Prove that a sequence give by a recurrence relation converges
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let $b_n$ be a real sequence which satisfies
$b_{n+1}=(frac{1}{n^2}-frac{1}{2})b_n -frac{1}{3n^2}$, $b_1=frac{2}{3}$
I want to show that $b_n$ converges to $0$, which seems to be true according to my simulation.
I have tried to use some general ways to proof the convergence of a sequence like ratio test, but could not finish.
sequences-and-series convergence recurrence-relations
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456 123
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up vote
1
down vote
favorite
let $b_n$ be a real sequence which satisfies
$b_{n+1}=(frac{1}{n^2}-frac{1}{2})b_n -frac{1}{3n^2}$, $b_1=frac{2}{3}$
I want to show that $b_n$ converges to $0$, which seems to be true according to my simulation.
I have tried to use some general ways to proof the convergence of a sequence like ratio test, but could not finish.
sequences-and-series convergence recurrence-relations
asked yesterday
456 123
1516
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
let $b_n$ be a real sequence which satisfies
$b_{n+1}=(frac{1}{n^2}-frac{1}{2})b_n -frac{1}{3n^2}$, $b_1=frac{2}{3}$
I want to show that $b_n$ converges to $0$, which seems to be true according to my simulation.
I have tried to use some general ways to proof the convergence of a sequence like ratio test, but could not finish.
sequences-and-series convergence recurrence-relations
asked yesterday
456 123
1516
let $b_n$ be a real sequence which satisfies
$b_{n+1}=(frac{1}{n^2}-frac{1}{2})b_n -frac{1}{3n^2}$, $b_1=frac{2}{3}$
I want to show that $b_n$ converges to $0$, which seems to be true according to my simulation.
I have tried to use some general ways to proof the convergence of a sequence like ratio test, but could not finish.
sequences-and-series convergence recurrence-relations
sequences-and-series convergence recurrence-relations
asked yesterday
456 123
1516
asked yesterday
456 123
1516
asked yesterday
456 123
1516
asked yesterday
456 123
1516
asked yesterday
456 123
1516
1516
add a comment |
add a comment |
1 Answer
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Note that for all $varepsilon>0$, we have $|b_{n+1}|leqfrac 23|b_n|+varepsilon$ eventually, hence
$$limsup |b_n|leq 3varepsilon$$
answered yesterday
Fabio Lucchini
7,49011226
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Note that for all $varepsilon>0$, we have $|b_{n+1}|leqfrac 23|b_n|+varepsilon$ eventually, hence
$$limsup |b_n|leq 3varepsilon$$
answered yesterday
Fabio Lucchini
7,49011226
add a comment |
up vote
0
down vote
accepted
Note that for all $varepsilon>0$, we have $|b_{n+1}|leqfrac 23|b_n|+varepsilon$ eventually, hence
$$limsup |b_n|leq 3varepsilon$$
answered yesterday
Fabio Lucchini
7,49011226
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Note that for all $varepsilon>0$, we have $|b_{n+1}|leqfrac 23|b_n|+varepsilon$ eventually, hence
$$limsup |b_n|leq 3varepsilon$$
answered yesterday
Fabio Lucchini
7,49011226
Note that for all $varepsilon>0$, we have $|b_{n+1}|leqfrac 23|b_n|+varepsilon$ eventually, hence
$$limsup |b_n|leq 3varepsilon$$
answered yesterday
Fabio Lucchini
7,49011226
edited yesterday
answered yesterday
Fabio Lucchini
7,49011226
answered yesterday
Fabio Lucchini
7,49011226
answered yesterday
Fabio Lucchini
7,49011226
7,49011226
add a comment |
add a comment |
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