If $ sum_{n=1}^{infty} a_n $ is convergent, then discuss the convergence or divergence of the following...
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If $ displaystylesum_{n=1}^{infty} a_n $ is convergent then discuss the convergence or divergence of the following series whose $n^{th}$ term is
$ a_n sin n $
$displaystyle frac {a_n}{1+| a_n |}$
If $displaystyle sum _{n=1}^{infty} a_n $ is absolutely convergent then then $displaystyle sum_{n=1}^{infty} a_n sin n $ and $displaystyle sum _{n=1}^{infty} dfrac {a_n}{1+|a_n|} $ are convergent. But what if $displaystyle sum _{n=1}^{infty} a_n $ is conditionally convergent.
A little hint would be appreciated.
Thanks in advance.
real-analysis
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show 1 more comment
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If $ displaystylesum_{n=1}^{infty} a_n $ is convergent then discuss the convergence or divergence of the following series whose $n^{th}$ term is
$ a_n sin n $
$displaystyle frac {a_n}{1+| a_n |}$
If $displaystyle sum _{n=1}^{infty} a_n $ is absolutely convergent then then $displaystyle sum_{n=1}^{infty} a_n sin n $ and $displaystyle sum _{n=1}^{infty} dfrac {a_n}{1+|a_n|} $ are convergent. But what if $displaystyle sum _{n=1}^{infty} a_n $ is conditionally convergent.
A little hint would be appreciated.
Thanks in advance.
real-analysis
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7
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@suchandaadhikari try $a_n = frac{sin(n)}{n}$, and you can see here for the proof that $frac{sin^2(n)}{n}$ doesn't converge
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– Jakobian
Dec 8 '18 at 13:09
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@ jakobian thank you so much ..nice example. .can u give a hint about the 2nd series. .
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– suchanda adhikari
Dec 8 '18 at 13:58
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I think other series is convergent as both have same nature as $nrightarrow infty$
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– neelkanth
Dec 8 '18 at 13:59
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If you please elaborate a little more...
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– suchanda adhikari
Dec 8 '18 at 14:06
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@Jakobian Why not an official answer?
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– Paul Frost
Dec 8 '18 at 16:13
|
show 1 more comment
$begingroup$
If $ displaystylesum_{n=1}^{infty} a_n $ is convergent then discuss the convergence or divergence of the following series whose $n^{th}$ term is
$ a_n sin n $
$displaystyle frac {a_n}{1+| a_n |}$
If $displaystyle sum _{n=1}^{infty} a_n $ is absolutely convergent then then $displaystyle sum_{n=1}^{infty} a_n sin n $ and $displaystyle sum _{n=1}^{infty} dfrac {a_n}{1+|a_n|} $ are convergent. But what if $displaystyle sum _{n=1}^{infty} a_n $ is conditionally convergent.
A little hint would be appreciated.
Thanks in advance.
real-analysis
$endgroup$
If $ displaystylesum_{n=1}^{infty} a_n $ is convergent then discuss the convergence or divergence of the following series whose $n^{th}$ term is
$ a_n sin n $
$displaystyle frac {a_n}{1+| a_n |}$
If $displaystyle sum _{n=1}^{infty} a_n $ is absolutely convergent then then $displaystyle sum_{n=1}^{infty} a_n sin n $ and $displaystyle sum _{n=1}^{infty} dfrac {a_n}{1+|a_n|} $ are convergent. But what if $displaystyle sum _{n=1}^{infty} a_n $ is conditionally convergent.
A little hint would be appreciated.
Thanks in advance.
real-analysis
real-analysis
edited Dec 8 '18 at 13:04
Yadati Kiran
1,769619
1,769619
asked Dec 8 '18 at 12:53
suchanda adhikarisuchanda adhikari
516
516
7
$begingroup$
@suchandaadhikari try $a_n = frac{sin(n)}{n}$, and you can see here for the proof that $frac{sin^2(n)}{n}$ doesn't converge
$endgroup$
– Jakobian
Dec 8 '18 at 13:09
$begingroup$
@ jakobian thank you so much ..nice example. .can u give a hint about the 2nd series. .
$endgroup$
– suchanda adhikari
Dec 8 '18 at 13:58
$begingroup$
I think other series is convergent as both have same nature as $nrightarrow infty$
$endgroup$
– neelkanth
Dec 8 '18 at 13:59
$begingroup$
If you please elaborate a little more...
$endgroup$
– suchanda adhikari
Dec 8 '18 at 14:06
$begingroup$
@Jakobian Why not an official answer?
$endgroup$
– Paul Frost
Dec 8 '18 at 16:13
|
show 1 more comment
7
$begingroup$
@suchandaadhikari try $a_n = frac{sin(n)}{n}$, and you can see here for the proof that $frac{sin^2(n)}{n}$ doesn't converge
$endgroup$
– Jakobian
Dec 8 '18 at 13:09
$begingroup$
@ jakobian thank you so much ..nice example. .can u give a hint about the 2nd series. .
$endgroup$
– suchanda adhikari
Dec 8 '18 at 13:58
$begingroup$
I think other series is convergent as both have same nature as $nrightarrow infty$
$endgroup$
– neelkanth
Dec 8 '18 at 13:59
$begingroup$
If you please elaborate a little more...
$endgroup$
– suchanda adhikari
Dec 8 '18 at 14:06
$begingroup$
@Jakobian Why not an official answer?
$endgroup$
– Paul Frost
Dec 8 '18 at 16:13
7
7
$begingroup$
@suchandaadhikari try $a_n = frac{sin(n)}{n}$, and you can see here for the proof that $frac{sin^2(n)}{n}$ doesn't converge
$endgroup$
– Jakobian
Dec 8 '18 at 13:09
$begingroup$
@suchandaadhikari try $a_n = frac{sin(n)}{n}$, and you can see here for the proof that $frac{sin^2(n)}{n}$ doesn't converge
$endgroup$
– Jakobian
Dec 8 '18 at 13:09
$begingroup$
@ jakobian thank you so much ..nice example. .can u give a hint about the 2nd series. .
$endgroup$
– suchanda adhikari
Dec 8 '18 at 13:58
$begingroup$
@ jakobian thank you so much ..nice example. .can u give a hint about the 2nd series. .
$endgroup$
– suchanda adhikari
Dec 8 '18 at 13:58
$begingroup$
I think other series is convergent as both have same nature as $nrightarrow infty$
$endgroup$
– neelkanth
Dec 8 '18 at 13:59
$begingroup$
I think other series is convergent as both have same nature as $nrightarrow infty$
$endgroup$
– neelkanth
Dec 8 '18 at 13:59
$begingroup$
If you please elaborate a little more...
$endgroup$
– suchanda adhikari
Dec 8 '18 at 14:06
$begingroup$
If you please elaborate a little more...
$endgroup$
– suchanda adhikari
Dec 8 '18 at 14:06
$begingroup$
@Jakobian Why not an official answer?
$endgroup$
– Paul Frost
Dec 8 '18 at 16:13
$begingroup$
@Jakobian Why not an official answer?
$endgroup$
– Paul Frost
Dec 8 '18 at 16:13
|
show 1 more comment
0
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7
$begingroup$
@suchandaadhikari try $a_n = frac{sin(n)}{n}$, and you can see here for the proof that $frac{sin^2(n)}{n}$ doesn't converge
$endgroup$
– Jakobian
Dec 8 '18 at 13:09
$begingroup$
@ jakobian thank you so much ..nice example. .can u give a hint about the 2nd series. .
$endgroup$
– suchanda adhikari
Dec 8 '18 at 13:58
$begingroup$
I think other series is convergent as both have same nature as $nrightarrow infty$
$endgroup$
– neelkanth
Dec 8 '18 at 13:59
$begingroup$
If you please elaborate a little more...
$endgroup$
– suchanda adhikari
Dec 8 '18 at 14:06
$begingroup$
@Jakobian Why not an official answer?
$endgroup$
– Paul Frost
Dec 8 '18 at 16:13