Mixing
A sequence defined inductively.
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Let $a_n$ be a sequence such that $a_1=1$ and $$a_n+1=a_n+frac1a_n^2.$$ Prove that $a_n$ is unbounded. Any hint?
sequences-and-series
asked 24 mins ago
bateman
1,817919
add a comment |Â
up vote
2
down vote
favorite
Let $a_n$ be a sequence such that $a_1=1$ and $$a_n+1=a_n+frac1a_n^2.$$ Prove that $a_n$ is unbounded. Any hint?
sequences-and-series
asked 24 mins ago
bateman
1,817919
1
What have you tried? Have you considered the opposite and supposed that there exists $L$ such that $L:=lim_ntoinftya_n$? What happens then?
– TheSimpliFire
22 mins ago
1
@TheSimpliFire I thought that the sequence is positive and monotone, so if I assume it's bounded, then it will be convergent. But I don't know how to get a contradiction from that.
– bateman
19 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $a_n$ be a sequence such that $a_1=1$ and $$a_n+1=a_n+frac1a_n^2.$$ Prove that $a_n$ is unbounded. Any hint?
sequences-and-series
asked 24 mins ago
bateman
1,817919
Let $a_n$ be a sequence such that $a_1=1$ and $$a_n+1=a_n+frac1a_n^2.$$ Prove that $a_n$ is unbounded. Any hint?
sequences-and-series
sequences-and-series
asked 24 mins ago
bateman
1,817919
asked 24 mins ago
bateman
1,817919
edited 18 mins ago
asked 24 mins ago
bateman
1,817919
asked 24 mins ago
bateman
1,817919
asked 24 mins ago
bateman
1,817919
1,817919
1
What have you tried? Have you considered the opposite and supposed that there exists $L$ such that $L:=lim_ntoinftya_n$? What happens then?
– TheSimpliFire
22 mins ago
1
@TheSimpliFire I thought that the sequence is positive and monotone, so if I assume it's bounded, then it will be convergent. But I don't know how to get a contradiction from that.
– bateman
19 mins ago
add a comment |Â
1
What have you tried? Have you considered the opposite and supposed that there exists $L$ such that $L:=lim_ntoinftya_n$? What happens then?
– TheSimpliFire
22 mins ago
1
@TheSimpliFire I thought that the sequence is positive and monotone, so if I assume it's bounded, then it will be convergent. But I don't know how to get a contradiction from that.
– bateman
19 mins ago
1
1
What have you tried? Have you considered the opposite and supposed that there exists $L$ such that $L:=lim_ntoinftya_n$? What happens then?
– TheSimpliFire
22 mins ago
What have you tried? Have you considered the opposite and supposed that there exists $L$ such that $L:=lim_ntoinftya_n$? What happens then?
– TheSimpliFire
22 mins ago
1
1
@TheSimpliFire I thought that the sequence is positive and monotone, so if I assume it's bounded, then it will be convergent. But I don't know how to get a contradiction from that.
– bateman
19 mins ago
@TheSimpliFire I thought that the sequence is positive and monotone, so if I assume it's bounded, then it will be convergent. But I don't know how to get a contradiction from that.
– bateman
19 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Suppose it is bounded then the limit exists since it is a strictly increasing sequence.So $L = L + 1/L^2$ . This shows $L = infty$ contradiction to it being bounded.
answered 17 mins ago
DeepSea
69.6k54285
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Suppose it is bounded then the limit exists since it is a strictly increasing sequence.So $L = L + 1/L^2$ . This shows $L = infty$ contradiction to it being bounded.
answered 17 mins ago
DeepSea
69.6k54285
add a comment |Â
up vote
4
down vote
accepted
Suppose it is bounded then the limit exists since it is a strictly increasing sequence.So $L = L + 1/L^2$ . This shows $L = infty$ contradiction to it being bounded.
answered 17 mins ago
DeepSea
69.6k54285
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Suppose it is bounded then the limit exists since it is a strictly increasing sequence.So $L = L + 1/L^2$ . This shows $L = infty$ contradiction to it being bounded.
answered 17 mins ago
DeepSea
69.6k54285
Suppose it is bounded then the limit exists since it is a strictly increasing sequence.So $L = L + 1/L^2$ . This shows $L = infty$ contradiction to it being bounded.
answered 17 mins ago
DeepSea
69.6k54285
edited 13 mins ago
answered 17 mins ago
DeepSea
69.6k54285
answered 17 mins ago
DeepSea
69.6k54285
answered 17 mins ago
DeepSea
69.6k54285
69.6k54285
add a comment |Â
add a comment |Â
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1
What have you tried? Have you considered the opposite and supposed that there exists $L$ such that $L:=lim_ntoinftya_n$? What happens then?
– TheSimpliFire
22 mins ago
1
@TheSimpliFire I thought that the sequence is positive and monotone, so if I assume it's bounded, then it will be convergent. But I don't know how to get a contradiction from that.
– bateman
19 mins ago