Find $limlimits_nrightarrowinftysumlimits_k=1^nfrac1n+sqrt(k^2-k+1)$Find the limit of $limlimits_ntoinftyfrac1n^2sumlimits_k=1^nkarctanbig(fracpk-p+1pnbig)$Show that: $lim limits_n rightarrow+infty int_0^1f(x^n)dx=f(0)$Evaluate $limlimits_xtoinftyfrac1sqrtxint_1^xln(1+frac1sqrtt)dt$prove that $lim limits_n rightarrow infty n sum limits_j=1^n fraccos(fracnj) f(fracnj)j^2$ exists and final.Find $ limlimits_n rightarrow infty int_0^1 left(1+ fracxnright)^n dx$How to find: $ limlimits_nrightarrowinfty leftlfloor frac-1nrightrfloor $Find $limlimits_nto +inftybig(frac1sqrtn^2 + 1 + frac1sqrtn^2 + 2 + cdots + frac1sqrtn^2 + nbig)$?Calculate $limlimits_ntoinftysumlimits_0leqslant kleqslant2nfrac kk+n^2$ using Riemann sumsHow would you calculate this limit? $limlimits_n rightarrowinftyfracpi2nsumlimits_k=1^ncosleft(fracpi2nkright)$$limlimits_n rightarrow +infty fracsumlimits_k=1^n sqrt[k] k n= 1$Find$ limlimits_nrightarrowinftyfracx_nn$
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Find $limlimits_nrightarrowinftysumlimits_k=1^nfrac1n+sqrt(k^2-k+1)$
Find the limit of $limlimits_ntoinftyfrac1n^2sumlimits_k=1^nkarctanbig(fracpk-p+1pnbig)$Show that: $lim limits_n rightarrow+infty int_0^1f(x^n)dx=f(0)$Evaluate $limlimits_xtoinftyfrac1sqrtxint_1^xln(1+frac1sqrtt)dt$prove that $lim limits_n rightarrow infty n sum limits_j=1^n fraccos(fracnj) f(fracnj)j^2$ exists and final.Find $ limlimits_n rightarrow infty int_0^1 left(1+ fracxnright)^n dx$How to find: $ limlimits_nrightarrowinfty leftlfloor frac-1nrightrfloor $Find $limlimits_nto +inftybig(frac1sqrtn^2 + 1 + frac1sqrtn^2 + 2 + cdots + frac1sqrtn^2 + nbig)$?Calculate $limlimits_ntoinftysumlimits_0leqslant kleqslant2nfrac kk+n^2$ using Riemann sumsHow would you calculate this limit? $limlimits_n rightarrowinftyfracpi2nsumlimits_k=1^ncosleft(fracpi2nkright)$$limlimits_n rightarrow +infty fracsumlimits_k=1^n sqrt[k] k n= 1$Find$ limlimits_nrightarrowinftyfracx_nn$
$begingroup$
Find $limlimits_nrightarrowinftysumlimits_k=1^nfrac1n+sqrt(k^2-k+1)$.
I observed it is a Riemann integral and can be written as $frac1nsumlimits_k=1^nfrac11+sqrtleft(left(fracknright)^2-frackn^2+frac1n^2right)$, and for $x_i=frackn$ this is a Riemann sum. I have problems with passing to the limit as I obtain $int_0^1frac11+sqrtx^2-fracxn+frac1n^2dx$. can I apply this limit for the integral as to reduce the $frac1n$ as n converges to $infty$?
integration limits definite-integrals
edited yesterday
rtybase
11.3k21533
asked yesterday
Jacob DeniculaJacob Denicula
384
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 |
$begingroup$
Find $limlimits_nrightarrowinftysumlimits_k=1^nfrac1n+sqrt(k^2-k+1)$.
I observed it is a Riemann integral and can be written as $frac1nsumlimits_k=1^nfrac11+sqrtleft(left(fracknright)^2-frackn^2+frac1n^2right)$, and for $x_i=frackn$ this is a Riemann sum. I have problems with passing to the limit as I obtain $int_0^1frac11+sqrtx^2-fracxn+frac1n^2dx$. can I apply this limit for the integral as to reduce the $frac1n$ as n converges to $infty$?
integration limits definite-integrals
edited yesterday
rtybase
11.3k21533
asked yesterday
Jacob DeniculaJacob Denicula
384
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.
$endgroup$
$begingroup$
This integral does have a solution but it's a rather long one. Are you su're you aren't missing anything ?
$endgroup$
– Rebellos
yesterday
add a comment |
$begingroup$
Find $limlimits_nrightarrowinftysumlimits_k=1^nfrac1n+sqrt(k^2-k+1)$.
I observed it is a Riemann integral and can be written as $frac1nsumlimits_k=1^nfrac11+sqrtleft(left(fracknright)^2-frackn^2+frac1n^2right)$, and for $x_i=frackn$ this is a Riemann sum. I have problems with passing to the limit as I obtain $int_0^1frac11+sqrtx^2-fracxn+frac1n^2dx$. can I apply this limit for the integral as to reduce the $frac1n$ as n converges to $infty$?
integration limits definite-integrals
edited yesterday
rtybase
11.3k21533
asked yesterday
Jacob DeniculaJacob Denicula
384
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.
$endgroup$
Find $limlimits_nrightarrowinftysumlimits_k=1^nfrac1n+sqrt(k^2-k+1)$.
I observed it is a Riemann integral and can be written as $frac1nsumlimits_k=1^nfrac11+sqrtleft(left(fracknright)^2-frackn^2+frac1n^2right)$, and for $x_i=frackn$ this is a Riemann sum. I have problems with passing to the limit as I obtain $int_0^1frac11+sqrtx^2-fracxn+frac1n^2dx$. can I apply this limit for the integral as to reduce the $frac1n$ as n converges to $infty$?
integration limits definite-integrals
integration limits definite-integrals
edited yesterday
rtybase
11.3k21533
asked yesterday
Jacob DeniculaJacob Denicula
384
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 yesterday
rtybase
11.3k21533
asked yesterday
Jacob DeniculaJacob Denicula
384
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 yesterday
rtybase
11.3k21533
edited yesterday
rtybase
11.3k21533
edited yesterday
rtybase
11.3k21533
11.3k21533
asked yesterday
Jacob DeniculaJacob Denicula
384
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
384
asked yesterday
Jacob DeniculaJacob Denicula
384
384
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.
$begingroup$
This integral does have a solution but it's a rather long one. Are you su're you aren't missing anything ?
$endgroup$
– Rebellos
yesterday
add a comment |
$begingroup$
This integral does have a solution but it's a rather long one. Are you su're you aren't missing anything ?
$endgroup$
– Rebellos
yesterday
$begingroup$
This integral does have a solution but it's a rather long one. Are you su're you aren't missing anything ?
$endgroup$
– Rebellos
yesterday
$begingroup$
This integral does have a solution but it's a rather long one. Are you su're you aren't missing anything ?
$endgroup$
– Rebellos
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you evaluate a limit as $n$ goes to infinity, then the result should not depend on $n$.
Instead, note that
$$frac1nsum_k=1^nfrac11+fracknleq frac1nsum_k=1^nfrac11+sqrtfrack^2n^2-frack-1n^2leq frac1nsum_k=1^nfrac11+sqrtfrac(k-1)^2n^2=frac1nsum_k=0^n-1frac11+frackn.$$
Now use the Riemann sum approach for the left-side and the right-side.
Can you take it from here?
answered yesterday
Robert ZRobert Z
100k1069140
$endgroup$
$begingroup$
Yes, and it is $int_0^1frac11+xdx=ln2$, no?
$endgroup$
– Jacob Denicula
18 hours ago
$begingroup$
@JacobDenicula Yes, you are correct!
$endgroup$
– Robert Z
18 hours ago
add a comment |
Your Answer
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1 Answer
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oldest
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1 Answer
1
active
oldest
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active
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votes
$begingroup$
If you evaluate a limit as $n$ goes to infinity, then the result should not depend on $n$.
Instead, note that
$$frac1nsum_k=1^nfrac11+fracknleq frac1nsum_k=1^nfrac11+sqrtfrack^2n^2-frack-1n^2leq frac1nsum_k=1^nfrac11+sqrtfrac(k-1)^2n^2=frac1nsum_k=0^n-1frac11+frackn.$$
Now use the Riemann sum approach for the left-side and the right-side.
Can you take it from here?
answered yesterday
Robert ZRobert Z
100k1069140
$endgroup$
$begingroup$
Yes, and it is $int_0^1frac11+xdx=ln2$, no?
$endgroup$
– Jacob Denicula
18 hours ago
$begingroup$
@JacobDenicula Yes, you are correct!
$endgroup$
– Robert Z
18 hours ago
add a comment |
$begingroup$
If you evaluate a limit as $n$ goes to infinity, then the result should not depend on $n$.
Instead, note that
$$frac1nsum_k=1^nfrac11+fracknleq frac1nsum_k=1^nfrac11+sqrtfrack^2n^2-frack-1n^2leq frac1nsum_k=1^nfrac11+sqrtfrac(k-1)^2n^2=frac1nsum_k=0^n-1frac11+frackn.$$
Now use the Riemann sum approach for the left-side and the right-side.
Can you take it from here?
answered yesterday
Robert ZRobert Z
100k1069140
$endgroup$
$begingroup$
Yes, and it is $int_0^1frac11+xdx=ln2$, no?
$endgroup$
– Jacob Denicula
18 hours ago
$begingroup$
@JacobDenicula Yes, you are correct!
$endgroup$
– Robert Z
18 hours ago
add a comment |
$begingroup$
If you evaluate a limit as $n$ goes to infinity, then the result should not depend on $n$.
Instead, note that
$$frac1nsum_k=1^nfrac11+fracknleq frac1nsum_k=1^nfrac11+sqrtfrack^2n^2-frack-1n^2leq frac1nsum_k=1^nfrac11+sqrtfrac(k-1)^2n^2=frac1nsum_k=0^n-1frac11+frackn.$$
Now use the Riemann sum approach for the left-side and the right-side.
Can you take it from here?
answered yesterday
Robert ZRobert Z
100k1069140
$endgroup$
If you evaluate a limit as $n$ goes to infinity, then the result should not depend on $n$.
Instead, note that
$$frac1nsum_k=1^nfrac11+fracknleq frac1nsum_k=1^nfrac11+sqrtfrack^2n^2-frack-1n^2leq frac1nsum_k=1^nfrac11+sqrtfrac(k-1)^2n^2=frac1nsum_k=0^n-1frac11+frackn.$$
Now use the Riemann sum approach for the left-side and the right-side.
Can you take it from here?
answered yesterday
Robert ZRobert Z
100k1069140
edited yesterday
answered yesterday
Robert ZRobert Z
100k1069140
answered yesterday
Robert ZRobert Z
100k1069140
answered yesterday
Robert ZRobert Z
100k1069140
100k1069140
$begingroup$
Yes, and it is $int_0^1frac11+xdx=ln2$, no?
$endgroup$
– Jacob Denicula
18 hours ago
$begingroup$
@JacobDenicula Yes, you are correct!
$endgroup$
– Robert Z
18 hours ago
add a comment |
$begingroup$
Yes, and it is $int_0^1frac11+xdx=ln2$, no?
$endgroup$
– Jacob Denicula
18 hours ago
$begingroup$
@JacobDenicula Yes, you are correct!
$endgroup$
– Robert Z
18 hours ago
$begingroup$
Yes, and it is $int_0^1frac11+xdx=ln2$, no?
$endgroup$
– Jacob Denicula
18 hours ago
$begingroup$
Yes, and it is $int_0^1frac11+xdx=ln2$, no?
$endgroup$
– Jacob Denicula
18 hours ago
$begingroup$
@JacobDenicula Yes, you are correct!
$endgroup$
– Robert Z
18 hours ago
$begingroup$
@JacobDenicula Yes, you are correct!
$endgroup$
– Robert Z
18 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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$begingroup$
This integral does have a solution but it's a rather long one. Are you su're you aren't missing anything ?
$endgroup$
– Rebellos
yesterday