Finding $p$ such that $sumlimits_{n=1}^infty n(1+n^2)^p$ converges. Check my work.












4












$begingroup$



Given series
$$
sumlimits_{n=1}^infty n(1+n^2)^p
$$
Find the values of $p$, such that the series is convergent.




To find $p$, I use integral test,
assume $f(x)=x(1+x^2)^p$,



begin{eqnarray}
intlimits_1^infty x(1+x^2)^pdx &=& limlimits_{btoinfty} intlimits_1^b x(1+x^2)^pdx\
&=& limlimits_{btoinfty} left[dfrac{1}{2(p+1)}left((1+b^2)^{p+1}-2^{p+1}right)right]
end{eqnarray}




  • If $p=1$, the integral is divergent,


  • If $p>-1$, the integral is divergent,


  • If $p<-1$, the integral is convergent to $-dfrac{1}{p+1}2^p$,




So, I conclude the series is convergent while $p<-1$.



This answer is correct or incorrect?







sequences-and-series convergence







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes, it is correct
    $endgroup$
    – caverac
    Dec 14 '18 at 2:26






  • 1




    $begingroup$
    Well done .....
    $endgroup$
    – Mark Viola
    Dec 14 '18 at 2:26










  • $begingroup$
    OK, thank you so much
    $endgroup$
    – Ongky Denny Wijaya
    Dec 14 '18 at 2:34










  • $begingroup$
    OK, thank you so much
    $endgroup$
    – Ongky Denny Wijaya
    Dec 14 '18 at 2:34
















4












$begingroup$



Given series
$$
sumlimits_{n=1}^infty n(1+n^2)^p
$$
Find the values of $p$, such that the series is convergent.




To find $p$, I use integral test,
assume $f(x)=x(1+x^2)^p$,



begin{eqnarray}
intlimits_1^infty x(1+x^2)^pdx &=& limlimits_{btoinfty} intlimits_1^b x(1+x^2)^pdx\
&=& limlimits_{btoinfty} left[dfrac{1}{2(p+1)}left((1+b^2)^{p+1}-2^{p+1}right)right]
end{eqnarray}




  • If $p=1$, the integral is divergent,


  • If $p>-1$, the integral is divergent,


  • If $p<-1$, the integral is convergent to $-dfrac{1}{p+1}2^p$,




So, I conclude the series is convergent while $p<-1$.



This answer is correct or incorrect?







sequences-and-series convergence







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes, it is correct
    $endgroup$
    – caverac
    Dec 14 '18 at 2:26






  • 1




    $begingroup$
    Well done .....
    $endgroup$
    – Mark Viola
    Dec 14 '18 at 2:26










  • $begingroup$
    OK, thank you so much
    $endgroup$
    – Ongky Denny Wijaya
    Dec 14 '18 at 2:34










  • $begingroup$
    OK, thank you so much
    $endgroup$
    – Ongky Denny Wijaya
    Dec 14 '18 at 2:34














4












4








4





$begingroup$



Given series
$$
sumlimits_{n=1}^infty n(1+n^2)^p
$$
Find the values of $p$, such that the series is convergent.




To find $p$, I use integral test,
assume $f(x)=x(1+x^2)^p$,



begin{eqnarray}
intlimits_1^infty x(1+x^2)^pdx &=& limlimits_{btoinfty} intlimits_1^b x(1+x^2)^pdx\
&=& limlimits_{btoinfty} left[dfrac{1}{2(p+1)}left((1+b^2)^{p+1}-2^{p+1}right)right]
end{eqnarray}




  • If $p=1$, the integral is divergent,


  • If $p>-1$, the integral is divergent,


  • If $p<-1$, the integral is convergent to $-dfrac{1}{p+1}2^p$,




So, I conclude the series is convergent while $p<-1$.



This answer is correct or incorrect?







sequences-and-series convergence







share|cite|improve this question











$endgroup$





Given series
$$
sumlimits_{n=1}^infty n(1+n^2)^p
$$
Find the values of $p$, such that the series is convergent.




To find $p$, I use integral test,
assume $f(x)=x(1+x^2)^p$,



begin{eqnarray}
intlimits_1^infty x(1+x^2)^pdx &=& limlimits_{btoinfty} intlimits_1^b x(1+x^2)^pdx\
&=& limlimits_{btoinfty} left[dfrac{1}{2(p+1)}left((1+b^2)^{p+1}-2^{p+1}right)right]
end{eqnarray}




  • If $p=1$, the integral is divergent,


  • If $p>-1$, the integral is divergent,


  • If $p<-1$, the integral is convergent to $-dfrac{1}{p+1}2^p$,




So, I conclude the series is convergent while $p<-1$.



This answer is correct or incorrect?







sequences-and-series convergence




sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 2:57









Blue

48.6k870156




48.6k870156










asked Dec 14 '18 at 2:20









Ongky Denny WijayaOngky Denny Wijaya

3678




3678








  • 1




    $begingroup$
    Yes, it is correct
    $endgroup$
    – caverac
    Dec 14 '18 at 2:26






  • 1




    $begingroup$
    Well done .....
    $endgroup$
    – Mark Viola
    Dec 14 '18 at 2:26










  • $begingroup$
    OK, thank you so much
    $endgroup$
    – Ongky Denny Wijaya
    Dec 14 '18 at 2:34










  • $begingroup$
    OK, thank you so much
    $endgroup$
    – Ongky Denny Wijaya
    Dec 14 '18 at 2:34














  • 1




    $begingroup$
    Yes, it is correct
    $endgroup$
    – caverac
    Dec 14 '18 at 2:26






  • 1




    $begingroup$
    Well done .....
    $endgroup$
    – Mark Viola
    Dec 14 '18 at 2:26










  • $begingroup$
    OK, thank you so much
    $endgroup$
    – Ongky Denny Wijaya
    Dec 14 '18 at 2:34










  • $begingroup$
    OK, thank you so much
    $endgroup$
    – Ongky Denny Wijaya
    Dec 14 '18 at 2:34








1




1




$begingroup$
Yes, it is correct
$endgroup$
– caverac
Dec 14 '18 at 2:26




$begingroup$
Yes, it is correct
$endgroup$
– caverac
Dec 14 '18 at 2:26




1




1




$begingroup$
Well done .....
$endgroup$
– Mark Viola
Dec 14 '18 at 2:26




$begingroup$
Well done .....
$endgroup$
– Mark Viola
Dec 14 '18 at 2:26












$begingroup$
OK, thank you so much
$endgroup$
– Ongky Denny Wijaya
Dec 14 '18 at 2:34




$begingroup$
OK, thank you so much
$endgroup$
– Ongky Denny Wijaya
Dec 14 '18 at 2:34












$begingroup$
OK, thank you so much
$endgroup$
– Ongky Denny Wijaya
Dec 14 '18 at 2:34




$begingroup$
OK, thank you so much
$endgroup$
– Ongky Denny Wijaya
Dec 14 '18 at 2:34










1 Answer
1






active

oldest

votes


















2












$begingroup$

Alternative:



If $pge -1$, then $(1+n^2)^p ge (1+n^2)^{-1}$



$$sum_{n=1}^infty n(1+n^2)^p gesum_{n=1}^infty frac{n}{1+n^2} ge sum_{n=1}^infty frac{n}{2n^2}= sum_{n=1}^infty frac{1}{2n}$$



Hence it diverges.



If $p<-1$, then $-p > 1$, $(1+n^2)^{-p} ge (n^2)^{-p}$. Also, we have $-2p-1>2-1=1$



$$sum_{n=1}^infty frac{n}{(1+n^2)^{-p}} le sum_{n=1}^infty frac{n}{n^{-2p}}= sum_{n=1}^infty frac{1}{n^{-2p-1}}$$



Hence, by $p$-series and comparison test, it converges.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much Siong Thye Goh
    $endgroup$
    – Ongky Denny Wijaya
    Dec 14 '18 at 3:29











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

Alternative:



If $pge -1$, then $(1+n^2)^p ge (1+n^2)^{-1}$



$$sum_{n=1}^infty n(1+n^2)^p gesum_{n=1}^infty frac{n}{1+n^2} ge sum_{n=1}^infty frac{n}{2n^2}= sum_{n=1}^infty frac{1}{2n}$$



Hence it diverges.



If $p<-1$, then $-p > 1$, $(1+n^2)^{-p} ge (n^2)^{-p}$. Also, we have $-2p-1>2-1=1$



$$sum_{n=1}^infty frac{n}{(1+n^2)^{-p}} le sum_{n=1}^infty frac{n}{n^{-2p}}= sum_{n=1}^infty frac{1}{n^{-2p-1}}$$



Hence, by $p$-series and comparison test, it converges.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much Siong Thye Goh
    $endgroup$
    – Ongky Denny Wijaya
    Dec 14 '18 at 3:29
















2












$begingroup$

Alternative:



If $pge -1$, then $(1+n^2)^p ge (1+n^2)^{-1}$



$$sum_{n=1}^infty n(1+n^2)^p gesum_{n=1}^infty frac{n}{1+n^2} ge sum_{n=1}^infty frac{n}{2n^2}= sum_{n=1}^infty frac{1}{2n}$$



Hence it diverges.



If $p<-1$, then $-p > 1$, $(1+n^2)^{-p} ge (n^2)^{-p}$. Also, we have $-2p-1>2-1=1$



$$sum_{n=1}^infty frac{n}{(1+n^2)^{-p}} le sum_{n=1}^infty frac{n}{n^{-2p}}= sum_{n=1}^infty frac{1}{n^{-2p-1}}$$



Hence, by $p$-series and comparison test, it converges.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much Siong Thye Goh
    $endgroup$
    – Ongky Denny Wijaya
    Dec 14 '18 at 3:29














2












2








2





$begingroup$

Alternative:



If $pge -1$, then $(1+n^2)^p ge (1+n^2)^{-1}$



$$sum_{n=1}^infty n(1+n^2)^p gesum_{n=1}^infty frac{n}{1+n^2} ge sum_{n=1}^infty frac{n}{2n^2}= sum_{n=1}^infty frac{1}{2n}$$



Hence it diverges.



If $p<-1$, then $-p > 1$, $(1+n^2)^{-p} ge (n^2)^{-p}$. Also, we have $-2p-1>2-1=1$



$$sum_{n=1}^infty frac{n}{(1+n^2)^{-p}} le sum_{n=1}^infty frac{n}{n^{-2p}}= sum_{n=1}^infty frac{1}{n^{-2p-1}}$$



Hence, by $p$-series and comparison test, it converges.






share|cite|improve this answer









$endgroup$



Alternative:



If $pge -1$, then $(1+n^2)^p ge (1+n^2)^{-1}$



$$sum_{n=1}^infty n(1+n^2)^p gesum_{n=1}^infty frac{n}{1+n^2} ge sum_{n=1}^infty frac{n}{2n^2}= sum_{n=1}^infty frac{1}{2n}$$



Hence it diverges.



If $p<-1$, then $-p > 1$, $(1+n^2)^{-p} ge (n^2)^{-p}$. Also, we have $-2p-1>2-1=1$



$$sum_{n=1}^infty frac{n}{(1+n^2)^{-p}} le sum_{n=1}^infty frac{n}{n^{-2p}}= sum_{n=1}^infty frac{1}{n^{-2p-1}}$$



Hence, by $p$-series and comparison test, it converges.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 14 '18 at 2:47









Siong Thye GohSiong Thye Goh

102k1466118




102k1466118












  • $begingroup$
    Thank you so much Siong Thye Goh
    $endgroup$
    – Ongky Denny Wijaya
    Dec 14 '18 at 3:29


















  • $begingroup$
    Thank you so much Siong Thye Goh
    $endgroup$
    – Ongky Denny Wijaya
    Dec 14 '18 at 3:29
















$begingroup$
Thank you so much Siong Thye Goh
$endgroup$
– Ongky Denny Wijaya
Dec 14 '18 at 3:29




$begingroup$
Thank you so much Siong Thye Goh
$endgroup$
– Ongky Denny Wijaya
Dec 14 '18 at 3:29


















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