Find the limit of $limlimits_ntoinftyfrac1n^2sumlimits_k=1^nkarctanbig(fracpk-p+1pnbig)$Find $limlimits_nrightarrowinftysumlimits_k=1^nfrac1n+sqrt(k^2-k+1)$evaluation of limit $limlimits_xto inftyleft(frac2arctan(x)piright)^x $How to evaluate the $limlimits_xto 0frac 2sin(x)-arctan(x)-xcos(x^2)x^5$, using power series?Evaluate $limlimits_xtoinftyx(fracpi2-arctan(x))$ without using L'HôpitalFind $lim limits_xto 0fraclogleft(cos xright)x^2$ without L'HopitalSolving $limlimits_x to infty fracsin(x)x-pi$ using L'Hôpital's ruleGiven $f(x) = fracsin x^alpha$ with $alpha >0$, find $limlimits_x to inftyf(x)$ and justify using limit definition at infinity.Show that $limlimits_n to inftysin n^2$ does not exist.Find $limlimits_n to infty sumlimits_k=1^inftyfrac1k^2sqrt[k]nsin^2left(fracn pikright)$How to solve the limit $limlimits_xto infty (x arctan x - fracxpi2)$Calculating $limlimits_ntoinftysumlimits_k=1^n arctan(frac1n+k)$
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Find the limit of $limlimits_ntoinftyfrac1n^2sumlimits_k=1^nkarctanbig(fracpk-p+1pnbig)$
Find $limlimits_nrightarrowinftysumlimits_k=1^nfrac1n+sqrt(k^2-k+1)$evaluation of limit $limlimits_xto inftyleft(frac2arctan(x)piright)^x $How to evaluate the $limlimits_xto 0frac 2sin(x)-arctan(x)-xcos(x^2)x^5$, using power series?Evaluate $limlimits_xtoinftyx(fracpi2-arctan(x))$ without using L'HôpitalFind $lim limits_xto 0fraclogleft(cos xright)x^2$ without L'HopitalSolving $limlimits_x to infty fracsin(x)x-pi$ using L'Hôpital's ruleGiven $f(x) = fracsin x$ with $alpha >0$, find $limlimits_x to inftyf(x)$ and justify using limit definition at infinity.Show that $limlimits_n to inftysin n^2$ does not exist.Find $limlimits_n to infty sumlimits_k=1^inftyfrac1k^2sqrt[k]nsin^2left(fracn pikright)$How to solve the limit $limlimits_xto infty (x arctan x - fracxpi2)$Calculating $limlimits_ntoinftysumlimits_k=1^n arctan(frac1n+k)$
$begingroup$
I am required to find the limits of two "siblings" using the same idea, they are:
$$lim_ntoinftysinleft(frac1pn+rright)sum_k=1^nsinleft(frac2pk-2p+r2pnright)$$ with $0<r<2p$ and
$$lim_ntoinftyfrac1n^2sum_k=1^nk arctanleft(fracpk-p+1pnright)$$ where $p>1$.
For the first I tried $sum_k=1^nsin(frac2pk+r2pn-frac1n)=cos(frac1n)sum_k=1^nsin(frac2pk+r2pn)-sin(frac1n)sum_k=1^ncos(frac2pk+r2pn)$ but I don`t know how to find the limit using the ,,squeeze" theorem.
Well, for the second I stumble upon the same problem as to the first: $limlimits_ntoinftyfrac1nsumlimits_k=1^nfracknarctanleft(frackn+fracp-1pnright)$. Is this converging to $int_0^1xarctan(x)dx$ as $ntoinfty$? Is this the key ideea?
integration limits definite-integrals riemann-integration riemann-sum
asked yesterday
Jacob DeniculaJacob Denicula
454
New contributor
Jacob Denicula 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 |
$begingroup$
I am required to find the limits of two "siblings" using the same idea, they are:
$$lim_ntoinftysinleft(frac1pn+rright)sum_k=1^nsinleft(frac2pk-2p+r2pnright)$$ with $0<r<2p$ and
$$lim_ntoinftyfrac1n^2sum_k=1^nk arctanleft(fracpk-p+1pnright)$$ where $p>1$.
For the first I tried $sum_k=1^nsin(frac2pk+r2pn-frac1n)=cos(frac1n)sum_k=1^nsin(frac2pk+r2pn)-sin(frac1n)sum_k=1^ncos(frac2pk+r2pn)$ but I don`t know how to find the limit using the ,,squeeze" theorem.
Well, for the second I stumble upon the same problem as to the first: $limlimits_ntoinftyfrac1nsumlimits_k=1^nfracknarctanleft(frackn+fracp-1pnright)$. Is this converging to $int_0^1xarctan(x)dx$ as $ntoinfty$? Is this the key ideea?
integration limits definite-integrals riemann-integration riemann-sum
asked yesterday
Jacob DeniculaJacob Denicula
454
New contributor
Jacob Denicula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Try to use the same idea given for your previous question math.stackexchange.com/questions/3140464/…
$endgroup$
– Robert Z
yesterday
$begingroup$
I've tried but in this case I can`t use fractions. I obtained an upper bound for the second sum but no lower one.
$endgroup$
– Jacob Denicula
yesterday
add a comment |
$begingroup$
I am required to find the limits of two "siblings" using the same idea, they are:
$$lim_ntoinftysinleft(frac1pn+rright)sum_k=1^nsinleft(frac2pk-2p+r2pnright)$$ with $0<r<2p$ and
$$lim_ntoinftyfrac1n^2sum_k=1^nk arctanleft(fracpk-p+1pnright)$$ where $p>1$.
For the first I tried $sum_k=1^nsin(frac2pk+r2pn-frac1n)=cos(frac1n)sum_k=1^nsin(frac2pk+r2pn)-sin(frac1n)sum_k=1^ncos(frac2pk+r2pn)$ but I don`t know how to find the limit using the ,,squeeze" theorem.
Well, for the second I stumble upon the same problem as to the first: $limlimits_ntoinftyfrac1nsumlimits_k=1^nfracknarctanleft(frackn+fracp-1pnright)$. Is this converging to $int_0^1xarctan(x)dx$ as $ntoinfty$? Is this the key ideea?
integration limits definite-integrals riemann-integration riemann-sum
asked yesterday
Jacob DeniculaJacob Denicula
454
New contributor
Jacob Denicula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am required to find the limits of two "siblings" using the same idea, they are:
$$lim_ntoinftysinleft(frac1pn+rright)sum_k=1^nsinleft(frac2pk-2p+r2pnright)$$ with $0<r<2p$ and
$$lim_ntoinftyfrac1n^2sum_k=1^nk arctanleft(fracpk-p+1pnright)$$ where $p>1$.
For the first I tried $sum_k=1^nsin(frac2pk+r2pn-frac1n)=cos(frac1n)sum_k=1^nsin(frac2pk+r2pn)-sin(frac1n)sum_k=1^ncos(frac2pk+r2pn)$ but I don`t know how to find the limit using the ,,squeeze" theorem.
Well, for the second I stumble upon the same problem as to the first: $limlimits_ntoinftyfrac1nsumlimits_k=1^nfracknarctanleft(frackn+fracp-1pnright)$. Is this converging to $int_0^1xarctan(x)dx$ as $ntoinfty$? Is this the key ideea?
integration limits definite-integrals riemann-integration riemann-sum
integration limits definite-integrals riemann-integration riemann-sum
asked yesterday
Jacob DeniculaJacob Denicula
454
New contributor
Jacob Denicula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Jacob DeniculaJacob Denicula
454
New contributor
Jacob Denicula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 15 hours ago
Jacob Denicula
asked yesterday
Jacob DeniculaJacob Denicula
454
New contributor
Jacob Denicula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Jacob DeniculaJacob Denicula
454
asked yesterday
Jacob DeniculaJacob Denicula
454
454
New contributor
Jacob Denicula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jacob Denicula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jacob Denicula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Try to use the same idea given for your previous question math.stackexchange.com/questions/3140464/…
$endgroup$
– Robert Z
yesterday
$begingroup$
I've tried but in this case I can`t use fractions. I obtained an upper bound for the second sum but no lower one.
$endgroup$
– Jacob Denicula
yesterday
add a comment |
2
$begingroup$
Try to use the same idea given for your previous question math.stackexchange.com/questions/3140464/…
$endgroup$
– Robert Z
yesterday
$begingroup$
I've tried but in this case I can`t use fractions. I obtained an upper bound for the second sum but no lower one.
$endgroup$
– Jacob Denicula
yesterday
2
2
$begingroup$
Try to use the same idea given for your previous question math.stackexchange.com/questions/3140464/…
$endgroup$
– Robert Z
yesterday
$begingroup$
Try to use the same idea given for your previous question math.stackexchange.com/questions/3140464/…
$endgroup$
– Robert Z
yesterday
$begingroup$
I've tried but in this case I can`t use fractions. I obtained an upper bound for the second sum but no lower one.
$endgroup$
– Jacob Denicula
yesterday
$begingroup$
I've tried but in this case I can`t use fractions. I obtained an upper bound for the second sum but no lower one.
$endgroup$
– Jacob Denicula
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint. As regards the second sum, note that,
$$frac1nsum_k=1^nfracknarctanbig(frack-1nbig)leq frac1n^2sum_k=1^nkarctanbig(fracpk-p+1pnbig)leq frac1nsum_k=1^nfracknarctanbig(fracknbig)$$
The left side can be written as
$$frac1ncdotfrac1nsum_k=0^n-1arctanbig(fracknbig)+
frac1nsum_k=0^n-1fracknarctanbig(fracknbig).$$
Can you take it form here?
Use a similar approach for the first sum.
answered yesterday
Robert ZRobert Z
100k1069140
$endgroup$
$begingroup$
I obtained the result $int_0^1xarctgxdx$ for the second sum but for the first this approach does not work. My ideea is to use $sin(x-y)=sinxcosy-sinycosx$. So, $sum_k=1^nsin(frac2pk+r2pn-frac1n)=cos(frac1n)sum_k=1^nsin(frac2pk+r2pn)-sin(frac1n)sum_k=1^ncos(frac2pk+r2pn)$. How can I continue from here?
$endgroup$
– Jacob Denicula
18 hours ago
$begingroup$
@JacobDenicula Are you able to find $lim_ntoinftyfrac1nsum_k=1^nsinleft(frac2pk-2p+r2pnright)$? Note that $sin$ is increasing in $[0,1]$.
$endgroup$
– Robert Z
16 hours ago
$begingroup$
I am unsure I can apply Riemann integration because I remain with an $fracr2pn$. Does that reduce when $n$ goes to $infty$?
$endgroup$
– Jacob Denicula
16 hours ago
$begingroup$
@JacobDenicula Please edit your question with your attempt (detailed) for the first sum.
$endgroup$
– Robert Z
15 hours ago
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint. As regards the second sum, note that,
$$frac1nsum_k=1^nfracknarctanbig(frack-1nbig)leq frac1n^2sum_k=1^nkarctanbig(fracpk-p+1pnbig)leq frac1nsum_k=1^nfracknarctanbig(fracknbig)$$
The left side can be written as
$$frac1ncdotfrac1nsum_k=0^n-1arctanbig(fracknbig)+
frac1nsum_k=0^n-1fracknarctanbig(fracknbig).$$
Can you take it form here?
Use a similar approach for the first sum.
answered yesterday
Robert ZRobert Z
100k1069140
$endgroup$
$begingroup$
I obtained the result $int_0^1xarctgxdx$ for the second sum but for the first this approach does not work. My ideea is to use $sin(x-y)=sinxcosy-sinycosx$. So, $sum_k=1^nsin(frac2pk+r2pn-frac1n)=cos(frac1n)sum_k=1^nsin(frac2pk+r2pn)-sin(frac1n)sum_k=1^ncos(frac2pk+r2pn)$. How can I continue from here?
$endgroup$
– Jacob Denicula
18 hours ago
$begingroup$
@JacobDenicula Are you able to find $lim_ntoinftyfrac1nsum_k=1^nsinleft(frac2pk-2p+r2pnright)$? Note that $sin$ is increasing in $[0,1]$.
$endgroup$
– Robert Z
16 hours ago
$begingroup$
I am unsure I can apply Riemann integration because I remain with an $fracr2pn$. Does that reduce when $n$ goes to $infty$?
$endgroup$
– Jacob Denicula
16 hours ago
$begingroup$
@JacobDenicula Please edit your question with your attempt (detailed) for the first sum.
$endgroup$
– Robert Z
15 hours ago
add a comment |
$begingroup$
Hint. As regards the second sum, note that,
$$frac1nsum_k=1^nfracknarctanbig(frack-1nbig)leq frac1n^2sum_k=1^nkarctanbig(fracpk-p+1pnbig)leq frac1nsum_k=1^nfracknarctanbig(fracknbig)$$
The left side can be written as
$$frac1ncdotfrac1nsum_k=0^n-1arctanbig(fracknbig)+
frac1nsum_k=0^n-1fracknarctanbig(fracknbig).$$
Can you take it form here?
Use a similar approach for the first sum.
answered yesterday
Robert ZRobert Z
100k1069140
$endgroup$
$begingroup$
I obtained the result $int_0^1xarctgxdx$ for the second sum but for the first this approach does not work. My ideea is to use $sin(x-y)=sinxcosy-sinycosx$. So, $sum_k=1^nsin(frac2pk+r2pn-frac1n)=cos(frac1n)sum_k=1^nsin(frac2pk+r2pn)-sin(frac1n)sum_k=1^ncos(frac2pk+r2pn)$. How can I continue from here?
$endgroup$
– Jacob Denicula
18 hours ago
$begingroup$
@JacobDenicula Are you able to find $lim_ntoinftyfrac1nsum_k=1^nsinleft(frac2pk-2p+r2pnright)$? Note that $sin$ is increasing in $[0,1]$.
$endgroup$
– Robert Z
16 hours ago
$begingroup$
I am unsure I can apply Riemann integration because I remain with an $fracr2pn$. Does that reduce when $n$ goes to $infty$?
$endgroup$
– Jacob Denicula
16 hours ago
$begingroup$
@JacobDenicula Please edit your question with your attempt (detailed) for the first sum.
$endgroup$
– Robert Z
15 hours ago
add a comment |
$begingroup$
Hint. As regards the second sum, note that,
$$frac1nsum_k=1^nfracknarctanbig(frack-1nbig)leq frac1n^2sum_k=1^nkarctanbig(fracpk-p+1pnbig)leq frac1nsum_k=1^nfracknarctanbig(fracknbig)$$
The left side can be written as
$$frac1ncdotfrac1nsum_k=0^n-1arctanbig(fracknbig)+
frac1nsum_k=0^n-1fracknarctanbig(fracknbig).$$
Can you take it form here?
Use a similar approach for the first sum.
answered yesterday
Robert ZRobert Z
100k1069140
$endgroup$
Hint. As regards the second sum, note that,
$$frac1nsum_k=1^nfracknarctanbig(frack-1nbig)leq frac1n^2sum_k=1^nkarctanbig(fracpk-p+1pnbig)leq frac1nsum_k=1^nfracknarctanbig(fracknbig)$$
The left side can be written as
$$frac1ncdotfrac1nsum_k=0^n-1arctanbig(fracknbig)+
frac1nsum_k=0^n-1fracknarctanbig(fracknbig).$$
Can you take it form here?
Use a similar approach for the first sum.
answered yesterday
Robert ZRobert Z
100k1069140
answered yesterday
Robert ZRobert Z
100k1069140
answered yesterday
Robert ZRobert Z
100k1069140
answered yesterday
Robert ZRobert Z
100k1069140
100k1069140
$begingroup$
I obtained the result $int_0^1xarctgxdx$ for the second sum but for the first this approach does not work. My ideea is to use $sin(x-y)=sinxcosy-sinycosx$. So, $sum_k=1^nsin(frac2pk+r2pn-frac1n)=cos(frac1n)sum_k=1^nsin(frac2pk+r2pn)-sin(frac1n)sum_k=1^ncos(frac2pk+r2pn)$. How can I continue from here?
$endgroup$
– Jacob Denicula
18 hours ago
$begingroup$
@JacobDenicula Are you able to find $lim_ntoinftyfrac1nsum_k=1^nsinleft(frac2pk-2p+r2pnright)$? Note that $sin$ is increasing in $[0,1]$.
$endgroup$
– Robert Z
16 hours ago
$begingroup$
I am unsure I can apply Riemann integration because I remain with an $fracr2pn$. Does that reduce when $n$ goes to $infty$?
$endgroup$
– Jacob Denicula
16 hours ago
$begingroup$
@JacobDenicula Please edit your question with your attempt (detailed) for the first sum.
$endgroup$
– Robert Z
15 hours ago
add a comment |
$begingroup$
I obtained the result $int_0^1xarctgxdx$ for the second sum but for the first this approach does not work. My ideea is to use $sin(x-y)=sinxcosy-sinycosx$. So, $sum_k=1^nsin(frac2pk+r2pn-frac1n)=cos(frac1n)sum_k=1^nsin(frac2pk+r2pn)-sin(frac1n)sum_k=1^ncos(frac2pk+r2pn)$. How can I continue from here?
$endgroup$
– Jacob Denicula
18 hours ago
$begingroup$
@JacobDenicula Are you able to find $lim_ntoinftyfrac1nsum_k=1^nsinleft(frac2pk-2p+r2pnright)$? Note that $sin$ is increasing in $[0,1]$.
$endgroup$
– Robert Z
16 hours ago
$begingroup$
I am unsure I can apply Riemann integration because I remain with an $fracr2pn$. Does that reduce when $n$ goes to $infty$?
$endgroup$
– Jacob Denicula
16 hours ago
$begingroup$
@JacobDenicula Please edit your question with your attempt (detailed) for the first sum.
$endgroup$
– Robert Z
15 hours ago
$begingroup$
I obtained the result $int_0^1xarctgxdx$ for the second sum but for the first this approach does not work. My ideea is to use $sin(x-y)=sinxcosy-sinycosx$. So, $sum_k=1^nsin(frac2pk+r2pn-frac1n)=cos(frac1n)sum_k=1^nsin(frac2pk+r2pn)-sin(frac1n)sum_k=1^ncos(frac2pk+r2pn)$. How can I continue from here?
$endgroup$
– Jacob Denicula
18 hours ago
$begingroup$
I obtained the result $int_0^1xarctgxdx$ for the second sum but for the first this approach does not work. My ideea is to use $sin(x-y)=sinxcosy-sinycosx$. So, $sum_k=1^nsin(frac2pk+r2pn-frac1n)=cos(frac1n)sum_k=1^nsin(frac2pk+r2pn)-sin(frac1n)sum_k=1^ncos(frac2pk+r2pn)$. How can I continue from here?
$endgroup$
– Jacob Denicula
18 hours ago
$begingroup$
@JacobDenicula Are you able to find $lim_ntoinftyfrac1nsum_k=1^nsinleft(frac2pk-2p+r2pnright)$? Note that $sin$ is increasing in $[0,1]$.
$endgroup$
– Robert Z
16 hours ago
$begingroup$
@JacobDenicula Are you able to find $lim_ntoinftyfrac1nsum_k=1^nsinleft(frac2pk-2p+r2pnright)$? Note that $sin$ is increasing in $[0,1]$.
$endgroup$
– Robert Z
16 hours ago
$begingroup$
I am unsure I can apply Riemann integration because I remain with an $fracr2pn$. Does that reduce when $n$ goes to $infty$?
$endgroup$
– Jacob Denicula
16 hours ago
$begingroup$
I am unsure I can apply Riemann integration because I remain with an $fracr2pn$. Does that reduce when $n$ goes to $infty$?
$endgroup$
– Jacob Denicula
16 hours ago
$begingroup$
@JacobDenicula Please edit your question with your attempt (detailed) for the first sum.
$endgroup$
– Robert Z
15 hours ago
$begingroup$
@JacobDenicula Please edit your question with your attempt (detailed) for the first sum.
$endgroup$
– Robert Z
15 hours ago
add a comment |
Jacob Denicula is a new contributor. Be nice, and check out our Code of Conduct.
Jacob Denicula is a new contributor. Be nice, and check out our Code of Conduct.
Jacob Denicula is a new contributor. Be nice, and check out our Code of Conduct.
Jacob Denicula is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Try to use the same idea given for your previous question math.stackexchange.com/questions/3140464/…
$endgroup$
– Robert Z
yesterday
$begingroup$
I've tried but in this case I can`t use fractions. I obtained an upper bound for the second sum but no lower one.
$endgroup$
– Jacob Denicula
yesterday