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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$

1

$begingroup$

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

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edited yesterday

rtybase

11.3k21533

asked yesterday

Jacob DeniculaJacob Denicula

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New contributor

Jacob Denicula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

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.

$endgroup$

share|cite|improve this question

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.

share|cite|improve this question

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

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

1 Answer
1

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5

$begingroup$

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?

share|cite|improve this answer

edited yesterday

answered yesterday

Robert ZRobert Z

100k1069140

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, and it is $int_0^1frac11+xdx=ln2$, no?
  $endgroup$
  – Jacob Denicula
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JacobDenicula Yes, you are correct!
  $endgroup$
  – Robert Z
  20 hours ago
  

  
  
  
  

add a comment | 

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5

$begingroup$

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?

share|cite|improve this answer

edited yesterday

answered yesterday

Robert ZRobert Z

100k1069140

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, and it is $int_0^1frac11+xdx=ln2$, no?
  $endgroup$
  – Jacob Denicula
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JacobDenicula Yes, you are correct!
  $endgroup$
  – Robert Z
  20 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

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?

share|cite|improve this answer

edited yesterday

answered yesterday

Robert ZRobert Z

100k1069140

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, and it is $int_0^1frac11+xdx=ln2$, no?
  $endgroup$
  – Jacob Denicula
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JacobDenicula Yes, you are correct!
  $endgroup$
  – Robert Z
  20 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?

share|cite|improve this answer

edited yesterday

answered yesterday

Robert ZRobert Z

100k1069140

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

Robert ZRobert Z

100k1069140

answered yesterday

Robert ZRobert Z

100k1069140

answered yesterday

Robert ZRobert Z

100k1069140