Principle of proof of infinitesimal sequence
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I have such tasks:
Prove by definition that a sequence ($a_n$) is infinitely small if:
1) $a_n=frac{3}{n}$
2) $a_n=frac{(-1)^n}{sqrt[3]{n}}$
3) $a_n=frac{1}{2n-1}$
4) $a_n=frac{2}{sqrt{n}+1}$
Tell me the sources for which I can learn to solve such tasks
analysis
asked Jan 6 at 10:45
yoloyyoloy
1033
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add a comment |
$begingroup$
I have such tasks:
Prove by definition that a sequence ($a_n$) is infinitely small if:
1) $a_n=frac{3}{n}$
2) $a_n=frac{(-1)^n}{sqrt[3]{n}}$
3) $a_n=frac{1}{2n-1}$
4) $a_n=frac{2}{sqrt{n}+1}$
Tell me the sources for which I can learn to solve such tasks
analysis
asked Jan 6 at 10:45
yoloyyoloy
1033
$endgroup$
add a comment |
$begingroup$
I have such tasks:
Prove by definition that a sequence ($a_n$) is infinitely small if:
1) $a_n=frac{3}{n}$
2) $a_n=frac{(-1)^n}{sqrt[3]{n}}$
3) $a_n=frac{1}{2n-1}$
4) $a_n=frac{2}{sqrt{n}+1}$
Tell me the sources for which I can learn to solve such tasks
analysis
asked Jan 6 at 10:45
yoloyyoloy
1033
$endgroup$
I have such tasks:
Prove by definition that a sequence ($a_n$) is infinitely small if:
1) $a_n=frac{3}{n}$
2) $a_n=frac{(-1)^n}{sqrt[3]{n}}$
3) $a_n=frac{1}{2n-1}$
4) $a_n=frac{2}{sqrt{n}+1}$
Tell me the sources for which I can learn to solve such tasks
analysis
analysis
asked Jan 6 at 10:45
yoloyyoloy
1033
asked Jan 6 at 10:45
yoloyyoloy
1033
asked Jan 6 at 10:45
yoloyyoloy
1033
asked Jan 6 at 10:45
yoloyyoloy
1033
asked Jan 6 at 10:45
yoloyyoloy
1033
1033
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Let $varepsilon>0$, you have to prove that $exists n_{varepsilon}$ such that $|a_{n}-0|<varepsilon$ $forall n>n_{varepsilon}$.
1) $frac{3}{n}<varepsiloniff 3<varepsilon n$.
$mathbb{R}$ is archimedean, so exists $n_{varepsilon}$ such that $3<varepsilon n_{varepsilon}$.
It's easy to prove that the sequence is strictly decreasing, and that finishes the proof.
2) It's quite the same: $|a_{n}-0|<varepsilon iff 1<varepsilon sqrt[3]{n} iff 1<varepsilon^{3} n$, so you can complete the proof in the same way as above.
3) $2n-1>n$, so you can repeat the same strategy used in the first case
4) You can reason as in the second case
I think you can learn how to solve exercise like this in any analysis 1 book you like.
answered Jan 6 at 13:56
ecrinecrin
3477
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add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Let $varepsilon>0$, you have to prove that $exists n_{varepsilon}$ such that $|a_{n}-0|<varepsilon$ $forall n>n_{varepsilon}$.
1) $frac{3}{n}<varepsiloniff 3<varepsilon n$.
$mathbb{R}$ is archimedean, so exists $n_{varepsilon}$ such that $3<varepsilon n_{varepsilon}$.
It's easy to prove that the sequence is strictly decreasing, and that finishes the proof.
2) It's quite the same: $|a_{n}-0|<varepsilon iff 1<varepsilon sqrt[3]{n} iff 1<varepsilon^{3} n$, so you can complete the proof in the same way as above.
3) $2n-1>n$, so you can repeat the same strategy used in the first case
4) You can reason as in the second case
I think you can learn how to solve exercise like this in any analysis 1 book you like.
answered Jan 6 at 13:56
ecrinecrin
3477
$endgroup$
add a comment |
$begingroup$
Let $varepsilon>0$, you have to prove that $exists n_{varepsilon}$ such that $|a_{n}-0|<varepsilon$ $forall n>n_{varepsilon}$.
1) $frac{3}{n}<varepsiloniff 3<varepsilon n$.
$mathbb{R}$ is archimedean, so exists $n_{varepsilon}$ such that $3<varepsilon n_{varepsilon}$.
It's easy to prove that the sequence is strictly decreasing, and that finishes the proof.
2) It's quite the same: $|a_{n}-0|<varepsilon iff 1<varepsilon sqrt[3]{n} iff 1<varepsilon^{3} n$, so you can complete the proof in the same way as above.
3) $2n-1>n$, so you can repeat the same strategy used in the first case
4) You can reason as in the second case
I think you can learn how to solve exercise like this in any analysis 1 book you like.
answered Jan 6 at 13:56
ecrinecrin
3477
$endgroup$
add a comment |
$begingroup$
Let $varepsilon>0$, you have to prove that $exists n_{varepsilon}$ such that $|a_{n}-0|<varepsilon$ $forall n>n_{varepsilon}$.
1) $frac{3}{n}<varepsiloniff 3<varepsilon n$.
$mathbb{R}$ is archimedean, so exists $n_{varepsilon}$ such that $3<varepsilon n_{varepsilon}$.
It's easy to prove that the sequence is strictly decreasing, and that finishes the proof.
2) It's quite the same: $|a_{n}-0|<varepsilon iff 1<varepsilon sqrt[3]{n} iff 1<varepsilon^{3} n$, so you can complete the proof in the same way as above.
3) $2n-1>n$, so you can repeat the same strategy used in the first case
4) You can reason as in the second case
I think you can learn how to solve exercise like this in any analysis 1 book you like.
answered Jan 6 at 13:56
ecrinecrin
3477
$endgroup$
Let $varepsilon>0$, you have to prove that $exists n_{varepsilon}$ such that $|a_{n}-0|<varepsilon$ $forall n>n_{varepsilon}$.
1) $frac{3}{n}<varepsiloniff 3<varepsilon n$.
$mathbb{R}$ is archimedean, so exists $n_{varepsilon}$ such that $3<varepsilon n_{varepsilon}$.
It's easy to prove that the sequence is strictly decreasing, and that finishes the proof.
2) It's quite the same: $|a_{n}-0|<varepsilon iff 1<varepsilon sqrt[3]{n} iff 1<varepsilon^{3} n$, so you can complete the proof in the same way as above.
3) $2n-1>n$, so you can repeat the same strategy used in the first case
4) You can reason as in the second case
I think you can learn how to solve exercise like this in any analysis 1 book you like.
answered Jan 6 at 13:56
ecrinecrin
3477
answered Jan 6 at 13:56
ecrinecrin
3477
answered Jan 6 at 13:56
ecrinecrin
3477
answered Jan 6 at 13:56
ecrinecrin
3477
3477
add a comment |
add a comment |
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