If $ sum_{n=1}^{infty} a_n $ is convergent, then discuss the convergence or divergence of the following...












3












$begingroup$



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.










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$endgroup$








  • 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
















3












$begingroup$



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.










share|cite|improve this question











$endgroup$








  • 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














3












3








3





$begingroup$



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.










share|cite|improve this question











$endgroup$





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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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














  • 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










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