Determine this limitIs my solution for divergence of $int_0^infty fracsin^10x ln xsqrtx$ correct?Limit involving the Riemann zeta function, why is this identity trivial?Show by comparison that $sumlimits_n=1^infty sin(frac1n)$ diverges?Determine con-/divergence of $sumlimits_n=1^infty left[ sqrtn^3+1 - n^frac32 right]$ by (limit) comparison testcomparison or limit comparison test..?Incorrect comparison test on seriesDetermine whether the series converges or diverges.How to determine the convergence of $sum frac 1 n!left(frac n eright)^n$ using Raabe's testLimit Comparison Test for integralWhy do I have to use L'Hopital in limit comparison test for $sum_n=1^infty sinleft(frac1nright)$Find whether a series is divergent or convergent with using the Limit Comparison Test
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Determine this limit
Is my solution for divergence of $int_0^infty fracsin^10x ln xsqrtx$ correct?Limit involving the Riemann zeta function, why is this identity trivial?Show by comparison that $sumlimits_n=1^infty sin(frac1n)$ diverges?Determine con-/divergence of $sumlimits_n=1^infty left[ sqrtn^3+1 - n^frac32 right]$ by (limit) comparison testcomparison or limit comparison test..?Incorrect comparison test on seriesDetermine whether the series converges or diverges.How to determine the convergence of $sum frac 1 n!left(frac n eright)^n$ using Raabe's testLimit Comparison Test for integralWhy do I have to use L'Hopital in limit comparison test for $sum_n=1^infty sinleft(frac1nright)$Find whether a series is divergent or convergent with using the Limit Comparison Test
$begingroup$
how can I determine the following limit? $$lim_ntoinfty fraclnleft(frac3pi4 + 2nright)-lnleft(fracpi4+2nright)ln(2n+2)-ln(2n).$$
This question stems from this question. The proof presented there is incorrect, and it would be trivial to show that the mentioned integral diverges if the above limit is $>0$ by using the comparison test (anyone feel free to do this by the way).
sequences-and-series limits
edited 1 hour ago
El Ectric
353214
asked 2 hours ago
Acacia MarshAcacia Marsh
155
New contributor
Acacia Marsh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
how can I determine the following limit? $$lim_ntoinfty fraclnleft(frac3pi4 + 2nright)-lnleft(fracpi4+2nright)ln(2n+2)-ln(2n).$$
This question stems from this question. The proof presented there is incorrect, and it would be trivial to show that the mentioned integral diverges if the above limit is $>0$ by using the comparison test (anyone feel free to do this by the way).
sequences-and-series limits
edited 1 hour ago
El Ectric
353214
asked 2 hours ago
Acacia MarshAcacia Marsh
155
New contributor
Acacia Marsh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Remember... $ln(a)-ln(b)=ln(frac ab)$.
$endgroup$
– manooooh
1 hour ago
1
$begingroup$
Welcome to math.SE!!
$endgroup$
– manooooh
1 hour ago
add a comment |
$begingroup$
how can I determine the following limit? $$lim_ntoinfty fraclnleft(frac3pi4 + 2nright)-lnleft(fracpi4+2nright)ln(2n+2)-ln(2n).$$
This question stems from this question. The proof presented there is incorrect, and it would be trivial to show that the mentioned integral diverges if the above limit is $>0$ by using the comparison test (anyone feel free to do this by the way).
sequences-and-series limits
edited 1 hour ago
El Ectric
353214
asked 2 hours ago
Acacia MarshAcacia Marsh
155
New contributor
Acacia Marsh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
how can I determine the following limit? $$lim_ntoinfty fraclnleft(frac3pi4 + 2nright)-lnleft(fracpi4+2nright)ln(2n+2)-ln(2n).$$
This question stems from this question. The proof presented there is incorrect, and it would be trivial to show that the mentioned integral diverges if the above limit is $>0$ by using the comparison test (anyone feel free to do this by the way).
sequences-and-series limits
sequences-and-series limits
edited 1 hour ago
El Ectric
353214
asked 2 hours ago
Acacia MarshAcacia Marsh
155
New contributor
Acacia Marsh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
El Ectric
353214
asked 2 hours ago
Acacia MarshAcacia Marsh
155
New contributor
Acacia Marsh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
El Ectric
353214
edited 1 hour ago
El Ectric
353214
edited 1 hour ago
El Ectric
353214
353214
asked 2 hours ago
Acacia MarshAcacia Marsh
155
New contributor
Acacia Marsh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Acacia MarshAcacia Marsh
155
asked 2 hours ago
Acacia MarshAcacia Marsh
155
155
New contributor
Acacia Marsh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Acacia Marsh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Remember... $ln(a)-ln(b)=ln(frac ab)$.
$endgroup$
– manooooh
1 hour ago
1
$begingroup$
Welcome to math.SE!!
$endgroup$
– manooooh
1 hour ago
add a comment |
1
$begingroup$
Remember... $ln(a)-ln(b)=ln(frac ab)$.
$endgroup$
– manooooh
1 hour ago
1
$begingroup$
Welcome to math.SE!!
$endgroup$
– manooooh
1 hour ago
1
1
$begingroup$
Remember... $ln(a)-ln(b)=ln(frac ab)$.
$endgroup$
– manooooh
1 hour ago
$begingroup$
Remember... $ln(a)-ln(b)=ln(frac ab)$.
$endgroup$
– manooooh
1 hour ago
1
1
$begingroup$
Welcome to math.SE!!
$endgroup$
– manooooh
1 hour ago
$begingroup$
Welcome to math.SE!!
$endgroup$
– manooooh
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Use l'Hôpital (and some logarithm identities) to get:
beginequation
beginsplit
lim_ntoinfty fracln(frac34 pi + 2n)-ln(fracpi4+2n)ln(2n+2)-ln(2n)
&= lim_ntoinfty fracln(3-frac16n8n+pi)ln(1+frac1n) \
&oversettextl'Hôpital= 16pi lim_ntoinfty fracn(n+1)(8n+pi)(8n+3pi) \
&oversettextlinearity of the limit=16pi bigg(lim_ntoinftyfrac n8n+pibigg)bigg(lim_ntoinftyfracn+18n+3pibigg) \
&= 16picdotfrac18cdotfrac18=fracpi4.
endsplit
endequation
answered 1 hour ago
Maximilian JanischMaximilian Janisch
1,24017
$endgroup$
$begingroup$
That's nice! Thanks
$endgroup$
– Acacia Marsh
1 hour ago
add a comment |
$begingroup$
Let $f(x)=lnleft(frac3pi3+xright)-lnleft(fracpi3+xright)$ and $g(x)=ln(x+2)-ln x$. Thenbeginalignlim_xtoinftyfracf(x)g(x)&=lim_xtoinftyfracf'(x)g'(x)\&=lim_xtoinftyfrac-frac8 pi (4 x+pi ) (4 x+3 pi )-frac2x^2+2 x\&=lim_xtoinftyfrac4pi(x^2+2x)(4 x+pi ) (4 x+3 pi )\&=fracpi4.endalignSo, your limit is $fracpi4$.
answered 1 hour ago
José Carlos SantosJosé Carlos Santos
186k24145259
$endgroup$
$begingroup$
Thank you! I like it
$endgroup$
– Acacia Marsh
1 hour ago
add a comment |
$begingroup$
Using the rules of logarithm so write $$fraclnleft(fracfrac5pi4+2nfracpi4+2nright)lnleft(frac2n+22nright)$$
edited 1 hour ago
user10354138
10.1k21025
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.1k42867
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use l'Hôpital (and some logarithm identities) to get:
beginequation
beginsplit
lim_ntoinfty fracln(frac34 pi + 2n)-ln(fracpi4+2n)ln(2n+2)-ln(2n)
&= lim_ntoinfty fracln(3-frac16n8n+pi)ln(1+frac1n) \
&oversettextl'Hôpital= 16pi lim_ntoinfty fracn(n+1)(8n+pi)(8n+3pi) \
&oversettextlinearity of the limit=16pi bigg(lim_ntoinftyfrac n8n+pibigg)bigg(lim_ntoinftyfracn+18n+3pibigg) \
&= 16picdotfrac18cdotfrac18=fracpi4.
endsplit
endequation
answered 1 hour ago
Maximilian JanischMaximilian Janisch
1,24017
$endgroup$
$begingroup$
That's nice! Thanks
$endgroup$
– Acacia Marsh
1 hour ago
add a comment |
$begingroup$
Use l'Hôpital (and some logarithm identities) to get:
beginequation
beginsplit
lim_ntoinfty fracln(frac34 pi + 2n)-ln(fracpi4+2n)ln(2n+2)-ln(2n)
&= lim_ntoinfty fracln(3-frac16n8n+pi)ln(1+frac1n) \
&oversettextl'Hôpital= 16pi lim_ntoinfty fracn(n+1)(8n+pi)(8n+3pi) \
&oversettextlinearity of the limit=16pi bigg(lim_ntoinftyfrac n8n+pibigg)bigg(lim_ntoinftyfracn+18n+3pibigg) \
&= 16picdotfrac18cdotfrac18=fracpi4.
endsplit
endequation
answered 1 hour ago
Maximilian JanischMaximilian Janisch
1,24017
$endgroup$
$begingroup$
That's nice! Thanks
$endgroup$
– Acacia Marsh
1 hour ago
add a comment |
$begingroup$
Use l'Hôpital (and some logarithm identities) to get:
beginequation
beginsplit
lim_ntoinfty fracln(frac34 pi + 2n)-ln(fracpi4+2n)ln(2n+2)-ln(2n)
&= lim_ntoinfty fracln(3-frac16n8n+pi)ln(1+frac1n) \
&oversettextl'Hôpital= 16pi lim_ntoinfty fracn(n+1)(8n+pi)(8n+3pi) \
&oversettextlinearity of the limit=16pi bigg(lim_ntoinftyfrac n8n+pibigg)bigg(lim_ntoinftyfracn+18n+3pibigg) \
&= 16picdotfrac18cdotfrac18=fracpi4.
endsplit
endequation
answered 1 hour ago
Maximilian JanischMaximilian Janisch
1,24017
$endgroup$
Use l'Hôpital (and some logarithm identities) to get:
beginequation
beginsplit
lim_ntoinfty fracln(frac34 pi + 2n)-ln(fracpi4+2n)ln(2n+2)-ln(2n)
&= lim_ntoinfty fracln(3-frac16n8n+pi)ln(1+frac1n) \
&oversettextl'Hôpital= 16pi lim_ntoinfty fracn(n+1)(8n+pi)(8n+3pi) \
&oversettextlinearity of the limit=16pi bigg(lim_ntoinftyfrac n8n+pibigg)bigg(lim_ntoinftyfracn+18n+3pibigg) \
&= 16picdotfrac18cdotfrac18=fracpi4.
endsplit
endequation
answered 1 hour ago
Maximilian JanischMaximilian Janisch
1,24017
answered 1 hour ago
Maximilian JanischMaximilian Janisch
1,24017
answered 1 hour ago
Maximilian JanischMaximilian Janisch
1,24017
answered 1 hour ago
Maximilian JanischMaximilian Janisch
1,24017
1,24017
$begingroup$
That's nice! Thanks
$endgroup$
– Acacia Marsh
1 hour ago
add a comment |
$begingroup$
That's nice! Thanks
$endgroup$
– Acacia Marsh
1 hour ago
$begingroup$
That's nice! Thanks
$endgroup$
– Acacia Marsh
1 hour ago
$begingroup$
That's nice! Thanks
$endgroup$
– Acacia Marsh
1 hour ago
add a comment |
$begingroup$
Let $f(x)=lnleft(frac3pi3+xright)-lnleft(fracpi3+xright)$ and $g(x)=ln(x+2)-ln x$. Thenbeginalignlim_xtoinftyfracf(x)g(x)&=lim_xtoinftyfracf'(x)g'(x)\&=lim_xtoinftyfrac-frac8 pi (4 x+pi ) (4 x+3 pi )-frac2x^2+2 x\&=lim_xtoinftyfrac4pi(x^2+2x)(4 x+pi ) (4 x+3 pi )\&=fracpi4.endalignSo, your limit is $fracpi4$.
answered 1 hour ago
José Carlos SantosJosé Carlos Santos
186k24145259
$endgroup$
$begingroup$
Thank you! I like it
$endgroup$
– Acacia Marsh
1 hour ago
add a comment |
$begingroup$
Let $f(x)=lnleft(frac3pi3+xright)-lnleft(fracpi3+xright)$ and $g(x)=ln(x+2)-ln x$. Thenbeginalignlim_xtoinftyfracf(x)g(x)&=lim_xtoinftyfracf'(x)g'(x)\&=lim_xtoinftyfrac-frac8 pi (4 x+pi ) (4 x+3 pi )-frac2x^2+2 x\&=lim_xtoinftyfrac4pi(x^2+2x)(4 x+pi ) (4 x+3 pi )\&=fracpi4.endalignSo, your limit is $fracpi4$.
answered 1 hour ago
José Carlos SantosJosé Carlos Santos
186k24145259
$endgroup$
$begingroup$
Thank you! I like it
$endgroup$
– Acacia Marsh
1 hour ago
add a comment |
$begingroup$
Let $f(x)=lnleft(frac3pi3+xright)-lnleft(fracpi3+xright)$ and $g(x)=ln(x+2)-ln x$. Thenbeginalignlim_xtoinftyfracf(x)g(x)&=lim_xtoinftyfracf'(x)g'(x)\&=lim_xtoinftyfrac-frac8 pi (4 x+pi ) (4 x+3 pi )-frac2x^2+2 x\&=lim_xtoinftyfrac4pi(x^2+2x)(4 x+pi ) (4 x+3 pi )\&=fracpi4.endalignSo, your limit is $fracpi4$.
answered 1 hour ago
José Carlos SantosJosé Carlos Santos
186k24145259
$endgroup$
Let $f(x)=lnleft(frac3pi3+xright)-lnleft(fracpi3+xright)$ and $g(x)=ln(x+2)-ln x$. Thenbeginalignlim_xtoinftyfracf(x)g(x)&=lim_xtoinftyfracf'(x)g'(x)\&=lim_xtoinftyfrac-frac8 pi (4 x+pi ) (4 x+3 pi )-frac2x^2+2 x\&=lim_xtoinftyfrac4pi(x^2+2x)(4 x+pi ) (4 x+3 pi )\&=fracpi4.endalignSo, your limit is $fracpi4$.
answered 1 hour ago
José Carlos SantosJosé Carlos Santos
186k24145259
answered 1 hour ago
José Carlos SantosJosé Carlos Santos
186k24145259
answered 1 hour ago
José Carlos SantosJosé Carlos Santos
186k24145259
answered 1 hour ago
José Carlos SantosJosé Carlos Santos
186k24145259
186k24145259
$begingroup$
Thank you! I like it
$endgroup$
– Acacia Marsh
1 hour ago
add a comment |
$begingroup$
Thank you! I like it
$endgroup$
– Acacia Marsh
1 hour ago
$begingroup$
Thank you! I like it
$endgroup$
– Acacia Marsh
1 hour ago
$begingroup$
Thank you! I like it
$endgroup$
– Acacia Marsh
1 hour ago
add a comment |
$begingroup$
Using the rules of logarithm so write $$fraclnleft(fracfrac5pi4+2nfracpi4+2nright)lnleft(frac2n+22nright)$$
edited 1 hour ago
user10354138
10.1k21025
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.1k42867
$endgroup$
add a comment |
$begingroup$
Using the rules of logarithm so write $$fraclnleft(fracfrac5pi4+2nfracpi4+2nright)lnleft(frac2n+22nright)$$
edited 1 hour ago
user10354138
10.1k21025
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.1k42867
$endgroup$
add a comment |
$begingroup$
Using the rules of logarithm so write $$fraclnleft(fracfrac5pi4+2nfracpi4+2nright)lnleft(frac2n+22nright)$$
edited 1 hour ago
user10354138
10.1k21025
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.1k42867
$endgroup$
Using the rules of logarithm so write $$fraclnleft(fracfrac5pi4+2nfracpi4+2nright)lnleft(frac2n+22nright)$$
edited 1 hour ago
user10354138
10.1k21025
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.1k42867
edited 1 hour ago
user10354138
10.1k21025
edited 1 hour ago
user10354138
10.1k21025
edited 1 hour ago
user10354138
10.1k21025
10.1k21025
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.1k42867
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.1k42867
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
81.1k42867
81.1k42867
add a comment |
add a comment |
Acacia Marsh is a new contributor. Be nice, and check out our Code of Conduct.
Acacia Marsh is a new contributor. Be nice, and check out our Code of Conduct.
Acacia Marsh is a new contributor. Be nice, and check out our Code of Conduct.
Acacia Marsh is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
Remember... $ln(a)-ln(b)=ln(frac ab)$.
$endgroup$
– manooooh
1 hour ago
1
$begingroup$
Welcome to math.SE!!
$endgroup$
– manooooh
1 hour ago