: $a_n<a_{n+1}<3$ and $a_{n+1}>0$
Then $a_{n+1}=3-frac{1}{a_n}$ with $a_{n+2}=3-frac{a_n}{3a_n-1}$
But not really sure how to go any further even if this is the right way to go in the first place? Any input would be greatly be appreciated!
sequences-and-series limits inequality convergence induction
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add a comment |
$begingroup$
I have a sequence defined by $a_1=1$ and $a_{n+1}=3-frac{1}{a_n}$. I need to use induction to show that the sequence is increasing and $a_n<3$ for all $n$.
Also to deduce that $a_n$ is convergence and find its limit.
So far I have found: $a_n<a_{n+1}<3$ and $a_{n+1}>0$
Then $a_{n+1}=3-frac{1}{a_n}$ with $a_{n+2}=3-frac{a_n}{3a_n-1}$
But not really sure how to go any further even if this is the right way to go in the first place? Any input would be greatly be appreciated!
sequences-and-series limits inequality convergence induction
$endgroup$
add a comment |
$begingroup$
I have a sequence defined by $a_1=1$ and $a_{n+1}=3-frac{1}{a_n}$. I need to use induction to show that the sequence is increasing and $a_n<3$ for all $n$.
Also to deduce that $a_n$ is convergence and find its limit.
So far I have found: $a_n<a_{n+1}<3$ and $a_{n+1}>0$
Then $a_{n+1}=3-frac{1}{a_n}$ with $a_{n+2}=3-frac{a_n}{3a_n-1}$
But not really sure how to go any further even if this is the right way to go in the first place? Any input would be greatly be appreciated!
sequences-and-series limits inequality convergence induction
$endgroup$
I have a sequence defined by $a_1=1$ and $a_{n+1}=3-frac{1}{a_n}$. I need to use induction to show that the sequence is increasing and $a_n<3$ for all $n$.
Also to deduce that $a_n$ is convergence and find its limit.
So far I have found: $a_n<a_{n+1}<3$ and $a_{n+1}>0$
Then $a_{n+1}=3-frac{1}{a_n}$ with $a_{n+2}=3-frac{a_n}{3a_n-1}$
But not really sure how to go any further even if this is the right way to go in the first place? Any input would be greatly be appreciated!
sequences-and-series limits inequality convergence induction
sequences-and-series limits inequality convergence induction
edited Apr 20 at 12:35
YuiTo Cheng
2,82141139
2,82141139
asked Apr 20 at 11:01
James odareJames odare
40114
40114
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2 Answers
2
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You have already done most of the work. By showing that $1 leq a_n < a_{n+1} < 3$ where $n in mathbb{Z}_+$, you have demonstrated that the sequence is increasing and bounded above by $3$. Now the Monotone Convergence Theorem says that any increasing sequence that is bounded above converges; indeed it converges to the supremum of ${a_n}_{n in mathbb{Z}_+}$. So you now know that $limlimits_{n to infty} a_n = alpha in mathbb{R}$ for some $$boxed{1 leq alpha leq 3}$$
To find $alpha$, note that $f(x) = 3 - frac{1}{x}$ is a continuous real valued function for real $x > 0$. So, by continuity, and the fact that $alpha > 0$, we can put limits inside the function like so: $$limlimits_{n to infty}f(a_n) = f(limlimits_{n to infty}a_n)$$ But the LHS is just $limlimits_{n to infty}f(a_n) = limlimits_{n to infty}(3 - frac{1}{a_n}) = limlimits_{n to infty} a_{n + 1} = alpha$ while the RHS is just $f(limlimits_{n to infty}a_n) = f(alpha) = 3 - frac{1}{alpha}$. So, putting LHS $=$ RHS we get
$$alpha = 3 - frac{1}{alpha}$$
$$boxed{alpha^2 - 3 alpha + 1 = 0}$$ Now, solve this quadratic equation to get the value of $alpha$. You will get two values but only one of them is the answer. See if you can eliminate the wrong one using the properties of $alpha$.
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$begingroup$
Thank you very much!
$endgroup$
– James odare
Apr 20 at 11:54
$begingroup$
No problem, you're welcome.
$endgroup$
– ZeroXLR
Apr 20 at 11:56
add a comment |
$begingroup$
Let $P(n)$ be the assertion “$a_ninleft[1,frac{3+sqrt5}2right)$ and $a_{n+1}>a_n$”.
If $n=1$, then this is true, since $a_1=1$ and $a_2=2>1=a_1$.
Let $ninmathbb N$ and assume that $P(n)$ holds. Then:
$a_{n+1}=3-frac1{a_n}<3$. since $a_ngeqslant1>0$;
$a_{n+1}geqslant1$, since $a_{n}geqslant1$ and so $3-frac1{a_n}geqslant3-frac11=2$.
$a_{n+2}-a_{n+1}=3-dfrac1{a_{n+1}}-a_{n+1}=dfrac{3a_{n+1}-1-{a_{n+1}}^2}{{a_{n+1}}^2}=dfrac{varphi(a_{n+1})}{{a_{n+1}}^2}$, where $varphi(x)=-x^2+3x-1$. But the roots of the quadratic polynomial $varphi(x)$ are $dfrac{3pmsqrt5}2$ and so, since $a_{n+1}$ is between them and the coefficient of $x^2$ in $varphi(x)$ is negative, $varphi(a_{n+1})>0$.
So, $P(n+1)$ is proved.
What I wrote above is an inductive proof of the fact that we always have $P(n)$. And it is very easy now to conclude that $lim_{ntoinfty}a_n=dfrac{3+sqrt5}2$.
$endgroup$
$begingroup$
Great explanation thank you
$endgroup$
– James odare
Apr 20 at 11:54
add a comment |
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2 Answers
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$begingroup$
You have already done most of the work. By showing that $1 leq a_n < a_{n+1} < 3$ where $n in mathbb{Z}_+$, you have demonstrated that the sequence is increasing and bounded above by $3$. Now the Monotone Convergence Theorem says that any increasing sequence that is bounded above converges; indeed it converges to the supremum of ${a_n}_{n in mathbb{Z}_+}$. So you now know that $limlimits_{n to infty} a_n = alpha in mathbb{R}$ for some $$boxed{1 leq alpha leq 3}$$
To find $alpha$, note that $f(x) = 3 - frac{1}{x}$ is a continuous real valued function for real $x > 0$. So, by continuity, and the fact that $alpha > 0$, we can put limits inside the function like so: $$limlimits_{n to infty}f(a_n) = f(limlimits_{n to infty}a_n)$$ But the LHS is just $limlimits_{n to infty}f(a_n) = limlimits_{n to infty}(3 - frac{1}{a_n}) = limlimits_{n to infty} a_{n + 1} = alpha$ while the RHS is just $f(limlimits_{n to infty}a_n) = f(alpha) = 3 - frac{1}{alpha}$. So, putting LHS $=$ RHS we get
$$alpha = 3 - frac{1}{alpha}$$
$$boxed{alpha^2 - 3 alpha + 1 = 0}$$ Now, solve this quadratic equation to get the value of $alpha$. You will get two values but only one of them is the answer. See if you can eliminate the wrong one using the properties of $alpha$.
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– James odare
Apr 20 at 11:54
$begingroup$
No problem, you're welcome.
$endgroup$
– ZeroXLR
Apr 20 at 11:56
add a comment |
$begingroup$
You have already done most of the work. By showing that $1 leq a_n < a_{n+1} < 3$ where $n in mathbb{Z}_+$, you have demonstrated that the sequence is increasing and bounded above by $3$. Now the Monotone Convergence Theorem says that any increasing sequence that is bounded above converges; indeed it converges to the supremum of ${a_n}_{n in mathbb{Z}_+}$. So you now know that $limlimits_{n to infty} a_n = alpha in mathbb{R}$ for some $$boxed{1 leq alpha leq 3}$$
To find $alpha$, note that $f(x) = 3 - frac{1}{x}$ is a continuous real valued function for real $x > 0$. So, by continuity, and the fact that $alpha > 0$, we can put limits inside the function like so: $$limlimits_{n to infty}f(a_n) = f(limlimits_{n to infty}a_n)$$ But the LHS is just $limlimits_{n to infty}f(a_n) = limlimits_{n to infty}(3 - frac{1}{a_n}) = limlimits_{n to infty} a_{n + 1} = alpha$ while the RHS is just $f(limlimits_{n to infty}a_n) = f(alpha) = 3 - frac{1}{alpha}$. So, putting LHS $=$ RHS we get
$$alpha = 3 - frac{1}{alpha}$$
$$boxed{alpha^2 - 3 alpha + 1 = 0}$$ Now, solve this quadratic equation to get the value of $alpha$. You will get two values but only one of them is the answer. See if you can eliminate the wrong one using the properties of $alpha$.
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– James odare
Apr 20 at 11:54
$begingroup$
No problem, you're welcome.
$endgroup$
– ZeroXLR
Apr 20 at 11:56
add a comment |
$begingroup$
You have already done most of the work. By showing that $1 leq a_n < a_{n+1} < 3$ where $n in mathbb{Z}_+$, you have demonstrated that the sequence is increasing and bounded above by $3$. Now the Monotone Convergence Theorem says that any increasing sequence that is bounded above converges; indeed it converges to the supremum of ${a_n}_{n in mathbb{Z}_+}$. So you now know that $limlimits_{n to infty} a_n = alpha in mathbb{R}$ for some $$boxed{1 leq alpha leq 3}$$
To find $alpha$, note that $f(x) = 3 - frac{1}{x}$ is a continuous real valued function for real $x > 0$. So, by continuity, and the fact that $alpha > 0$, we can put limits inside the function like so: $$limlimits_{n to infty}f(a_n) = f(limlimits_{n to infty}a_n)$$ But the LHS is just $limlimits_{n to infty}f(a_n) = limlimits_{n to infty}(3 - frac{1}{a_n}) = limlimits_{n to infty} a_{n + 1} = alpha$ while the RHS is just $f(limlimits_{n to infty}a_n) = f(alpha) = 3 - frac{1}{alpha}$. So, putting LHS $=$ RHS we get
$$alpha = 3 - frac{1}{alpha}$$
$$boxed{alpha^2 - 3 alpha + 1 = 0}$$ Now, solve this quadratic equation to get the value of $alpha$. You will get two values but only one of them is the answer. See if you can eliminate the wrong one using the properties of $alpha$.
$endgroup$
You have already done most of the work. By showing that $1 leq a_n < a_{n+1} < 3$ where $n in mathbb{Z}_+$, you have demonstrated that the sequence is increasing and bounded above by $3$. Now the Monotone Convergence Theorem says that any increasing sequence that is bounded above converges; indeed it converges to the supremum of ${a_n}_{n in mathbb{Z}_+}$. So you now know that $limlimits_{n to infty} a_n = alpha in mathbb{R}$ for some $$boxed{1 leq alpha leq 3}$$
To find $alpha$, note that $f(x) = 3 - frac{1}{x}$ is a continuous real valued function for real $x > 0$. So, by continuity, and the fact that $alpha > 0$, we can put limits inside the function like so: $$limlimits_{n to infty}f(a_n) = f(limlimits_{n to infty}a_n)$$ But the LHS is just $limlimits_{n to infty}f(a_n) = limlimits_{n to infty}(3 - frac{1}{a_n}) = limlimits_{n to infty} a_{n + 1} = alpha$ while the RHS is just $f(limlimits_{n to infty}a_n) = f(alpha) = 3 - frac{1}{alpha}$. So, putting LHS $=$ RHS we get
$$alpha = 3 - frac{1}{alpha}$$
$$boxed{alpha^2 - 3 alpha + 1 = 0}$$ Now, solve this quadratic equation to get the value of $alpha$. You will get two values but only one of them is the answer. See if you can eliminate the wrong one using the properties of $alpha$.
edited Apr 20 at 12:28
answered Apr 20 at 11:29
ZeroXLRZeroXLR
1,821620
1,821620
$begingroup$
Thank you very much!
$endgroup$
– James odare
Apr 20 at 11:54
$begingroup$
No problem, you're welcome.
$endgroup$
– ZeroXLR
Apr 20 at 11:56
add a comment |
$begingroup$
Thank you very much!
$endgroup$
– James odare
Apr 20 at 11:54
$begingroup$
No problem, you're welcome.
$endgroup$
– ZeroXLR
Apr 20 at 11:56
$begingroup$
Thank you very much!
$endgroup$
– James odare
Apr 20 at 11:54
$begingroup$
Thank you very much!
$endgroup$
– James odare
Apr 20 at 11:54
$begingroup$
No problem, you're welcome.
$endgroup$
– ZeroXLR
Apr 20 at 11:56
$begingroup$
No problem, you're welcome.
$endgroup$
– ZeroXLR
Apr 20 at 11:56
add a comment |
$begingroup$
Let $P(n)$ be the assertion “$a_ninleft[1,frac{3+sqrt5}2right)$ and $a_{n+1}>a_n$”.
If $n=1$, then this is true, since $a_1=1$ and $a_2=2>1=a_1$.
Let $ninmathbb N$ and assume that $P(n)$ holds. Then:
$a_{n+1}=3-frac1{a_n}<3$. since $a_ngeqslant1>0$;
$a_{n+1}geqslant1$, since $a_{n}geqslant1$ and so $3-frac1{a_n}geqslant3-frac11=2$.
$a_{n+2}-a_{n+1}=3-dfrac1{a_{n+1}}-a_{n+1}=dfrac{3a_{n+1}-1-{a_{n+1}}^2}{{a_{n+1}}^2}=dfrac{varphi(a_{n+1})}{{a_{n+1}}^2}$, where $varphi(x)=-x^2+3x-1$. But the roots of the quadratic polynomial $varphi(x)$ are $dfrac{3pmsqrt5}2$ and so, since $a_{n+1}$ is between them and the coefficient of $x^2$ in $varphi(x)$ is negative, $varphi(a_{n+1})>0$.
So, $P(n+1)$ is proved.
What I wrote above is an inductive proof of the fact that we always have $P(n)$. And it is very easy now to conclude that $lim_{ntoinfty}a_n=dfrac{3+sqrt5}2$.
$endgroup$
$begingroup$
Great explanation thank you
$endgroup$
– James odare
Apr 20 at 11:54
add a comment |
$begingroup$
Let $P(n)$ be the assertion “$a_ninleft[1,frac{3+sqrt5}2right)$ and $a_{n+1}>a_n$”.
If $n=1$, then this is true, since $a_1=1$ and $a_2=2>1=a_1$.
Let $ninmathbb N$ and assume that $P(n)$ holds. Then:
$a_{n+1}=3-frac1{a_n}<3$. since $a_ngeqslant1>0$;
$a_{n+1}geqslant1$, since $a_{n}geqslant1$ and so $3-frac1{a_n}geqslant3-frac11=2$.
$a_{n+2}-a_{n+1}=3-dfrac1{a_{n+1}}-a_{n+1}=dfrac{3a_{n+1}-1-{a_{n+1}}^2}{{a_{n+1}}^2}=dfrac{varphi(a_{n+1})}{{a_{n+1}}^2}$, where $varphi(x)=-x^2+3x-1$. But the roots of the quadratic polynomial $varphi(x)$ are $dfrac{3pmsqrt5}2$ and so, since $a_{n+1}$ is between them and the coefficient of $x^2$ in $varphi(x)$ is negative, $varphi(a_{n+1})>0$.
So, $P(n+1)$ is proved.
What I wrote above is an inductive proof of the fact that we always have $P(n)$. And it is very easy now to conclude that $lim_{ntoinfty}a_n=dfrac{3+sqrt5}2$.
$endgroup$
$begingroup$
Great explanation thank you
$endgroup$
– James odare
Apr 20 at 11:54
add a comment |
$begingroup$
Let $P(n)$ be the assertion “$a_ninleft[1,frac{3+sqrt5}2right)$ and $a_{n+1}>a_n$”.
If $n=1$, then this is true, since $a_1=1$ and $a_2=2>1=a_1$.
Let $ninmathbb N$ and assume that $P(n)$ holds. Then:
$a_{n+1}=3-frac1{a_n}<3$. since $a_ngeqslant1>0$;
$a_{n+1}geqslant1$, since $a_{n}geqslant1$ and so $3-frac1{a_n}geqslant3-frac11=2$.
$a_{n+2}-a_{n+1}=3-dfrac1{a_{n+1}}-a_{n+1}=dfrac{3a_{n+1}-1-{a_{n+1}}^2}{{a_{n+1}}^2}=dfrac{varphi(a_{n+1})}{{a_{n+1}}^2}$, where $varphi(x)=-x^2+3x-1$. But the roots of the quadratic polynomial $varphi(x)$ are $dfrac{3pmsqrt5}2$ and so, since $a_{n+1}$ is between them and the coefficient of $x^2$ in $varphi(x)$ is negative, $varphi(a_{n+1})>0$.
So, $P(n+1)$ is proved.
What I wrote above is an inductive proof of the fact that we always have $P(n)$. And it is very easy now to conclude that $lim_{ntoinfty}a_n=dfrac{3+sqrt5}2$.
$endgroup$
Let $P(n)$ be the assertion “$a_ninleft[1,frac{3+sqrt5}2right)$ and $a_{n+1}>a_n$”.
If $n=1$, then this is true, since $a_1=1$ and $a_2=2>1=a_1$.
Let $ninmathbb N$ and assume that $P(n)$ holds. Then:
$a_{n+1}=3-frac1{a_n}<3$. since $a_ngeqslant1>0$;
$a_{n+1}geqslant1$, since $a_{n}geqslant1$ and so $3-frac1{a_n}geqslant3-frac11=2$.
$a_{n+2}-a_{n+1}=3-dfrac1{a_{n+1}}-a_{n+1}=dfrac{3a_{n+1}-1-{a_{n+1}}^2}{{a_{n+1}}^2}=dfrac{varphi(a_{n+1})}{{a_{n+1}}^2}$, where $varphi(x)=-x^2+3x-1$. But the roots of the quadratic polynomial $varphi(x)$ are $dfrac{3pmsqrt5}2$ and so, since $a_{n+1}$ is between them and the coefficient of $x^2$ in $varphi(x)$ is negative, $varphi(a_{n+1})>0$.
So, $P(n+1)$ is proved.
What I wrote above is an inductive proof of the fact that we always have $P(n)$. And it is very easy now to conclude that $lim_{ntoinfty}a_n=dfrac{3+sqrt5}2$.
answered Apr 20 at 11:33
José Carlos SantosJosé Carlos Santos
177k24138251
177k24138251
$begingroup$
Great explanation thank you
$endgroup$
– James odare
Apr 20 at 11:54
add a comment |
$begingroup$
Great explanation thank you
$endgroup$
– James odare
Apr 20 at 11:54
$begingroup$
Great explanation thank you
$endgroup$
– James odare
Apr 20 at 11:54
$begingroup$
Great explanation thank you
$endgroup$
– James odare
Apr 20 at 11:54
add a comment |
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