Show that the sequence $a_{n+1}=sqrt{2+a_n}$ is convergent. [duplicate]
This question already has an answer here:
Evaluating the limit of a sequence given by recurrence relation $a_1=sqrt2$, $a_{n+1}=sqrt{2+a_n}$. Is my solution correct? [duplicate]
2 answers
Let $a_{n+1}=sqrt{2+a_n}, a_1=sqrt{3}$. Show that it is convergent.
I know that this is a classic nested square root sequence but how do I prove it's convergence? To know it's limit I can just take the limit on both sides and find $L$, or rewrite $a_n=sqrt{2+sqrt{2+sqrt{2+sqrt{2+sqrt{...}}}}}$
sequences-and-series convergence
edited Nov 25 '18 at 9:16
Nosrati
26.5k62353
asked Nov 25 '18 at 9:12
C. Cristi
1,476218
marked as duplicate by Nosrati, Community♦ Nov 25 '18 at 9:27
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Evaluating the limit of a sequence given by recurrence relation $a_1=sqrt2$, $a_{n+1}=sqrt{2+a_n}$. Is my solution correct? [duplicate]
2 answers
Let $a_{n+1}=sqrt{2+a_n}, a_1=sqrt{3}$. Show that it is convergent.
I know that this is a classic nested square root sequence but how do I prove it's convergence? To know it's limit I can just take the limit on both sides and find $L$, or rewrite $a_n=sqrt{2+sqrt{2+sqrt{2+sqrt{2+sqrt{...}}}}}$
sequences-and-series convergence
edited Nov 25 '18 at 9:16
Nosrati
26.5k62353
asked Nov 25 '18 at 9:12
C. Cristi
1,476218
marked as duplicate by Nosrati, Community♦ Nov 25 '18 at 9:27
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Bounded and increasing.
– Yves Daoust
Nov 25 '18 at 9:14
1
$$a_n<2implies a_n+2<4implies a_{n+1}<2.$$ $$frac12<a_n<2implies a_n^2-a_n-2=left(a_n-frac12right)^2-frac94le0implies a_nlesqrt{a_n+2}=a_{n+1}.$$ Hence the sequence is bounded and monotonous.
– Yves Daoust
Nov 25 '18 at 9:28
add a comment |
This question already has an answer here:
Evaluating the limit of a sequence given by recurrence relation $a_1=sqrt2$, $a_{n+1}=sqrt{2+a_n}$. Is my solution correct? [duplicate]
2 answers
Let $a_{n+1}=sqrt{2+a_n}, a_1=sqrt{3}$. Show that it is convergent.
I know that this is a classic nested square root sequence but how do I prove it's convergence? To know it's limit I can just take the limit on both sides and find $L$, or rewrite $a_n=sqrt{2+sqrt{2+sqrt{2+sqrt{2+sqrt{...}}}}}$
sequences-and-series convergence
edited Nov 25 '18 at 9:16
Nosrati
26.5k62353
asked Nov 25 '18 at 9:12
C. Cristi
1,476218
This question already has an answer here:
Evaluating the limit of a sequence given by recurrence relation $a_1=sqrt2$, $a_{n+1}=sqrt{2+a_n}$. Is my solution correct? [duplicate]
2 answers
Let $a_{n+1}=sqrt{2+a_n}, a_1=sqrt{3}$. Show that it is convergent.
I know that this is a classic nested square root sequence but how do I prove it's convergence? To know it's limit I can just take the limit on both sides and find $L$, or rewrite $a_n=sqrt{2+sqrt{2+sqrt{2+sqrt{2+sqrt{...}}}}}$
This question already has an answer here:
Evaluating the limit of a sequence given by recurrence relation $a_1=sqrt2$, $a_{n+1}=sqrt{2+a_n}$. Is my solution correct? [duplicate]
2 answers
sequences-and-series convergence
sequences-and-series convergence
edited Nov 25 '18 at 9:16
Nosrati
26.5k62353
asked Nov 25 '18 at 9:12
C. Cristi
1,476218
edited Nov 25 '18 at 9:16
Nosrati
26.5k62353
asked Nov 25 '18 at 9:12
C. Cristi
1,476218
edited Nov 25 '18 at 9:16
Nosrati
26.5k62353
edited Nov 25 '18 at 9:16
Nosrati
26.5k62353
edited Nov 25 '18 at 9:16
Nosrati
26.5k62353
26.5k62353
asked Nov 25 '18 at 9:12
C. Cristi
1,476218
asked Nov 25 '18 at 9:12
C. Cristi
1,476218
asked Nov 25 '18 at 9:12
C. Cristi
1,476218
1,476218
marked as duplicate by Nosrati, Community♦ Nov 25 '18 at 9:27
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Nosrati, Community♦ Nov 25 '18 at 9:27
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Bounded and increasing.
– Yves Daoust
Nov 25 '18 at 9:14
1
$$a_n<2implies a_n+2<4implies a_{n+1}<2.$$ $$frac12<a_n<2implies a_n^2-a_n-2=left(a_n-frac12right)^2-frac94le0implies a_nlesqrt{a_n+2}=a_{n+1}.$$ Hence the sequence is bounded and monotonous.
– Yves Daoust
Nov 25 '18 at 9:28
add a comment |
Bounded and increasing.
– Yves Daoust
Nov 25 '18 at 9:14
1
$$a_n<2implies a_n+2<4implies a_{n+1}<2.$$ $$frac12<a_n<2implies a_n^2-a_n-2=left(a_n-frac12right)^2-frac94le0implies a_nlesqrt{a_n+2}=a_{n+1}.$$ Hence the sequence is bounded and monotonous.
– Yves Daoust
Nov 25 '18 at 9:28
Bounded and increasing.
– Yves Daoust
Nov 25 '18 at 9:14
Bounded and increasing.
– Yves Daoust
Nov 25 '18 at 9:14
1
1
$$a_n<2implies a_n+2<4implies a_{n+1}<2.$$ $$frac12<a_n<2implies a_n^2-a_n-2=left(a_n-frac12right)^2-frac94le0implies a_nlesqrt{a_n+2}=a_{n+1}.$$ Hence the sequence is bounded and monotonous.
– Yves Daoust
Nov 25 '18 at 9:28
$$a_n<2implies a_n+2<4implies a_{n+1}<2.$$ $$frac12<a_n<2implies a_n^2-a_n-2=left(a_n-frac12right)^2-frac94le0implies a_nlesqrt{a_n+2}=a_{n+1}.$$ Hence the sequence is bounded and monotonous.
– Yves Daoust
Nov 25 '18 at 9:28
add a comment |
1 Answer
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Define $$b_n=a_n-2$$therefore $$b_{n+1}=a_{n+1}-2=sqrt{a_n+2}-2={a_n-2over 2+sqrt{a_n+2}}={b_nover 2+sqrt {a_n+2}}$$therefore $$|b_{n+1}|=|{b_nover 2+sqrt {a_n+2}}|le |{b_n over 2}|le cdots le {|b_1|over 2^n}$$therefore $b_nto 0$ when $nto infty$ which means that $a_nto 2$
answered Nov 25 '18 at 9:20
Mostafa Ayaz
13.6k3836
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Define $$b_n=a_n-2$$therefore $$b_{n+1}=a_{n+1}-2=sqrt{a_n+2}-2={a_n-2over 2+sqrt{a_n+2}}={b_nover 2+sqrt {a_n+2}}$$therefore $$|b_{n+1}|=|{b_nover 2+sqrt {a_n+2}}|le |{b_n over 2}|le cdots le {|b_1|over 2^n}$$therefore $b_nto 0$ when $nto infty$ which means that $a_nto 2$
answered Nov 25 '18 at 9:20
Mostafa Ayaz
13.6k3836
add a comment |
Define $$b_n=a_n-2$$therefore $$b_{n+1}=a_{n+1}-2=sqrt{a_n+2}-2={a_n-2over 2+sqrt{a_n+2}}={b_nover 2+sqrt {a_n+2}}$$therefore $$|b_{n+1}|=|{b_nover 2+sqrt {a_n+2}}|le |{b_n over 2}|le cdots le {|b_1|over 2^n}$$therefore $b_nto 0$ when $nto infty$ which means that $a_nto 2$
answered Nov 25 '18 at 9:20
Mostafa Ayaz
13.6k3836
add a comment |
Define $$b_n=a_n-2$$therefore $$b_{n+1}=a_{n+1}-2=sqrt{a_n+2}-2={a_n-2over 2+sqrt{a_n+2}}={b_nover 2+sqrt {a_n+2}}$$therefore $$|b_{n+1}|=|{b_nover 2+sqrt {a_n+2}}|le |{b_n over 2}|le cdots le {|b_1|over 2^n}$$therefore $b_nto 0$ when $nto infty$ which means that $a_nto 2$
answered Nov 25 '18 at 9:20
Mostafa Ayaz
13.6k3836
Define $$b_n=a_n-2$$therefore $$b_{n+1}=a_{n+1}-2=sqrt{a_n+2}-2={a_n-2over 2+sqrt{a_n+2}}={b_nover 2+sqrt {a_n+2}}$$therefore $$|b_{n+1}|=|{b_nover 2+sqrt {a_n+2}}|le |{b_n over 2}|le cdots le {|b_1|over 2^n}$$therefore $b_nto 0$ when $nto infty$ which means that $a_nto 2$
answered Nov 25 '18 at 9:20
Mostafa Ayaz
13.6k3836
answered Nov 25 '18 at 9:20
Mostafa Ayaz
13.6k3836
answered Nov 25 '18 at 9:20
Mostafa Ayaz
13.6k3836
answered Nov 25 '18 at 9:20
Mostafa Ayaz
13.6k3836
13.6k3836
add a comment |
add a comment |
Bounded and increasing.
– Yves Daoust
Nov 25 '18 at 9:14
1
$$a_n<2implies a_n+2<4implies a_{n+1}<2.$$ $$frac12<a_n<2implies a_n^2-a_n-2=left(a_n-frac12right)^2-frac94le0implies a_nlesqrt{a_n+2}=a_{n+1}.$$ Hence the sequence is bounded and monotonous.
– Yves Daoust
Nov 25 '18 at 9:28