Limit $ a_{n+1} = frac{1}{3}(2a_n + frac{5}{a_n^{2}}), a_1 > 0 $











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Let $ a_{n+1} = frac{1}{3}(2a_n + frac{5}{a_n^{2}}), a_1 > 0 $. Show the sequence converge and find it's limit.




I did the following:



We see that $a_n$ is positive for every n obviously (considering $a_1 > 0$). Then:



$$ a_{n+1} - a_n = frac{2a_n^{3} + 5}{3a_n^{2}} - a_n = frac{5 - a_n^{3}}{3a_n^{2}}$$



We see that for each $a_n > ^3sqrt{5}$ the sequence is decreasing.
That means the proof collapses to prove $a_n > ^3sqrt{5}$ in order for $a_n$ to be decreasing. Suppose $ a_n > ^3sqrt{5}$ then obviously $a_{n+1} = frac{2a_n^{3} + 5}{3a_n^{2}} > frac{2(sqrt[3]{5})^{3} + 5}{3(sqrt[3]{5})^{2}} = sqrt[3]{5}$ so we deduced $a_{n+1} > ^sqrt[3]{5} $.



(It's important to note $a_1$ might be less then $sqrt[3]{5}$, hence the sequence will decrease from the second element.)
So we observed $a_n$ is always positive and decreasig hence it's bounded below so it's bounded. As for it's limit:



$$begin{align} L = lim_{n to infty} a_{n+1} = lim_{n to infty} frac{1}{3}(2a_n + frac{5}{a_n^{2}}) = frac{1}{3}(2L + frac{5}{L^{2}}) = sqrt[3]{5}end{align}$$



I'm not sure as of the correctness of the solution. I'm aware that it's not a contradiction because it only implies it's in the epsilon-environment of $sqrt[3]{5}$, or rather in this case $(sqrt[3]{5}, sqrt[3]{5}+epsilon)$.



Is it correct?



EDIT: If anyone encounter a question in the likes, to sum it up you obviously should show the sequence is montonic and bounded. In this case for example we can show it using AM-GM inequality to see the sequence is bounded by $sqrt[3]{5}$ and then in the difference of two consequective elements we see the difference is negative due to every element being greater then the third root of 5. From here we know the sequence is bounded below and decreasing hence it's bounded. If we plug in the limit of the sequence to itself we will see it's the third root of five. And that's it.










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  • You can typeset $sqrt[3]{5}$ as sqrt[3]{5}
    – Martin R
    Nov 22 at 7:59















up vote
2
down vote

favorite













Let $ a_{n+1} = frac{1}{3}(2a_n + frac{5}{a_n^{2}}), a_1 > 0 $. Show the sequence converge and find it's limit.




I did the following:



We see that $a_n$ is positive for every n obviously (considering $a_1 > 0$). Then:



$$ a_{n+1} - a_n = frac{2a_n^{3} + 5}{3a_n^{2}} - a_n = frac{5 - a_n^{3}}{3a_n^{2}}$$



We see that for each $a_n > ^3sqrt{5}$ the sequence is decreasing.
That means the proof collapses to prove $a_n > ^3sqrt{5}$ in order for $a_n$ to be decreasing. Suppose $ a_n > ^3sqrt{5}$ then obviously $a_{n+1} = frac{2a_n^{3} + 5}{3a_n^{2}} > frac{2(sqrt[3]{5})^{3} + 5}{3(sqrt[3]{5})^{2}} = sqrt[3]{5}$ so we deduced $a_{n+1} > ^sqrt[3]{5} $.



(It's important to note $a_1$ might be less then $sqrt[3]{5}$, hence the sequence will decrease from the second element.)
So we observed $a_n$ is always positive and decreasig hence it's bounded below so it's bounded. As for it's limit:



$$begin{align} L = lim_{n to infty} a_{n+1} = lim_{n to infty} frac{1}{3}(2a_n + frac{5}{a_n^{2}}) = frac{1}{3}(2L + frac{5}{L^{2}}) = sqrt[3]{5}end{align}$$



I'm not sure as of the correctness of the solution. I'm aware that it's not a contradiction because it only implies it's in the epsilon-environment of $sqrt[3]{5}$, or rather in this case $(sqrt[3]{5}, sqrt[3]{5}+epsilon)$.



Is it correct?



EDIT: If anyone encounter a question in the likes, to sum it up you obviously should show the sequence is montonic and bounded. In this case for example we can show it using AM-GM inequality to see the sequence is bounded by $sqrt[3]{5}$ and then in the difference of two consequective elements we see the difference is negative due to every element being greater then the third root of 5. From here we know the sequence is bounded below and decreasing hence it's bounded. If we plug in the limit of the sequence to itself we will see it's the third root of five. And that's it.










share|cite|improve this question
























  • You can typeset $sqrt[3]{5}$ as sqrt[3]{5}
    – Martin R
    Nov 22 at 7:59













up vote
2
down vote

favorite









up vote
2
down vote

favorite












Let $ a_{n+1} = frac{1}{3}(2a_n + frac{5}{a_n^{2}}), a_1 > 0 $. Show the sequence converge and find it's limit.




I did the following:



We see that $a_n$ is positive for every n obviously (considering $a_1 > 0$). Then:



$$ a_{n+1} - a_n = frac{2a_n^{3} + 5}{3a_n^{2}} - a_n = frac{5 - a_n^{3}}{3a_n^{2}}$$



We see that for each $a_n > ^3sqrt{5}$ the sequence is decreasing.
That means the proof collapses to prove $a_n > ^3sqrt{5}$ in order for $a_n$ to be decreasing. Suppose $ a_n > ^3sqrt{5}$ then obviously $a_{n+1} = frac{2a_n^{3} + 5}{3a_n^{2}} > frac{2(sqrt[3]{5})^{3} + 5}{3(sqrt[3]{5})^{2}} = sqrt[3]{5}$ so we deduced $a_{n+1} > ^sqrt[3]{5} $.



(It's important to note $a_1$ might be less then $sqrt[3]{5}$, hence the sequence will decrease from the second element.)
So we observed $a_n$ is always positive and decreasig hence it's bounded below so it's bounded. As for it's limit:



$$begin{align} L = lim_{n to infty} a_{n+1} = lim_{n to infty} frac{1}{3}(2a_n + frac{5}{a_n^{2}}) = frac{1}{3}(2L + frac{5}{L^{2}}) = sqrt[3]{5}end{align}$$



I'm not sure as of the correctness of the solution. I'm aware that it's not a contradiction because it only implies it's in the epsilon-environment of $sqrt[3]{5}$, or rather in this case $(sqrt[3]{5}, sqrt[3]{5}+epsilon)$.



Is it correct?



EDIT: If anyone encounter a question in the likes, to sum it up you obviously should show the sequence is montonic and bounded. In this case for example we can show it using AM-GM inequality to see the sequence is bounded by $sqrt[3]{5}$ and then in the difference of two consequective elements we see the difference is negative due to every element being greater then the third root of 5. From here we know the sequence is bounded below and decreasing hence it's bounded. If we plug in the limit of the sequence to itself we will see it's the third root of five. And that's it.










share|cite|improve this question
















Let $ a_{n+1} = frac{1}{3}(2a_n + frac{5}{a_n^{2}}), a_1 > 0 $. Show the sequence converge and find it's limit.




I did the following:



We see that $a_n$ is positive for every n obviously (considering $a_1 > 0$). Then:



$$ a_{n+1} - a_n = frac{2a_n^{3} + 5}{3a_n^{2}} - a_n = frac{5 - a_n^{3}}{3a_n^{2}}$$



We see that for each $a_n > ^3sqrt{5}$ the sequence is decreasing.
That means the proof collapses to prove $a_n > ^3sqrt{5}$ in order for $a_n$ to be decreasing. Suppose $ a_n > ^3sqrt{5}$ then obviously $a_{n+1} = frac{2a_n^{3} + 5}{3a_n^{2}} > frac{2(sqrt[3]{5})^{3} + 5}{3(sqrt[3]{5})^{2}} = sqrt[3]{5}$ so we deduced $a_{n+1} > ^sqrt[3]{5} $.



(It's important to note $a_1$ might be less then $sqrt[3]{5}$, hence the sequence will decrease from the second element.)
So we observed $a_n$ is always positive and decreasig hence it's bounded below so it's bounded. As for it's limit:



$$begin{align} L = lim_{n to infty} a_{n+1} = lim_{n to infty} frac{1}{3}(2a_n + frac{5}{a_n^{2}}) = frac{1}{3}(2L + frac{5}{L^{2}}) = sqrt[3]{5}end{align}$$



I'm not sure as of the correctness of the solution. I'm aware that it's not a contradiction because it only implies it's in the epsilon-environment of $sqrt[3]{5}$, or rather in this case $(sqrt[3]{5}, sqrt[3]{5}+epsilon)$.



Is it correct?



EDIT: If anyone encounter a question in the likes, to sum it up you obviously should show the sequence is montonic and bounded. In this case for example we can show it using AM-GM inequality to see the sequence is bounded by $sqrt[3]{5}$ and then in the difference of two consequective elements we see the difference is negative due to every element being greater then the third root of 5. From here we know the sequence is bounded below and decreasing hence it's bounded. If we plug in the limit of the sequence to itself we will see it's the third root of five. And that's it.







calculus sequences-and-series analysis recursion






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edited Nov 22 at 10:18









Christian Blatter

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asked Nov 21 at 16:28









Moshe

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206












  • You can typeset $sqrt[3]{5}$ as sqrt[3]{5}
    – Martin R
    Nov 22 at 7:59


















  • You can typeset $sqrt[3]{5}$ as sqrt[3]{5}
    – Martin R
    Nov 22 at 7:59
















You can typeset $sqrt[3]{5}$ as sqrt[3]{5}
– Martin R
Nov 22 at 7:59




You can typeset $sqrt[3]{5}$ as sqrt[3]{5}
– Martin R
Nov 22 at 7:59










2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










First observe that
$$
a_{n+1}=frac{1}{3}left(2a_n+frac{5}{a_n^2}right)=frac{1}{3}left(a_n+a_n+frac{5}{a_n^2}right)ge left(a_ncdot a_ncdot frac{5}{a_n^2}right)^{1/3}=5^{1/3}.
$$

Hence, $a_nge 5^{1/3}$, for all $n>1$.



Next observe that
$$
f(x)=frac{1}{3}left(2x+frac{5}{x^2}right)<x,
$$

for all $x$ in $[5^{1/3},infty)$, since
$$
x-f(x)=frac{1}{3}left(x-frac{5}{x^2}right)=frac{x^3-5}{3x^2}ge 0.
$$



Hence
$$
5^{1/3}le a_{n+1}=f(a_n)le a_n, quad text{for all $n>1$.}
$$

Thus ${a_n}$ is bounded and decreasing, and therefore it is convergent.



Also, if $a_nto a$, then, as $f$ is continuous in $(0,infty)$, then
$$
a=lim a_n=lim a_{n+1}=lim f(a_n)=lim f(a)
$$

and hence the limit $a$ satisfies
$$
a=f(a)=frac{1}{3}left(2a+frac{5}{a^2}right)
$$

and thus $a^3=5$, which implies that $a_nto 5^{1/3}$.






share|cite|improve this answer























  • Thank you, that's that exact solution I intended.
    – Moshe
    Nov 22 at 11:11


















up vote
1
down vote













In order to for the sequence to converge:
$$lim_{ntoinfty}(a_{n+1}-a_{n})=0$$
Then, we can plug in the recursion formula for $a_{n+1}$ and we get
$$lim_{ntoinfty}(a_{n+1}-a_n)=lim_{ntoinfty}(frac{2}{3}a_n+frac{5}{3a_n^2}-a_n)=lim_{ntoinfty}(frac{1}{3}+frac{5}{3a_n^2})=3 lim_{ntoinfty}(-a_n+frac{5}{a_n^2})=3 lim_{ntoinfty}frac{5-a_n^3}{a_n^3}=0$$
As such, $lim_{ntoinfty} a_n=sqrt[3]5$






share|cite|improve this answer








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MoKo19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • The first statement is false.
    – lhf
    Nov 22 at 10:13












  • It is true that $a_2geqroot3of 5$ in any case, but you have not shown this.
    – Christian Blatter
    Nov 22 at 10:27











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










First observe that
$$
a_{n+1}=frac{1}{3}left(2a_n+frac{5}{a_n^2}right)=frac{1}{3}left(a_n+a_n+frac{5}{a_n^2}right)ge left(a_ncdot a_ncdot frac{5}{a_n^2}right)^{1/3}=5^{1/3}.
$$

Hence, $a_nge 5^{1/3}$, for all $n>1$.



Next observe that
$$
f(x)=frac{1}{3}left(2x+frac{5}{x^2}right)<x,
$$

for all $x$ in $[5^{1/3},infty)$, since
$$
x-f(x)=frac{1}{3}left(x-frac{5}{x^2}right)=frac{x^3-5}{3x^2}ge 0.
$$



Hence
$$
5^{1/3}le a_{n+1}=f(a_n)le a_n, quad text{for all $n>1$.}
$$

Thus ${a_n}$ is bounded and decreasing, and therefore it is convergent.



Also, if $a_nto a$, then, as $f$ is continuous in $(0,infty)$, then
$$
a=lim a_n=lim a_{n+1}=lim f(a_n)=lim f(a)
$$

and hence the limit $a$ satisfies
$$
a=f(a)=frac{1}{3}left(2a+frac{5}{a^2}right)
$$

and thus $a^3=5$, which implies that $a_nto 5^{1/3}$.






share|cite|improve this answer























  • Thank you, that's that exact solution I intended.
    – Moshe
    Nov 22 at 11:11















up vote
1
down vote



accepted










First observe that
$$
a_{n+1}=frac{1}{3}left(2a_n+frac{5}{a_n^2}right)=frac{1}{3}left(a_n+a_n+frac{5}{a_n^2}right)ge left(a_ncdot a_ncdot frac{5}{a_n^2}right)^{1/3}=5^{1/3}.
$$

Hence, $a_nge 5^{1/3}$, for all $n>1$.



Next observe that
$$
f(x)=frac{1}{3}left(2x+frac{5}{x^2}right)<x,
$$

for all $x$ in $[5^{1/3},infty)$, since
$$
x-f(x)=frac{1}{3}left(x-frac{5}{x^2}right)=frac{x^3-5}{3x^2}ge 0.
$$



Hence
$$
5^{1/3}le a_{n+1}=f(a_n)le a_n, quad text{for all $n>1$.}
$$

Thus ${a_n}$ is bounded and decreasing, and therefore it is convergent.



Also, if $a_nto a$, then, as $f$ is continuous in $(0,infty)$, then
$$
a=lim a_n=lim a_{n+1}=lim f(a_n)=lim f(a)
$$

and hence the limit $a$ satisfies
$$
a=f(a)=frac{1}{3}left(2a+frac{5}{a^2}right)
$$

and thus $a^3=5$, which implies that $a_nto 5^{1/3}$.






share|cite|improve this answer























  • Thank you, that's that exact solution I intended.
    – Moshe
    Nov 22 at 11:11













up vote
1
down vote



accepted







up vote
1
down vote



accepted






First observe that
$$
a_{n+1}=frac{1}{3}left(2a_n+frac{5}{a_n^2}right)=frac{1}{3}left(a_n+a_n+frac{5}{a_n^2}right)ge left(a_ncdot a_ncdot frac{5}{a_n^2}right)^{1/3}=5^{1/3}.
$$

Hence, $a_nge 5^{1/3}$, for all $n>1$.



Next observe that
$$
f(x)=frac{1}{3}left(2x+frac{5}{x^2}right)<x,
$$

for all $x$ in $[5^{1/3},infty)$, since
$$
x-f(x)=frac{1}{3}left(x-frac{5}{x^2}right)=frac{x^3-5}{3x^2}ge 0.
$$



Hence
$$
5^{1/3}le a_{n+1}=f(a_n)le a_n, quad text{for all $n>1$.}
$$

Thus ${a_n}$ is bounded and decreasing, and therefore it is convergent.



Also, if $a_nto a$, then, as $f$ is continuous in $(0,infty)$, then
$$
a=lim a_n=lim a_{n+1}=lim f(a_n)=lim f(a)
$$

and hence the limit $a$ satisfies
$$
a=f(a)=frac{1}{3}left(2a+frac{5}{a^2}right)
$$

and thus $a^3=5$, which implies that $a_nto 5^{1/3}$.






share|cite|improve this answer














First observe that
$$
a_{n+1}=frac{1}{3}left(2a_n+frac{5}{a_n^2}right)=frac{1}{3}left(a_n+a_n+frac{5}{a_n^2}right)ge left(a_ncdot a_ncdot frac{5}{a_n^2}right)^{1/3}=5^{1/3}.
$$

Hence, $a_nge 5^{1/3}$, for all $n>1$.



Next observe that
$$
f(x)=frac{1}{3}left(2x+frac{5}{x^2}right)<x,
$$

for all $x$ in $[5^{1/3},infty)$, since
$$
x-f(x)=frac{1}{3}left(x-frac{5}{x^2}right)=frac{x^3-5}{3x^2}ge 0.
$$



Hence
$$
5^{1/3}le a_{n+1}=f(a_n)le a_n, quad text{for all $n>1$.}
$$

Thus ${a_n}$ is bounded and decreasing, and therefore it is convergent.



Also, if $a_nto a$, then, as $f$ is continuous in $(0,infty)$, then
$$
a=lim a_n=lim a_{n+1}=lim f(a_n)=lim f(a)
$$

and hence the limit $a$ satisfies
$$
a=f(a)=frac{1}{3}left(2a+frac{5}{a^2}right)
$$

and thus $a^3=5$, which implies that $a_nto 5^{1/3}$.







share|cite|improve this answer














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edited Nov 22 at 11:45

























answered Nov 22 at 10:44









Yiorgos S. Smyrlis

61.7k1383161




61.7k1383161












  • Thank you, that's that exact solution I intended.
    – Moshe
    Nov 22 at 11:11


















  • Thank you, that's that exact solution I intended.
    – Moshe
    Nov 22 at 11:11
















Thank you, that's that exact solution I intended.
– Moshe
Nov 22 at 11:11




Thank you, that's that exact solution I intended.
– Moshe
Nov 22 at 11:11










up vote
1
down vote













In order to for the sequence to converge:
$$lim_{ntoinfty}(a_{n+1}-a_{n})=0$$
Then, we can plug in the recursion formula for $a_{n+1}$ and we get
$$lim_{ntoinfty}(a_{n+1}-a_n)=lim_{ntoinfty}(frac{2}{3}a_n+frac{5}{3a_n^2}-a_n)=lim_{ntoinfty}(frac{1}{3}+frac{5}{3a_n^2})=3 lim_{ntoinfty}(-a_n+frac{5}{a_n^2})=3 lim_{ntoinfty}frac{5-a_n^3}{a_n^3}=0$$
As such, $lim_{ntoinfty} a_n=sqrt[3]5$






share|cite|improve this answer








New contributor




MoKo19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • The first statement is false.
    – lhf
    Nov 22 at 10:13












  • It is true that $a_2geqroot3of 5$ in any case, but you have not shown this.
    – Christian Blatter
    Nov 22 at 10:27















up vote
1
down vote













In order to for the sequence to converge:
$$lim_{ntoinfty}(a_{n+1}-a_{n})=0$$
Then, we can plug in the recursion formula for $a_{n+1}$ and we get
$$lim_{ntoinfty}(a_{n+1}-a_n)=lim_{ntoinfty}(frac{2}{3}a_n+frac{5}{3a_n^2}-a_n)=lim_{ntoinfty}(frac{1}{3}+frac{5}{3a_n^2})=3 lim_{ntoinfty}(-a_n+frac{5}{a_n^2})=3 lim_{ntoinfty}frac{5-a_n^3}{a_n^3}=0$$
As such, $lim_{ntoinfty} a_n=sqrt[3]5$






share|cite|improve this answer








New contributor




MoKo19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • The first statement is false.
    – lhf
    Nov 22 at 10:13












  • It is true that $a_2geqroot3of 5$ in any case, but you have not shown this.
    – Christian Blatter
    Nov 22 at 10:27













up vote
1
down vote










up vote
1
down vote









In order to for the sequence to converge:
$$lim_{ntoinfty}(a_{n+1}-a_{n})=0$$
Then, we can plug in the recursion formula for $a_{n+1}$ and we get
$$lim_{ntoinfty}(a_{n+1}-a_n)=lim_{ntoinfty}(frac{2}{3}a_n+frac{5}{3a_n^2}-a_n)=lim_{ntoinfty}(frac{1}{3}+frac{5}{3a_n^2})=3 lim_{ntoinfty}(-a_n+frac{5}{a_n^2})=3 lim_{ntoinfty}frac{5-a_n^3}{a_n^3}=0$$
As such, $lim_{ntoinfty} a_n=sqrt[3]5$






share|cite|improve this answer








New contributor




MoKo19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









In order to for the sequence to converge:
$$lim_{ntoinfty}(a_{n+1}-a_{n})=0$$
Then, we can plug in the recursion formula for $a_{n+1}$ and we get
$$lim_{ntoinfty}(a_{n+1}-a_n)=lim_{ntoinfty}(frac{2}{3}a_n+frac{5}{3a_n^2}-a_n)=lim_{ntoinfty}(frac{1}{3}+frac{5}{3a_n^2})=3 lim_{ntoinfty}(-a_n+frac{5}{a_n^2})=3 lim_{ntoinfty}frac{5-a_n^3}{a_n^3}=0$$
As such, $lim_{ntoinfty} a_n=sqrt[3]5$







share|cite|improve this answer








New contributor




MoKo19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this answer



share|cite|improve this answer






New contributor




MoKo19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









answered Nov 21 at 17:20









MoKo19

663




663




New contributor




MoKo19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





MoKo19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






MoKo19 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • The first statement is false.
    – lhf
    Nov 22 at 10:13












  • It is true that $a_2geqroot3of 5$ in any case, but you have not shown this.
    – Christian Blatter
    Nov 22 at 10:27


















  • The first statement is false.
    – lhf
    Nov 22 at 10:13












  • It is true that $a_2geqroot3of 5$ in any case, but you have not shown this.
    – Christian Blatter
    Nov 22 at 10:27
















The first statement is false.
– lhf
Nov 22 at 10:13






The first statement is false.
– lhf
Nov 22 at 10:13














It is true that $a_2geqroot3of 5$ in any case, but you have not shown this.
– Christian Blatter
Nov 22 at 10:27




It is true that $a_2geqroot3of 5$ in any case, but you have not shown this.
– Christian Blatter
Nov 22 at 10:27


















 

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