Specify whether the series $sum_{n=n_0}^{+ infty } a_n cdot b_n$ must be convergent in cases a) or b)
$begingroup$
Specify whether the series $sum_{n=n_0}^{+ infty } a_n cdot b_n$ must be convergent when:
a) $sum a_n$ convergent, $sum b_n$ convergent
b)$sum a_n$ convergent absolutely, $sum b_n$ convergent.
I need to check my reasoning and I need help to guide them further:
If $sum b_n$ convergent, then in particular a sequence of partial sum $sum b_n$ is convergent, so the second condition of Dirichlet's test is met. That is why we should take care of $a_n$. The first condition of Dirichlet's test is $a_n$ monotonic and $a_n$ convergent to $0$. So I think in the case of a) the series $sum_{n=n_0}^{+ infty } a_n cdot b_n$ not always is convergent and b) must be convergement, but there are only my thoughts and I do not know how to prove it.
real-analysis
asked Dec 20 '18 at 14:12
MP3129MP3129
35310
$endgroup$
add a comment |
$begingroup$
Specify whether the series $sum_{n=n_0}^{+ infty } a_n cdot b_n$ must be convergent when:
a) $sum a_n$ convergent, $sum b_n$ convergent
b)$sum a_n$ convergent absolutely, $sum b_n$ convergent.
I need to check my reasoning and I need help to guide them further:
If $sum b_n$ convergent, then in particular a sequence of partial sum $sum b_n$ is convergent, so the second condition of Dirichlet's test is met. That is why we should take care of $a_n$. The first condition of Dirichlet's test is $a_n$ monotonic and $a_n$ convergent to $0$. So I think in the case of a) the series $sum_{n=n_0}^{+ infty } a_n cdot b_n$ not always is convergent and b) must be convergement, but there are only my thoughts and I do not know how to prove it.
real-analysis
asked Dec 20 '18 at 14:12
MP3129MP3129
35310
$endgroup$
add a comment |
$begingroup$
Specify whether the series $sum_{n=n_0}^{+ infty } a_n cdot b_n$ must be convergent when:
a) $sum a_n$ convergent, $sum b_n$ convergent
b)$sum a_n$ convergent absolutely, $sum b_n$ convergent.
I need to check my reasoning and I need help to guide them further:
If $sum b_n$ convergent, then in particular a sequence of partial sum $sum b_n$ is convergent, so the second condition of Dirichlet's test is met. That is why we should take care of $a_n$. The first condition of Dirichlet's test is $a_n$ monotonic and $a_n$ convergent to $0$. So I think in the case of a) the series $sum_{n=n_0}^{+ infty } a_n cdot b_n$ not always is convergent and b) must be convergement, but there are only my thoughts and I do not know how to prove it.
real-analysis
asked Dec 20 '18 at 14:12
MP3129MP3129
35310
$endgroup$
Specify whether the series $sum_{n=n_0}^{+ infty } a_n cdot b_n$ must be convergent when:
a) $sum a_n$ convergent, $sum b_n$ convergent
b)$sum a_n$ convergent absolutely, $sum b_n$ convergent.
I need to check my reasoning and I need help to guide them further:
If $sum b_n$ convergent, then in particular a sequence of partial sum $sum b_n$ is convergent, so the second condition of Dirichlet's test is met. That is why we should take care of $a_n$. The first condition of Dirichlet's test is $a_n$ monotonic and $a_n$ convergent to $0$. So I think in the case of a) the series $sum_{n=n_0}^{+ infty } a_n cdot b_n$ not always is convergent and b) must be convergement, but there are only my thoughts and I do not know how to prove it.
real-analysis
real-analysis
asked Dec 20 '18 at 14:12
MP3129MP3129
35310
asked Dec 20 '18 at 14:12
MP3129MP3129
35310
asked Dec 20 '18 at 14:12
MP3129MP3129
35310
asked Dec 20 '18 at 14:12
MP3129MP3129
35310
asked Dec 20 '18 at 14:12
MP3129MP3129
35310
35310
add a comment |
add a comment |
1 Answer
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$begingroup$
For the first case a counter-example is given by
$$a_n=b_n=frac{(-1)^n}{sqrt n}$$
For the second and since $sum b_n$ is convergent so for sufficiently large $n$ we get
$$vert b_nvert le 1$$
hence
$$vert a_n b_nvertle vert a_nvert$$
hence the series $sum a_n b_n$ is absolutely convergent.
answered Dec 20 '18 at 14:17
user296113user296113
6,955928
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1 Answer
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1 Answer
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$begingroup$
For the first case a counter-example is given by
$$a_n=b_n=frac{(-1)^n}{sqrt n}$$
For the second and since $sum b_n$ is convergent so for sufficiently large $n$ we get
$$vert b_nvert le 1$$
hence
$$vert a_n b_nvertle vert a_nvert$$
hence the series $sum a_n b_n$ is absolutely convergent.
answered Dec 20 '18 at 14:17
user296113user296113
6,955928
$endgroup$
add a comment |
$begingroup$
For the first case a counter-example is given by
$$a_n=b_n=frac{(-1)^n}{sqrt n}$$
For the second and since $sum b_n$ is convergent so for sufficiently large $n$ we get
$$vert b_nvert le 1$$
hence
$$vert a_n b_nvertle vert a_nvert$$
hence the series $sum a_n b_n$ is absolutely convergent.
answered Dec 20 '18 at 14:17
user296113user296113
6,955928
$endgroup$
add a comment |
$begingroup$
For the first case a counter-example is given by
$$a_n=b_n=frac{(-1)^n}{sqrt n}$$
For the second and since $sum b_n$ is convergent so for sufficiently large $n$ we get
$$vert b_nvert le 1$$
hence
$$vert a_n b_nvertle vert a_nvert$$
hence the series $sum a_n b_n$ is absolutely convergent.
answered Dec 20 '18 at 14:17
user296113user296113
6,955928
$endgroup$
For the first case a counter-example is given by
$$a_n=b_n=frac{(-1)^n}{sqrt n}$$
For the second and since $sum b_n$ is convergent so for sufficiently large $n$ we get
$$vert b_nvert le 1$$
hence
$$vert a_n b_nvertle vert a_nvert$$
hence the series $sum a_n b_n$ is absolutely convergent.
answered Dec 20 '18 at 14:17
user296113user296113
6,955928
answered Dec 20 '18 at 14:17
user296113user296113
6,955928
answered Dec 20 '18 at 14:17
user296113user296113
6,955928
answered Dec 20 '18 at 14:17
user296113user296113
6,955928
6,955928
add a comment |
add a comment |
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