Mixing
Convergence and divergence of $sum_{n=2}^infty {1over{sqrt[3]{n^2-1}}}$ and $sum_{n=1}^infty...
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I need to decide if these series are convergent or divergent, using the comparison test, the quotient test, the integral test, or the Leibniz Theorem:
$$sum_{n=2}^infty {1over{sqrt[3]{n^2-1}}}$$ $$sum_{n=1}^infty {1over{sqrt[3]{n^2+1}}}$$
For the first one, I think that when $n$ is big, $1over{sqrt[3]{n^2-1}}$ is almost equal to $1over{sqrt[3]{n^2}}$, right? I think I can use this to do a comparison test. $1over{sqrt[3]{n^2}}$ diverges because it is equal to $1over{n^{2/3}}$ and ${2over3}<1$? How do I solve this; is my reasoning correct? Thanks.
calculus sequences-and-series convergence
edited Mar 6 at 4:37
Robert Howard
2,2933935
asked Jan 30 at 3:25
davidllerenavdavidllerenav
3128
$endgroup$
add a comment |
$begingroup$
I need to decide if these series are convergent or divergent, using the comparison test, the quotient test, the integral test, or the Leibniz Theorem:
$$sum_{n=2}^infty {1over{sqrt[3]{n^2-1}}}$$ $$sum_{n=1}^infty {1over{sqrt[3]{n^2+1}}}$$
For the first one, I think that when $n$ is big, $1over{sqrt[3]{n^2-1}}$ is almost equal to $1over{sqrt[3]{n^2}}$, right? I think I can use this to do a comparison test. $1over{sqrt[3]{n^2}}$ diverges because it is equal to $1over{n^{2/3}}$ and ${2over3}<1$? How do I solve this; is my reasoning correct? Thanks.
calculus sequences-and-series convergence
edited Mar 6 at 4:37
Robert Howard
2,2933935
asked Jan 30 at 3:25
davidllerenavdavidllerenav
3128
$endgroup$
$begingroup$
Be careful not confusing a series and its general term: $frac{1}{n^{3/2}}$ DOES converge (to $0$). Convergent=to have a finite limit. It is true that $sumfrac{1}{n^{3/2}}$ diverges though
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– Taladris
Jan 30 at 7:54
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Yes, I always confuse that. It diverges because of the p-series right?
$endgroup$
– davidllerenav
Jan 31 at 2:49
add a comment |
$begingroup$
I need to decide if these series are convergent or divergent, using the comparison test, the quotient test, the integral test, or the Leibniz Theorem:
$$sum_{n=2}^infty {1over{sqrt[3]{n^2-1}}}$$ $$sum_{n=1}^infty {1over{sqrt[3]{n^2+1}}}$$
For the first one, I think that when $n$ is big, $1over{sqrt[3]{n^2-1}}$ is almost equal to $1over{sqrt[3]{n^2}}$, right? I think I can use this to do a comparison test. $1over{sqrt[3]{n^2}}$ diverges because it is equal to $1over{n^{2/3}}$ and ${2over3}<1$? How do I solve this; is my reasoning correct? Thanks.
calculus sequences-and-series convergence
edited Mar 6 at 4:37
Robert Howard
2,2933935
asked Jan 30 at 3:25
davidllerenavdavidllerenav
3128
$endgroup$
I need to decide if these series are convergent or divergent, using the comparison test, the quotient test, the integral test, or the Leibniz Theorem:
$$sum_{n=2}^infty {1over{sqrt[3]{n^2-1}}}$$ $$sum_{n=1}^infty {1over{sqrt[3]{n^2+1}}}$$
For the first one, I think that when $n$ is big, $1over{sqrt[3]{n^2-1}}$ is almost equal to $1over{sqrt[3]{n^2}}$, right? I think I can use this to do a comparison test. $1over{sqrt[3]{n^2}}$ diverges because it is equal to $1over{n^{2/3}}$ and ${2over3}<1$? How do I solve this; is my reasoning correct? Thanks.
calculus sequences-and-series convergence
calculus sequences-and-series convergence
edited Mar 6 at 4:37
Robert Howard
2,2933935
asked Jan 30 at 3:25
davidllerenavdavidllerenav
3128
edited Mar 6 at 4:37
Robert Howard
2,2933935
asked Jan 30 at 3:25
davidllerenavdavidllerenav
3128
edited Mar 6 at 4:37
Robert Howard
2,2933935
edited Mar 6 at 4:37
Robert Howard
2,2933935
edited Mar 6 at 4:37
Robert Howard
2,2933935
2,2933935
asked Jan 30 at 3:25
davidllerenavdavidllerenav
3128
asked Jan 30 at 3:25
davidllerenavdavidllerenav
3128
asked Jan 30 at 3:25
davidllerenavdavidllerenav
3128
3128
$begingroup$
Be careful not confusing a series and its general term: $frac{1}{n^{3/2}}$ DOES converge (to $0$). Convergent=to have a finite limit. It is true that $sumfrac{1}{n^{3/2}}$ diverges though
$endgroup$
– Taladris
Jan 30 at 7:54
$begingroup$
Yes, I always confuse that. It diverges because of the p-series right?
$endgroup$
– davidllerenav
Jan 31 at 2:49
add a comment |
$begingroup$
Be careful not confusing a series and its general term: $frac{1}{n^{3/2}}$ DOES converge (to $0$). Convergent=to have a finite limit. It is true that $sumfrac{1}{n^{3/2}}$ diverges though
$endgroup$
– Taladris
Jan 30 at 7:54
$begingroup$
Yes, I always confuse that. It diverges because of the p-series right?
$endgroup$
– davidllerenav
Jan 31 at 2:49
$begingroup$
Be careful not confusing a series and its general term: $frac{1}{n^{3/2}}$ DOES converge (to $0$). Convergent=to have a finite limit. It is true that $sumfrac{1}{n^{3/2}}$ diverges though
$endgroup$
– Taladris
Jan 30 at 7:54
$begingroup$
Be careful not confusing a series and its general term: $frac{1}{n^{3/2}}$ DOES converge (to $0$). Convergent=to have a finite limit. It is true that $sumfrac{1}{n^{3/2}}$ diverges though
$endgroup$
– Taladris
Jan 30 at 7:54
$begingroup$
Yes, I always confuse that. It diverges because of the p-series right?
$endgroup$
– davidllerenav
Jan 31 at 2:49
$begingroup$
Yes, I always confuse that. It diverges because of the p-series right?
$endgroup$
– davidllerenav
Jan 31 at 2:49
add a comment |
3 Answers
3
active
oldest
votes
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$1over{sqrt[3]{n^2-1}}$ >${1over n^{2/3}}=u_n$ so you can use the comparison test in fact the limit of the quotient of the general term of both series with $u_n$ is $1$, so they diverge.
Write $sqrt[3]{n^2-1}=n^{2/3}sqrt[3]{1-{1over n^{2/3}}}$
$sqrt[3]{n^2+1}=n^{2/3}sqrt[3]{1+{1over n^{2/3}}}$.
answered Jan 30 at 3:30
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
$begingroup$
What do you mean by "in fact the limit of the quotient of the general term of both series with $u_n$ is $1$"? And what does $sqrt[3]{n^2-1}=n^{2/3}sqrt[3]{1-{1over n^{2/3}}}$ does?
$endgroup$
– davidllerenav
Jan 30 at 3:50
add a comment |
$begingroup$
$sqrt [3] {n^{2}-1} leq n^{2/3}$ for all $n >1$. Hence the first series dominates $sum frac 1 {n^{2/3}}$ which is divergent. Second series can be handled in a similar way. [$sqrt [3] {n^{2}+1} leq 2n^{2/3}$].
answered Jan 30 at 5:56
Kavi Rama MurthyKavi Rama Murthy
71.8k53170
$endgroup$
$begingroup$
So $1over{n^{2/3}} leq 1over{sqrt[3]{n^2-1}}$, right? Why on the second one it is $2n^{2/3}$?
$endgroup$
– davidllerenav
Jan 30 at 6:38
$begingroup$
$sqrt [3] {n^{2}+1} > n^{2/3}$ and this is of no use here. But $sqrt [3] {n^{2}+1} leq 2n^{2/3}$ and this is useful. In both cases use the fact that $0 <aleq b$ implies $frac 1 a geq frac 1 b$.
$endgroup$
– Kavi Rama Murthy
Jan 30 at 6:41
$begingroup$
Where did $sqrt [3] {n^{2}+1} leq 2n^{2/3}$ came from? Can you explain it a little bit more? I understood that for the first one $sqrt [3] {n^{2}-1} leq n^{2/3}$ since its easy to see that $n^2 -1 < n^2$.
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– davidllerenav
Jan 31 at 2:46
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@davidllerenav $sqrt [3] {n^{2}+1} leq sqrt [3] {n^{2}+n^{2}}=2^{1/3} n^{2/3} leq 2n^{2/3}$.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 5:00
$begingroup$
Oh I see, thanks. O ne question. Every time I want to use the comparison test, I need to find a series that is bigger than the one I will compare it to, right? Like you did with $sqrt [3] {n^{2}-1} leq n^{2/3}$, but it doesn't matter that it will end up being $1over{sqrt [3] {n^{2}-1}} geq 1over{n^{2/3}}$?
$endgroup$
– davidllerenav
Jan 31 at 5:09
add a comment |
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Let $n ge 2:$
$dfrac{1}{2sqrt[3]{n^2}} lt dfrac{1}{sqrt[3]{n^2+n^2}} lt$
$dfrac{1}{sqrt[3]{n^2+1}} lt dfrac{1}{sqrt[3]{n^2-1}}.$
$(1/2)sum dfrac{1}{sqrt[3]{n^2}}$ diverges.
Hence?
answered Jan 30 at 7:04
Peter SzilasPeter Szilas
11.7k2822
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Why did you set $n>2$? I understand what you did, but why did you set $n>2$ sisnce the beginning
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– davidllerenav
Jan 31 at 2:48
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davidleranow.n >1, since denominator cannot be zero $(n^2-1)$, so $n ge 2$.
$endgroup$
– Peter Szilas
Jan 31 at 6:37
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$1over{sqrt[3]{n^2-1}}$ >${1over n^{2/3}}=u_n$ so you can use the comparison test in fact the limit of the quotient of the general term of both series with $u_n$ is $1$, so they diverge.
Write $sqrt[3]{n^2-1}=n^{2/3}sqrt[3]{1-{1over n^{2/3}}}$
$sqrt[3]{n^2+1}=n^{2/3}sqrt[3]{1+{1over n^{2/3}}}$.
answered Jan 30 at 3:30
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
$begingroup$
What do you mean by "in fact the limit of the quotient of the general term of both series with $u_n$ is $1$"? And what does $sqrt[3]{n^2-1}=n^{2/3}sqrt[3]{1-{1over n^{2/3}}}$ does?
$endgroup$
– davidllerenav
Jan 30 at 3:50
add a comment |
$begingroup$
$1over{sqrt[3]{n^2-1}}$ >${1over n^{2/3}}=u_n$ so you can use the comparison test in fact the limit of the quotient of the general term of both series with $u_n$ is $1$, so they diverge.
Write $sqrt[3]{n^2-1}=n^{2/3}sqrt[3]{1-{1over n^{2/3}}}$
$sqrt[3]{n^2+1}=n^{2/3}sqrt[3]{1+{1over n^{2/3}}}$.
answered Jan 30 at 3:30
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
$begingroup$
What do you mean by "in fact the limit of the quotient of the general term of both series with $u_n$ is $1$"? And what does $sqrt[3]{n^2-1}=n^{2/3}sqrt[3]{1-{1over n^{2/3}}}$ does?
$endgroup$
– davidllerenav
Jan 30 at 3:50
add a comment |
$begingroup$
$1over{sqrt[3]{n^2-1}}$ >${1over n^{2/3}}=u_n$ so you can use the comparison test in fact the limit of the quotient of the general term of both series with $u_n$ is $1$, so they diverge.
Write $sqrt[3]{n^2-1}=n^{2/3}sqrt[3]{1-{1over n^{2/3}}}$
$sqrt[3]{n^2+1}=n^{2/3}sqrt[3]{1+{1over n^{2/3}}}$.
answered Jan 30 at 3:30
Tsemo AristideTsemo Aristide
60.1k11446
$endgroup$
$1over{sqrt[3]{n^2-1}}$ >${1over n^{2/3}}=u_n$ so you can use the comparison test in fact the limit of the quotient of the general term of both series with $u_n$ is $1$, so they diverge.
Write $sqrt[3]{n^2-1}=n^{2/3}sqrt[3]{1-{1over n^{2/3}}}$
$sqrt[3]{n^2+1}=n^{2/3}sqrt[3]{1+{1over n^{2/3}}}$.
answered Jan 30 at 3:30
Tsemo AristideTsemo Aristide
60.1k11446
edited Jan 30 at 3:35
answered Jan 30 at 3:30
Tsemo AristideTsemo Aristide
60.1k11446
answered Jan 30 at 3:30
Tsemo AristideTsemo Aristide
60.1k11446
answered Jan 30 at 3:30
Tsemo AristideTsemo Aristide
60.1k11446
60.1k11446
$begingroup$
What do you mean by "in fact the limit of the quotient of the general term of both series with $u_n$ is $1$"? And what does $sqrt[3]{n^2-1}=n^{2/3}sqrt[3]{1-{1over n^{2/3}}}$ does?
$endgroup$
– davidllerenav
Jan 30 at 3:50
add a comment |
$begingroup$
What do you mean by "in fact the limit of the quotient of the general term of both series with $u_n$ is $1$"? And what does $sqrt[3]{n^2-1}=n^{2/3}sqrt[3]{1-{1over n^{2/3}}}$ does?
$endgroup$
– davidllerenav
Jan 30 at 3:50
$begingroup$
What do you mean by "in fact the limit of the quotient of the general term of both series with $u_n$ is $1$"? And what does $sqrt[3]{n^2-1}=n^{2/3}sqrt[3]{1-{1over n^{2/3}}}$ does?
$endgroup$
– davidllerenav
Jan 30 at 3:50
$begingroup$
What do you mean by "in fact the limit of the quotient of the general term of both series with $u_n$ is $1$"? And what does $sqrt[3]{n^2-1}=n^{2/3}sqrt[3]{1-{1over n^{2/3}}}$ does?
$endgroup$
– davidllerenav
Jan 30 at 3:50
add a comment |
$begingroup$
$sqrt [3] {n^{2}-1} leq n^{2/3}$ for all $n >1$. Hence the first series dominates $sum frac 1 {n^{2/3}}$ which is divergent. Second series can be handled in a similar way. [$sqrt [3] {n^{2}+1} leq 2n^{2/3}$].
answered Jan 30 at 5:56
Kavi Rama MurthyKavi Rama Murthy
71.8k53170
$endgroup$
$begingroup$
So $1over{n^{2/3}} leq 1over{sqrt[3]{n^2-1}}$, right? Why on the second one it is $2n^{2/3}$?
$endgroup$
– davidllerenav
Jan 30 at 6:38
$begingroup$
$sqrt [3] {n^{2}+1} > n^{2/3}$ and this is of no use here. But $sqrt [3] {n^{2}+1} leq 2n^{2/3}$ and this is useful. In both cases use the fact that $0 <aleq b$ implies $frac 1 a geq frac 1 b$.
$endgroup$
– Kavi Rama Murthy
Jan 30 at 6:41
$begingroup$
Where did $sqrt [3] {n^{2}+1} leq 2n^{2/3}$ came from? Can you explain it a little bit more? I understood that for the first one $sqrt [3] {n^{2}-1} leq n^{2/3}$ since its easy to see that $n^2 -1 < n^2$.
$endgroup$
– davidllerenav
Jan 31 at 2:46
$begingroup$
@davidllerenav $sqrt [3] {n^{2}+1} leq sqrt [3] {n^{2}+n^{2}}=2^{1/3} n^{2/3} leq 2n^{2/3}$.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 5:00
$begingroup$
Oh I see, thanks. O ne question. Every time I want to use the comparison test, I need to find a series that is bigger than the one I will compare it to, right? Like you did with $sqrt [3] {n^{2}-1} leq n^{2/3}$, but it doesn't matter that it will end up being $1over{sqrt [3] {n^{2}-1}} geq 1over{n^{2/3}}$?
$endgroup$
– davidllerenav
Jan 31 at 5:09
add a comment |
$begingroup$
$sqrt [3] {n^{2}-1} leq n^{2/3}$ for all $n >1$. Hence the first series dominates $sum frac 1 {n^{2/3}}$ which is divergent. Second series can be handled in a similar way. [$sqrt [3] {n^{2}+1} leq 2n^{2/3}$].
answered Jan 30 at 5:56
Kavi Rama MurthyKavi Rama Murthy
71.8k53170
$endgroup$
$begingroup$
So $1over{n^{2/3}} leq 1over{sqrt[3]{n^2-1}}$, right? Why on the second one it is $2n^{2/3}$?
$endgroup$
– davidllerenav
Jan 30 at 6:38
$begingroup$
$sqrt [3] {n^{2}+1} > n^{2/3}$ and this is of no use here. But $sqrt [3] {n^{2}+1} leq 2n^{2/3}$ and this is useful. In both cases use the fact that $0 <aleq b$ implies $frac 1 a geq frac 1 b$.
$endgroup$
– Kavi Rama Murthy
Jan 30 at 6:41
$begingroup$
Where did $sqrt [3] {n^{2}+1} leq 2n^{2/3}$ came from? Can you explain it a little bit more? I understood that for the first one $sqrt [3] {n^{2}-1} leq n^{2/3}$ since its easy to see that $n^2 -1 < n^2$.
$endgroup$
– davidllerenav
Jan 31 at 2:46
$begingroup$
@davidllerenav $sqrt [3] {n^{2}+1} leq sqrt [3] {n^{2}+n^{2}}=2^{1/3} n^{2/3} leq 2n^{2/3}$.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 5:00
$begingroup$
Oh I see, thanks. O ne question. Every time I want to use the comparison test, I need to find a series that is bigger than the one I will compare it to, right? Like you did with $sqrt [3] {n^{2}-1} leq n^{2/3}$, but it doesn't matter that it will end up being $1over{sqrt [3] {n^{2}-1}} geq 1over{n^{2/3}}$?
$endgroup$
– davidllerenav
Jan 31 at 5:09
add a comment |
$begingroup$
$sqrt [3] {n^{2}-1} leq n^{2/3}$ for all $n >1$. Hence the first series dominates $sum frac 1 {n^{2/3}}$ which is divergent. Second series can be handled in a similar way. [$sqrt [3] {n^{2}+1} leq 2n^{2/3}$].
answered Jan 30 at 5:56
Kavi Rama MurthyKavi Rama Murthy
71.8k53170
$endgroup$
$sqrt [3] {n^{2}-1} leq n^{2/3}$ for all $n >1$. Hence the first series dominates $sum frac 1 {n^{2/3}}$ which is divergent. Second series can be handled in a similar way. [$sqrt [3] {n^{2}+1} leq 2n^{2/3}$].
answered Jan 30 at 5:56
Kavi Rama MurthyKavi Rama Murthy
71.8k53170
answered Jan 30 at 5:56
Kavi Rama MurthyKavi Rama Murthy
71.8k53170
answered Jan 30 at 5:56
Kavi Rama MurthyKavi Rama Murthy
71.8k53170
answered Jan 30 at 5:56
Kavi Rama MurthyKavi Rama Murthy
71.8k53170
71.8k53170
$begingroup$
So $1over{n^{2/3}} leq 1over{sqrt[3]{n^2-1}}$, right? Why on the second one it is $2n^{2/3}$?
$endgroup$
– davidllerenav
Jan 30 at 6:38
$begingroup$
$sqrt [3] {n^{2}+1} > n^{2/3}$ and this is of no use here. But $sqrt [3] {n^{2}+1} leq 2n^{2/3}$ and this is useful. In both cases use the fact that $0 <aleq b$ implies $frac 1 a geq frac 1 b$.
$endgroup$
– Kavi Rama Murthy
Jan 30 at 6:41
$begingroup$
Where did $sqrt [3] {n^{2}+1} leq 2n^{2/3}$ came from? Can you explain it a little bit more? I understood that for the first one $sqrt [3] {n^{2}-1} leq n^{2/3}$ since its easy to see that $n^2 -1 < n^2$.
$endgroup$
– davidllerenav
Jan 31 at 2:46
$begingroup$
@davidllerenav $sqrt [3] {n^{2}+1} leq sqrt [3] {n^{2}+n^{2}}=2^{1/3} n^{2/3} leq 2n^{2/3}$.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 5:00
$begingroup$
Oh I see, thanks. O ne question. Every time I want to use the comparison test, I need to find a series that is bigger than the one I will compare it to, right? Like you did with $sqrt [3] {n^{2}-1} leq n^{2/3}$, but it doesn't matter that it will end up being $1over{sqrt [3] {n^{2}-1}} geq 1over{n^{2/3}}$?
$endgroup$
– davidllerenav
Jan 31 at 5:09
add a comment |
$begingroup$
So $1over{n^{2/3}} leq 1over{sqrt[3]{n^2-1}}$, right? Why on the second one it is $2n^{2/3}$?
$endgroup$
– davidllerenav
Jan 30 at 6:38
$begingroup$
$sqrt [3] {n^{2}+1} > n^{2/3}$ and this is of no use here. But $sqrt [3] {n^{2}+1} leq 2n^{2/3}$ and this is useful. In both cases use the fact that $0 <aleq b$ implies $frac 1 a geq frac 1 b$.
$endgroup$
– Kavi Rama Murthy
Jan 30 at 6:41
$begingroup$
Where did $sqrt [3] {n^{2}+1} leq 2n^{2/3}$ came from? Can you explain it a little bit more? I understood that for the first one $sqrt [3] {n^{2}-1} leq n^{2/3}$ since its easy to see that $n^2 -1 < n^2$.
$endgroup$
– davidllerenav
Jan 31 at 2:46
$begingroup$
@davidllerenav $sqrt [3] {n^{2}+1} leq sqrt [3] {n^{2}+n^{2}}=2^{1/3} n^{2/3} leq 2n^{2/3}$.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 5:00
$begingroup$
Oh I see, thanks. O ne question. Every time I want to use the comparison test, I need to find a series that is bigger than the one I will compare it to, right? Like you did with $sqrt [3] {n^{2}-1} leq n^{2/3}$, but it doesn't matter that it will end up being $1over{sqrt [3] {n^{2}-1}} geq 1over{n^{2/3}}$?
$endgroup$
– davidllerenav
Jan 31 at 5:09
$begingroup$
So $1over{n^{2/3}} leq 1over{sqrt[3]{n^2-1}}$, right? Why on the second one it is $2n^{2/3}$?
$endgroup$
– davidllerenav
Jan 30 at 6:38
$begingroup$
So $1over{n^{2/3}} leq 1over{sqrt[3]{n^2-1}}$, right? Why on the second one it is $2n^{2/3}$?
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– davidllerenav
Jan 30 at 6:38
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$sqrt [3] {n^{2}+1} > n^{2/3}$ and this is of no use here. But $sqrt [3] {n^{2}+1} leq 2n^{2/3}$ and this is useful. In both cases use the fact that $0 <aleq b$ implies $frac 1 a geq frac 1 b$.
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– Kavi Rama Murthy
Jan 30 at 6:41
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$sqrt [3] {n^{2}+1} > n^{2/3}$ and this is of no use here. But $sqrt [3] {n^{2}+1} leq 2n^{2/3}$ and this is useful. In both cases use the fact that $0 <aleq b$ implies $frac 1 a geq frac 1 b$.
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– Kavi Rama Murthy
Jan 30 at 6:41
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Where did $sqrt [3] {n^{2}+1} leq 2n^{2/3}$ came from? Can you explain it a little bit more? I understood that for the first one $sqrt [3] {n^{2}-1} leq n^{2/3}$ since its easy to see that $n^2 -1 < n^2$.
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– davidllerenav
Jan 31 at 2:46
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Where did $sqrt [3] {n^{2}+1} leq 2n^{2/3}$ came from? Can you explain it a little bit more? I understood that for the first one $sqrt [3] {n^{2}-1} leq n^{2/3}$ since its easy to see that $n^2 -1 < n^2$.
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– davidllerenav
Jan 31 at 2:46
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@davidllerenav $sqrt [3] {n^{2}+1} leq sqrt [3] {n^{2}+n^{2}}=2^{1/3} n^{2/3} leq 2n^{2/3}$.
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– Kavi Rama Murthy
Jan 31 at 5:00
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@davidllerenav $sqrt [3] {n^{2}+1} leq sqrt [3] {n^{2}+n^{2}}=2^{1/3} n^{2/3} leq 2n^{2/3}$.
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– Kavi Rama Murthy
Jan 31 at 5:00
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Oh I see, thanks. O ne question. Every time I want to use the comparison test, I need to find a series that is bigger than the one I will compare it to, right? Like you did with $sqrt [3] {n^{2}-1} leq n^{2/3}$, but it doesn't matter that it will end up being $1over{sqrt [3] {n^{2}-1}} geq 1over{n^{2/3}}$?
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– davidllerenav
Jan 31 at 5:09
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Oh I see, thanks. O ne question. Every time I want to use the comparison test, I need to find a series that is bigger than the one I will compare it to, right? Like you did with $sqrt [3] {n^{2}-1} leq n^{2/3}$, but it doesn't matter that it will end up being $1over{sqrt [3] {n^{2}-1}} geq 1over{n^{2/3}}$?
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– davidllerenav
Jan 31 at 5:09
add a comment |
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Let $n ge 2:$
$dfrac{1}{2sqrt[3]{n^2}} lt dfrac{1}{sqrt[3]{n^2+n^2}} lt$
$dfrac{1}{sqrt[3]{n^2+1}} lt dfrac{1}{sqrt[3]{n^2-1}}.$
$(1/2)sum dfrac{1}{sqrt[3]{n^2}}$ diverges.
Hence?
answered Jan 30 at 7:04
Peter SzilasPeter Szilas
11.7k2822
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Why did you set $n>2$? I understand what you did, but why did you set $n>2$ sisnce the beginning
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– davidllerenav
Jan 31 at 2:48
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davidleranow.n >1, since denominator cannot be zero $(n^2-1)$, so $n ge 2$.
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– Peter Szilas
Jan 31 at 6:37
add a comment |
$begingroup$
Let $n ge 2:$
$dfrac{1}{2sqrt[3]{n^2}} lt dfrac{1}{sqrt[3]{n^2+n^2}} lt$
$dfrac{1}{sqrt[3]{n^2+1}} lt dfrac{1}{sqrt[3]{n^2-1}}.$
$(1/2)sum dfrac{1}{sqrt[3]{n^2}}$ diverges.
Hence?
answered Jan 30 at 7:04
Peter SzilasPeter Szilas
11.7k2822
$endgroup$
$begingroup$
Why did you set $n>2$? I understand what you did, but why did you set $n>2$ sisnce the beginning
$endgroup$
– davidllerenav
Jan 31 at 2:48
$begingroup$
davidleranow.n >1, since denominator cannot be zero $(n^2-1)$, so $n ge 2$.
$endgroup$
– Peter Szilas
Jan 31 at 6:37
add a comment |
$begingroup$
Let $n ge 2:$
$dfrac{1}{2sqrt[3]{n^2}} lt dfrac{1}{sqrt[3]{n^2+n^2}} lt$
$dfrac{1}{sqrt[3]{n^2+1}} lt dfrac{1}{sqrt[3]{n^2-1}}.$
$(1/2)sum dfrac{1}{sqrt[3]{n^2}}$ diverges.
Hence?
answered Jan 30 at 7:04
Peter SzilasPeter Szilas
11.7k2822
$endgroup$
Let $n ge 2:$
$dfrac{1}{2sqrt[3]{n^2}} lt dfrac{1}{sqrt[3]{n^2+n^2}} lt$
$dfrac{1}{sqrt[3]{n^2+1}} lt dfrac{1}{sqrt[3]{n^2-1}}.$
$(1/2)sum dfrac{1}{sqrt[3]{n^2}}$ diverges.
Hence?
answered Jan 30 at 7:04
Peter SzilasPeter Szilas
11.7k2822
edited Jan 31 at 6:37
answered Jan 30 at 7:04
Peter SzilasPeter Szilas
11.7k2822
answered Jan 30 at 7:04
Peter SzilasPeter Szilas
11.7k2822
answered Jan 30 at 7:04
Peter SzilasPeter Szilas
11.7k2822
11.7k2822
$begingroup$
Why did you set $n>2$? I understand what you did, but why did you set $n>2$ sisnce the beginning
$endgroup$
– davidllerenav
Jan 31 at 2:48
$begingroup$
davidleranow.n >1, since denominator cannot be zero $(n^2-1)$, so $n ge 2$.
$endgroup$
– Peter Szilas
Jan 31 at 6:37
add a comment |
$begingroup$
Why did you set $n>2$? I understand what you did, but why did you set $n>2$ sisnce the beginning
$endgroup$
– davidllerenav
Jan 31 at 2:48
$begingroup$
davidleranow.n >1, since denominator cannot be zero $(n^2-1)$, so $n ge 2$.
$endgroup$
– Peter Szilas
Jan 31 at 6:37
$begingroup$
Why did you set $n>2$? I understand what you did, but why did you set $n>2$ sisnce the beginning
$endgroup$
– davidllerenav
Jan 31 at 2:48
$begingroup$
Why did you set $n>2$? I understand what you did, but why did you set $n>2$ sisnce the beginning
$endgroup$
– davidllerenav
Jan 31 at 2:48
$begingroup$
davidleranow.n >1, since denominator cannot be zero $(n^2-1)$, so $n ge 2$.
$endgroup$
– Peter Szilas
Jan 31 at 6:37
$begingroup$
davidleranow.n >1, since denominator cannot be zero $(n^2-1)$, so $n ge 2$.
$endgroup$
– Peter Szilas
Jan 31 at 6:37
add a comment |
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Be careful not confusing a series and its general term: $frac{1}{n^{3/2}}$ DOES converge (to $0$). Convergent=to have a finite limit. It is true that $sumfrac{1}{n^{3/2}}$ diverges though
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– Taladris
Jan 30 at 7:54
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Yes, I always confuse that. It diverges because of the p-series right?
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– davidllerenav
Jan 31 at 2:49