Finding $p$ such that $sumlimits_{n=1}^infty n(1+n^2)^p$ converges. Check my work.
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
add a comment |
$begingroup$
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
$endgroup$
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
edited Dec 14 '18 at 2:57
Blue
48.3k870153
48.3k870153
asked Dec 14 '18 at 2:20
Ongky Denny WijayaOngky Denny Wijaya
1417
1417
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
Thank you so much Siong Thye Goh
$endgroup$
– Ongky Denny Wijaya
Dec 14 '18 at 3:29
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
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.
$endgroup$
$begingroup$
Thank you so much Siong Thye Goh
$endgroup$
– Ongky Denny Wijaya
Dec 14 '18 at 3:29
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
Thank you so much Siong Thye Goh
$endgroup$
– Ongky Denny Wijaya
Dec 14 '18 at 3:29
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Dec 14 '18 at 2:47
Siong Thye GohSiong Thye Goh
101k1466118
101k1466118
$begingroup$
Thank you so much Siong Thye Goh
$endgroup$
– Ongky Denny Wijaya
Dec 14 '18 at 3:29
add a comment |
$begingroup$
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
add a comment |
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1
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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