Sequence $lim_ntoinfty(-1)^n fracsin (n)n^2$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Using the Test for DivergenceShow by comparison that $sumlimits_n=1^infty sin(frac1n)$ diverges?Convergence tests $sum_n=1^infty fracsin(n)n$Determining whether $ displaystyle sum_n = 0^infty 4cos(2pi n)e^-3n $ divergesDoes $sum_n=1^inftyln(nsin(frac1n))$ converge?Does $sumlimits_n=1^inftysin(n)sinleft(fracpi2nright)$ converge?Convergence of $sum_n=1^inftyfrac1(3n+8)!$A Sequence converges or divergesDetermine whether $sum_n=1^infty frac2sqrtn+2 $ converges or diverges?Show if the series $sum _n=1^infty frac 39n+1$ converges or diverges.
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Sequence $lim_ntoinfty(-1)^n fracsin (n)n^2$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Using the Test for DivergenceShow by comparison that $sumlimits_n=1^infty sin(frac1n)$ diverges?Convergence tests $sum_n=1^infty fracsin(n)n$Determining whether $ displaystyle sum_n = 0^infty 4cos(2pi n)e^-3n $ divergesDoes $sum_n=1^inftyln(nsin(frac1n))$ converge?Does $sumlimits_n=1^inftysin(n)sinleft(fracpi2nright)$ converge?Convergence of $sum_n=1^inftyfrac1(3n+8)!$A Sequence converges or divergesDetermine whether $sum_n=1^infty frac2sqrtn+2 $ converges or diverges?Show if the series $sum _n=1^infty frac 39n+1$ converges or diverges.
$begingroup$
So I have this problem and I need some help on it.
$$lim_ntoinfty(-1)^n fracsin (n)n^2$$
So for this, I know that the sin value will always be between -1 < n < 1.
And the series only converges if the limit is equal to 0, otherwise, it diverges by Divergence Test.
The problem is that I am having trouble taking the limit of the function:
$$lim_ntoinftyfracsin(n)n^2$$
So if anyone has any suggestions on how I may accomplish this, that would be great.
sequences-and-series
edited Apr 1 at 18:39
Bernard
124k742117
asked Apr 1 at 18:37
Christian MartinezChristian Martinez
648
$endgroup$
add a comment |
$begingroup$
So I have this problem and I need some help on it.
$$lim_ntoinfty(-1)^n fracsin (n)n^2$$
So for this, I know that the sin value will always be between -1 < n < 1.
And the series only converges if the limit is equal to 0, otherwise, it diverges by Divergence Test.
The problem is that I am having trouble taking the limit of the function:
$$lim_ntoinftyfracsin(n)n^2$$
So if anyone has any suggestions on how I may accomplish this, that would be great.
sequences-and-series
edited Apr 1 at 18:39
Bernard
124k742117
asked Apr 1 at 18:37
Christian MartinezChristian Martinez
648
$endgroup$
$begingroup$
Actually, you don't have $-1 < n <1$ but rather $-1 le sin n le 1.$
$endgroup$
– CiaPan
Apr 1 at 18:45
$begingroup$
Ah, I see thanks for the clarification.
$endgroup$
– Christian Martinez
Apr 1 at 18:48
add a comment |
$begingroup$
So I have this problem and I need some help on it.
$$lim_ntoinfty(-1)^n fracsin (n)n^2$$
So for this, I know that the sin value will always be between -1 < n < 1.
And the series only converges if the limit is equal to 0, otherwise, it diverges by Divergence Test.
The problem is that I am having trouble taking the limit of the function:
$$lim_ntoinftyfracsin(n)n^2$$
So if anyone has any suggestions on how I may accomplish this, that would be great.
sequences-and-series
edited Apr 1 at 18:39
Bernard
124k742117
asked Apr 1 at 18:37
Christian MartinezChristian Martinez
648
$endgroup$
So I have this problem and I need some help on it.
$$lim_ntoinfty(-1)^n fracsin (n)n^2$$
So for this, I know that the sin value will always be between -1 < n < 1.
And the series only converges if the limit is equal to 0, otherwise, it diverges by Divergence Test.
The problem is that I am having trouble taking the limit of the function:
$$lim_ntoinftyfracsin(n)n^2$$
So if anyone has any suggestions on how I may accomplish this, that would be great.
sequences-and-series
sequences-and-series
edited Apr 1 at 18:39
Bernard
124k742117
asked Apr 1 at 18:37
Christian MartinezChristian Martinez
648
edited Apr 1 at 18:39
Bernard
124k742117
asked Apr 1 at 18:37
Christian MartinezChristian Martinez
648
edited Apr 1 at 18:39
Bernard
124k742117
edited Apr 1 at 18:39
Bernard
124k742117
edited Apr 1 at 18:39
Bernard
124k742117
124k742117
asked Apr 1 at 18:37
Christian MartinezChristian Martinez
648
asked Apr 1 at 18:37
Christian MartinezChristian Martinez
648
asked Apr 1 at 18:37
Christian MartinezChristian Martinez
648
648
$begingroup$
Actually, you don't have $-1 < n <1$ but rather $-1 le sin n le 1.$
$endgroup$
– CiaPan
Apr 1 at 18:45
$begingroup$
Ah, I see thanks for the clarification.
$endgroup$
– Christian Martinez
Apr 1 at 18:48
add a comment |
$begingroup$
Actually, you don't have $-1 < n <1$ but rather $-1 le sin n le 1.$
$endgroup$
– CiaPan
Apr 1 at 18:45
$begingroup$
Ah, I see thanks for the clarification.
$endgroup$
– Christian Martinez
Apr 1 at 18:48
$begingroup$
Actually, you don't have $-1 < n <1$ but rather $-1 le sin n le 1.$
$endgroup$
– CiaPan
Apr 1 at 18:45
$begingroup$
Actually, you don't have $-1 < n <1$ but rather $-1 le sin n le 1.$
$endgroup$
– CiaPan
Apr 1 at 18:45
$begingroup$
Ah, I see thanks for the clarification.
$endgroup$
– Christian Martinez
Apr 1 at 18:48
$begingroup$
Ah, I see thanks for the clarification.
$endgroup$
– Christian Martinez
Apr 1 at 18:48
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Guide:
- Note that
$$left|fracsin nn^2right| le frac1n^2$$
answered Apr 1 at 18:39
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
1
$begingroup$
So would I take the limit of $$frac1n^2$$ which is 0. So then the sequence converges -- this is squeeze theorem right?
$endgroup$
– Christian Martinez
Apr 1 at 18:41
1
$begingroup$
yup, after you show that the absolute value of $|a_n|$ converges to $0$, you can conclude that $a_n$ converges to $0$ as well. Alternatively, view it as $-frac1n^2 le fracsin nn^2 le frac1n^2$.
$endgroup$
– Siong Thye Goh
Apr 1 at 18:42
1
$begingroup$
Alright, thanks!
$endgroup$
– Christian Martinez
Apr 1 at 18:43
add a comment |
$begingroup$
We need to observe that $$frac-1n^2 leq frac(-1)^nsin(n)n^2 leq frac1n^2$$
Since $lim_n to infty frac-1n^2 = lim_n to infty frac1n^2 = 0$, it follows by the squeeze theorem that
$$lim_n to infty frac(-1)^nsin(n)n^2 = 0$$
answered Apr 1 at 18:43
user516079user516079
537311
$endgroup$
add a comment |
$begingroup$
If you know that for some positive $M$, $-Mleq f(x)leq M$ for all $x$, then $-Mleq lim_xrightarrowinftyf(x)leq M$.
Now use the fact that $-1 leq (-1)^nsin(n)leq 1$ and $lim_nrightarrowinftyfrac 1 n^2=0$ to conclude that the answer to your question is zero.
answered Apr 1 at 18:48
zornzorn
516
$endgroup$
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
$begingroup$
Guide:
- Note that
$$left|fracsin nn^2right| le frac1n^2$$
answered Apr 1 at 18:39
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
1
$begingroup$
So would I take the limit of $$frac1n^2$$ which is 0. So then the sequence converges -- this is squeeze theorem right?
$endgroup$
– Christian Martinez
Apr 1 at 18:41
1
$begingroup$
yup, after you show that the absolute value of $|a_n|$ converges to $0$, you can conclude that $a_n$ converges to $0$ as well. Alternatively, view it as $-frac1n^2 le fracsin nn^2 le frac1n^2$.
$endgroup$
– Siong Thye Goh
Apr 1 at 18:42
1
$begingroup$
Alright, thanks!
$endgroup$
– Christian Martinez
Apr 1 at 18:43
add a comment |
$begingroup$
Guide:
- Note that
$$left|fracsin nn^2right| le frac1n^2$$
answered Apr 1 at 18:39
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
1
$begingroup$
So would I take the limit of $$frac1n^2$$ which is 0. So then the sequence converges -- this is squeeze theorem right?
$endgroup$
– Christian Martinez
Apr 1 at 18:41
1
$begingroup$
yup, after you show that the absolute value of $|a_n|$ converges to $0$, you can conclude that $a_n$ converges to $0$ as well. Alternatively, view it as $-frac1n^2 le fracsin nn^2 le frac1n^2$.
$endgroup$
– Siong Thye Goh
Apr 1 at 18:42
1
$begingroup$
Alright, thanks!
$endgroup$
– Christian Martinez
Apr 1 at 18:43
add a comment |
$begingroup$
Guide:
- Note that
$$left|fracsin nn^2right| le frac1n^2$$
answered Apr 1 at 18:39
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
Guide:
- Note that
$$left|fracsin nn^2right| le frac1n^2$$
answered Apr 1 at 18:39
Siong Thye GohSiong Thye Goh
104k1468120
answered Apr 1 at 18:39
Siong Thye GohSiong Thye Goh
104k1468120
answered Apr 1 at 18:39
Siong Thye GohSiong Thye Goh
104k1468120
answered Apr 1 at 18:39
Siong Thye GohSiong Thye Goh
104k1468120
104k1468120
1
$begingroup$
So would I take the limit of $$frac1n^2$$ which is 0. So then the sequence converges -- this is squeeze theorem right?
$endgroup$
– Christian Martinez
Apr 1 at 18:41
1
$begingroup$
yup, after you show that the absolute value of $|a_n|$ converges to $0$, you can conclude that $a_n$ converges to $0$ as well. Alternatively, view it as $-frac1n^2 le fracsin nn^2 le frac1n^2$.
$endgroup$
– Siong Thye Goh
Apr 1 at 18:42
1
$begingroup$
Alright, thanks!
$endgroup$
– Christian Martinez
Apr 1 at 18:43
add a comment |
1
$begingroup$
So would I take the limit of $$frac1n^2$$ which is 0. So then the sequence converges -- this is squeeze theorem right?
$endgroup$
– Christian Martinez
Apr 1 at 18:41
1
$begingroup$
yup, after you show that the absolute value of $|a_n|$ converges to $0$, you can conclude that $a_n$ converges to $0$ as well. Alternatively, view it as $-frac1n^2 le fracsin nn^2 le frac1n^2$.
$endgroup$
– Siong Thye Goh
Apr 1 at 18:42
1
$begingroup$
Alright, thanks!
$endgroup$
– Christian Martinez
Apr 1 at 18:43
1
1
$begingroup$
So would I take the limit of $$frac1n^2$$ which is 0. So then the sequence converges -- this is squeeze theorem right?
$endgroup$
– Christian Martinez
Apr 1 at 18:41
$begingroup$
So would I take the limit of $$frac1n^2$$ which is 0. So then the sequence converges -- this is squeeze theorem right?
$endgroup$
– Christian Martinez
Apr 1 at 18:41
1
1
$begingroup$
yup, after you show that the absolute value of $|a_n|$ converges to $0$, you can conclude that $a_n$ converges to $0$ as well. Alternatively, view it as $-frac1n^2 le fracsin nn^2 le frac1n^2$.
$endgroup$
– Siong Thye Goh
Apr 1 at 18:42
$begingroup$
yup, after you show that the absolute value of $|a_n|$ converges to $0$, you can conclude that $a_n$ converges to $0$ as well. Alternatively, view it as $-frac1n^2 le fracsin nn^2 le frac1n^2$.
$endgroup$
– Siong Thye Goh
Apr 1 at 18:42
1
1
$begingroup$
Alright, thanks!
$endgroup$
– Christian Martinez
Apr 1 at 18:43
$begingroup$
Alright, thanks!
$endgroup$
– Christian Martinez
Apr 1 at 18:43
add a comment |
$begingroup$
We need to observe that $$frac-1n^2 leq frac(-1)^nsin(n)n^2 leq frac1n^2$$
Since $lim_n to infty frac-1n^2 = lim_n to infty frac1n^2 = 0$, it follows by the squeeze theorem that
$$lim_n to infty frac(-1)^nsin(n)n^2 = 0$$
answered Apr 1 at 18:43
user516079user516079
537311
$endgroup$
add a comment |
$begingroup$
We need to observe that $$frac-1n^2 leq frac(-1)^nsin(n)n^2 leq frac1n^2$$
Since $lim_n to infty frac-1n^2 = lim_n to infty frac1n^2 = 0$, it follows by the squeeze theorem that
$$lim_n to infty frac(-1)^nsin(n)n^2 = 0$$
answered Apr 1 at 18:43
user516079user516079
537311
$endgroup$
add a comment |
$begingroup$
We need to observe that $$frac-1n^2 leq frac(-1)^nsin(n)n^2 leq frac1n^2$$
Since $lim_n to infty frac-1n^2 = lim_n to infty frac1n^2 = 0$, it follows by the squeeze theorem that
$$lim_n to infty frac(-1)^nsin(n)n^2 = 0$$
answered Apr 1 at 18:43
user516079user516079
537311
$endgroup$
We need to observe that $$frac-1n^2 leq frac(-1)^nsin(n)n^2 leq frac1n^2$$
Since $lim_n to infty frac-1n^2 = lim_n to infty frac1n^2 = 0$, it follows by the squeeze theorem that
$$lim_n to infty frac(-1)^nsin(n)n^2 = 0$$
answered Apr 1 at 18:43
user516079user516079
537311
answered Apr 1 at 18:43
user516079user516079
537311
answered Apr 1 at 18:43
user516079user516079
537311
answered Apr 1 at 18:43
user516079user516079
537311
537311
add a comment |
add a comment |
$begingroup$
If you know that for some positive $M$, $-Mleq f(x)leq M$ for all $x$, then $-Mleq lim_xrightarrowinftyf(x)leq M$.
Now use the fact that $-1 leq (-1)^nsin(n)leq 1$ and $lim_nrightarrowinftyfrac 1 n^2=0$ to conclude that the answer to your question is zero.
answered Apr 1 at 18:48
zornzorn
516
$endgroup$
add a comment |
$begingroup$
If you know that for some positive $M$, $-Mleq f(x)leq M$ for all $x$, then $-Mleq lim_xrightarrowinftyf(x)leq M$.
Now use the fact that $-1 leq (-1)^nsin(n)leq 1$ and $lim_nrightarrowinftyfrac 1 n^2=0$ to conclude that the answer to your question is zero.
answered Apr 1 at 18:48
zornzorn
516
$endgroup$
add a comment |
$begingroup$
If you know that for some positive $M$, $-Mleq f(x)leq M$ for all $x$, then $-Mleq lim_xrightarrowinftyf(x)leq M$.
Now use the fact that $-1 leq (-1)^nsin(n)leq 1$ and $lim_nrightarrowinftyfrac 1 n^2=0$ to conclude that the answer to your question is zero.
answered Apr 1 at 18:48
zornzorn
516
$endgroup$
If you know that for some positive $M$, $-Mleq f(x)leq M$ for all $x$, then $-Mleq lim_xrightarrowinftyf(x)leq M$.
Now use the fact that $-1 leq (-1)^nsin(n)leq 1$ and $lim_nrightarrowinftyfrac 1 n^2=0$ to conclude that the answer to your question is zero.
answered Apr 1 at 18:48
zornzorn
516
answered Apr 1 at 18:48
zornzorn
516
answered Apr 1 at 18:48
zornzorn
516
answered Apr 1 at 18:48
zornzorn
516
516
add a comment |
add a comment |
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$begingroup$
Actually, you don't have $-1 < n <1$ but rather $-1 le sin n le 1.$
$endgroup$
– CiaPan
Apr 1 at 18:45
$begingroup$
Ah, I see thanks for the clarification.
$endgroup$
– Christian Martinez
Apr 1 at 18:48