Three series convergence problem [closed]
Having that $sum_{n=1}^{infty}a_n$ is a convergent series and we know that $a_n > 0$ can we say that $sum_{n=1}^{infty}a_n sin(a_n)$ also converges?
sequences-and-series convergence
asked Dec 18 at 12:54
avan1235
1776
closed as off-topic by Saad, Ben, user21820, user 170039, Greg Martin Dec 18 at 19:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Ben, user21820, user 170039, Greg Martin
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Having that $sum_{n=1}^{infty}a_n$ is a convergent series and we know that $a_n > 0$ can we say that $sum_{n=1}^{infty}a_n sin(a_n)$ also converges?
sequences-and-series convergence
asked Dec 18 at 12:54
avan1235
1776
closed as off-topic by Saad, Ben, user21820, user 170039, Greg Martin Dec 18 at 19:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Ben, user21820, user 170039, Greg Martin
If this question can be reworded to fit the rules in the help center, please edit the question.
2
What did you try?
– José Carlos Santos
Dec 18 at 12:56
@JoséCarlosSantos I'm not sure if I can use something like comparision for the series that $-a_n<a_n sin(a_n) < a_n$ but I tried only this method
– avan1235
Dec 18 at 12:57
add a comment |
Having that $sum_{n=1}^{infty}a_n$ is a convergent series and we know that $a_n > 0$ can we say that $sum_{n=1}^{infty}a_n sin(a_n)$ also converges?
sequences-and-series convergence
asked Dec 18 at 12:54
avan1235
1776
Having that $sum_{n=1}^{infty}a_n$ is a convergent series and we know that $a_n > 0$ can we say that $sum_{n=1}^{infty}a_n sin(a_n)$ also converges?
sequences-and-series convergence
sequences-and-series convergence
asked Dec 18 at 12:54
avan1235
1776
asked Dec 18 at 12:54
avan1235
1776
asked Dec 18 at 12:54
avan1235
1776
asked Dec 18 at 12:54
avan1235
1776
asked Dec 18 at 12:54
avan1235
1776
1776
closed as off-topic by Saad, Ben, user21820, user 170039, Greg Martin Dec 18 at 19:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Ben, user21820, user 170039, Greg Martin
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Ben, user21820, user 170039, Greg Martin Dec 18 at 19:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Ben, user21820, user 170039, Greg Martin
If this question can be reworded to fit the rules in the help center, please edit the question.
2
What did you try?
– José Carlos Santos
Dec 18 at 12:56
@JoséCarlosSantos I'm not sure if I can use something like comparision for the series that $-a_n<a_n sin(a_n) < a_n$ but I tried only this method
– avan1235
Dec 18 at 12:57
add a comment |
2
What did you try?
– José Carlos Santos
Dec 18 at 12:56
@JoséCarlosSantos I'm not sure if I can use something like comparision for the series that $-a_n<a_n sin(a_n) < a_n$ but I tried only this method
– avan1235
Dec 18 at 12:57
2
2
What did you try?
– José Carlos Santos
Dec 18 at 12:56
What did you try?
– José Carlos Santos
Dec 18 at 12:56
@JoséCarlosSantos I'm not sure if I can use something like comparision for the series that $-a_n<a_n sin(a_n) < a_n$ but I tried only this method
– avan1235
Dec 18 at 12:57
@JoséCarlosSantos I'm not sure if I can use something like comparision for the series that $-a_n<a_n sin(a_n) < a_n$ but I tried only this method
– avan1235
Dec 18 at 12:57
add a comment |
2 Answers
2
active
oldest
votes
Yes, it converges, by the comparaison test and because$$(forall ninmathbb{N}):bigllvert a_nsin(a_n)bigrrvertleqslantlvert a_nrvert=a_n.$$
answered Dec 18 at 12:59
José Carlos Santos
150k22120221
2
+1..Forgot about this.
– xbh
Dec 18 at 13:02
Note that, since $sum a_n$ is convergent, $lim_{ntoinfty}a_n=0$. Since $a_n>0$, this implies that $a_nin(0,frac{pi}{2})$ for $n$ large enough, so $sin(a_n)>0$ for $n$ large enough. Therefore, you don't need the absolute value. (Of course, it makes the proof longer...)
– Taladris
Dec 18 at 15:09
add a comment |
Since $sum a_n$ converges, for each $varepsilon >0$, there exists $N in mathbb N^*$ s.t. whenever $mathbb N^* ni n > N, p in mathbb N^*$, $sum_{n+1}^{n+p} a_k < varepsilon$. Then
$$
leftvert sum_{n+1}^{n+p} a_k sin (a_k) rightvert leqslant sum_{n+1}^{n+p} vert a_k sin (a_k ) vert leqslant sum_{n+1}^{n+p} a_k < varepsilon,
$$
hence $sum a_n sin(a_n)$ converges by Cauchy convergence principle.
answered Dec 18 at 13:02
xbh
5,6551522
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yes, it converges, by the comparaison test and because$$(forall ninmathbb{N}):bigllvert a_nsin(a_n)bigrrvertleqslantlvert a_nrvert=a_n.$$
answered Dec 18 at 12:59
José Carlos Santos
150k22120221
2
+1..Forgot about this.
– xbh
Dec 18 at 13:02
Note that, since $sum a_n$ is convergent, $lim_{ntoinfty}a_n=0$. Since $a_n>0$, this implies that $a_nin(0,frac{pi}{2})$ for $n$ large enough, so $sin(a_n)>0$ for $n$ large enough. Therefore, you don't need the absolute value. (Of course, it makes the proof longer...)
– Taladris
Dec 18 at 15:09
add a comment |
Yes, it converges, by the comparaison test and because$$(forall ninmathbb{N}):bigllvert a_nsin(a_n)bigrrvertleqslantlvert a_nrvert=a_n.$$
answered Dec 18 at 12:59
José Carlos Santos
150k22120221
2
+1..Forgot about this.
– xbh
Dec 18 at 13:02
Note that, since $sum a_n$ is convergent, $lim_{ntoinfty}a_n=0$. Since $a_n>0$, this implies that $a_nin(0,frac{pi}{2})$ for $n$ large enough, so $sin(a_n)>0$ for $n$ large enough. Therefore, you don't need the absolute value. (Of course, it makes the proof longer...)
– Taladris
Dec 18 at 15:09
add a comment |
Yes, it converges, by the comparaison test and because$$(forall ninmathbb{N}):bigllvert a_nsin(a_n)bigrrvertleqslantlvert a_nrvert=a_n.$$
answered Dec 18 at 12:59
José Carlos Santos
150k22120221
Yes, it converges, by the comparaison test and because$$(forall ninmathbb{N}):bigllvert a_nsin(a_n)bigrrvertleqslantlvert a_nrvert=a_n.$$
answered Dec 18 at 12:59
José Carlos Santos
150k22120221
answered Dec 18 at 12:59
José Carlos Santos
150k22120221
answered Dec 18 at 12:59
José Carlos Santos
150k22120221
answered Dec 18 at 12:59
José Carlos Santos
150k22120221
150k22120221
2
+1..Forgot about this.
– xbh
Dec 18 at 13:02
Note that, since $sum a_n$ is convergent, $lim_{ntoinfty}a_n=0$. Since $a_n>0$, this implies that $a_nin(0,frac{pi}{2})$ for $n$ large enough, so $sin(a_n)>0$ for $n$ large enough. Therefore, you don't need the absolute value. (Of course, it makes the proof longer...)
– Taladris
Dec 18 at 15:09
add a comment |
2
+1..Forgot about this.
– xbh
Dec 18 at 13:02
Note that, since $sum a_n$ is convergent, $lim_{ntoinfty}a_n=0$. Since $a_n>0$, this implies that $a_nin(0,frac{pi}{2})$ for $n$ large enough, so $sin(a_n)>0$ for $n$ large enough. Therefore, you don't need the absolute value. (Of course, it makes the proof longer...)
– Taladris
Dec 18 at 15:09
2
2
+1..Forgot about this.
– xbh
Dec 18 at 13:02
+1..Forgot about this.
– xbh
Dec 18 at 13:02
Note that, since $sum a_n$ is convergent, $lim_{ntoinfty}a_n=0$. Since $a_n>0$, this implies that $a_nin(0,frac{pi}{2})$ for $n$ large enough, so $sin(a_n)>0$ for $n$ large enough. Therefore, you don't need the absolute value. (Of course, it makes the proof longer...)
– Taladris
Dec 18 at 15:09
Note that, since $sum a_n$ is convergent, $lim_{ntoinfty}a_n=0$. Since $a_n>0$, this implies that $a_nin(0,frac{pi}{2})$ for $n$ large enough, so $sin(a_n)>0$ for $n$ large enough. Therefore, you don't need the absolute value. (Of course, it makes the proof longer...)
– Taladris
Dec 18 at 15:09
add a comment |
Since $sum a_n$ converges, for each $varepsilon >0$, there exists $N in mathbb N^*$ s.t. whenever $mathbb N^* ni n > N, p in mathbb N^*$, $sum_{n+1}^{n+p} a_k < varepsilon$. Then
$$
leftvert sum_{n+1}^{n+p} a_k sin (a_k) rightvert leqslant sum_{n+1}^{n+p} vert a_k sin (a_k ) vert leqslant sum_{n+1}^{n+p} a_k < varepsilon,
$$
hence $sum a_n sin(a_n)$ converges by Cauchy convergence principle.
answered Dec 18 at 13:02
xbh
5,6551522
add a comment |
Since $sum a_n$ converges, for each $varepsilon >0$, there exists $N in mathbb N^*$ s.t. whenever $mathbb N^* ni n > N, p in mathbb N^*$, $sum_{n+1}^{n+p} a_k < varepsilon$. Then
$$
leftvert sum_{n+1}^{n+p} a_k sin (a_k) rightvert leqslant sum_{n+1}^{n+p} vert a_k sin (a_k ) vert leqslant sum_{n+1}^{n+p} a_k < varepsilon,
$$
hence $sum a_n sin(a_n)$ converges by Cauchy convergence principle.
answered Dec 18 at 13:02
xbh
5,6551522
add a comment |
Since $sum a_n$ converges, for each $varepsilon >0$, there exists $N in mathbb N^*$ s.t. whenever $mathbb N^* ni n > N, p in mathbb N^*$, $sum_{n+1}^{n+p} a_k < varepsilon$. Then
$$
leftvert sum_{n+1}^{n+p} a_k sin (a_k) rightvert leqslant sum_{n+1}^{n+p} vert a_k sin (a_k ) vert leqslant sum_{n+1}^{n+p} a_k < varepsilon,
$$
hence $sum a_n sin(a_n)$ converges by Cauchy convergence principle.
answered Dec 18 at 13:02
xbh
5,6551522
Since $sum a_n$ converges, for each $varepsilon >0$, there exists $N in mathbb N^*$ s.t. whenever $mathbb N^* ni n > N, p in mathbb N^*$, $sum_{n+1}^{n+p} a_k < varepsilon$. Then
$$
leftvert sum_{n+1}^{n+p} a_k sin (a_k) rightvert leqslant sum_{n+1}^{n+p} vert a_k sin (a_k ) vert leqslant sum_{n+1}^{n+p} a_k < varepsilon,
$$
hence $sum a_n sin(a_n)$ converges by Cauchy convergence principle.
answered Dec 18 at 13:02
xbh
5,6551522
answered Dec 18 at 13:02
xbh
5,6551522
answered Dec 18 at 13:02
xbh
5,6551522
answered Dec 18 at 13:02
xbh
5,6551522
5,6551522
add a comment |
add a comment |
2
What did you try?
– José Carlos Santos
Dec 18 at 12:56
@JoséCarlosSantos I'm not sure if I can use something like comparision for the series that $-a_n<a_n sin(a_n) < a_n$ but I tried only this method
– avan1235
Dec 18 at 12:57