Ways to compute $sumlimits_{i=1}^{sqrt n}sqrt i$ [duplicate]












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  • How closely can we estimate $sum_{i=0}^n sqrt{i}$

    7 answers




I have the following sum $sumlimits_{i=1}^{lceil sqrt n rceil}sqrt i$.



This sum is $leq n$, but what is a good bound and what is the method to compute this type of sums?






sequences-and-series summation







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marked as duplicate by Aryabhata, Cesareo, José Carlos Santos sequences-and-series
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 Feb 1 at 9:26


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 2




    $begingroup$
    There is no general closed formula for these sums. (But what is a sum up to $sqrt n$ when $sqrt n$ is not an integer?)
    $endgroup$
    – Did
    Jan 30 at 10:08










  • $begingroup$
    @Did Thank you Did. I put the ceiling function.
    $endgroup$
    – Dingo13
    Jan 30 at 10:10








  • 1




    $begingroup$
    "Good bounds": $$int_0^{lceilsqrt nrceil}sqrt xdxleqslant S_nleqslantint_1^{lceilsqrt nrceil+1}sqrt xdx$$ hence $$frac23lceilsqrt nrceilsqrt{lceilsqrt nrceil}leqslant S_nleqslantfrac23(lceilsqrt nrceil+1)left(sqrt{lceilsqrt nrceil+1}-1right)$$ which can be simplified into $$frac23n^{3/4}leqslant S_nleqslantfrac23(sqrt n+2)^{3/2}$$ in particular, $$S_n=frac23n^{3/4}+O(n^{1/4})$$
    $endgroup$
    – Did
    Jan 30 at 10:19












  • $begingroup$
    @Did Thank you very much for your detailed comment.
    $endgroup$
    – Dingo13
    Jan 30 at 10:21










  • $begingroup$
    @Did, please make that comment an answer.
    $endgroup$
    – lhf
    Jan 30 at 10:53
















0












$begingroup$



This question already has an answer here:




  • How closely can we estimate $sum_{i=0}^n sqrt{i}$

    7 answers




I have the following sum $sumlimits_{i=1}^{lceil sqrt n rceil}sqrt i$.



This sum is $leq n$, but what is a good bound and what is the method to compute this type of sums?






sequences-and-series summation







share|cite|improve this question











$endgroup$



marked as duplicate by Aryabhata, Cesareo, José Carlos Santos sequences-and-series
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 Feb 1 at 9:26


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 2




    $begingroup$
    There is no general closed formula for these sums. (But what is a sum up to $sqrt n$ when $sqrt n$ is not an integer?)
    $endgroup$
    – Did
    Jan 30 at 10:08










  • $begingroup$
    @Did Thank you Did. I put the ceiling function.
    $endgroup$
    – Dingo13
    Jan 30 at 10:10








  • 1




    $begingroup$
    "Good bounds": $$int_0^{lceilsqrt nrceil}sqrt xdxleqslant S_nleqslantint_1^{lceilsqrt nrceil+1}sqrt xdx$$ hence $$frac23lceilsqrt nrceilsqrt{lceilsqrt nrceil}leqslant S_nleqslantfrac23(lceilsqrt nrceil+1)left(sqrt{lceilsqrt nrceil+1}-1right)$$ which can be simplified into $$frac23n^{3/4}leqslant S_nleqslantfrac23(sqrt n+2)^{3/2}$$ in particular, $$S_n=frac23n^{3/4}+O(n^{1/4})$$
    $endgroup$
    – Did
    Jan 30 at 10:19












  • $begingroup$
    @Did Thank you very much for your detailed comment.
    $endgroup$
    – Dingo13
    Jan 30 at 10:21










  • $begingroup$
    @Did, please make that comment an answer.
    $endgroup$
    – lhf
    Jan 30 at 10:53














0












0








0





$begingroup$



This question already has an answer here:




  • How closely can we estimate $sum_{i=0}^n sqrt{i}$

    7 answers




I have the following sum $sumlimits_{i=1}^{lceil sqrt n rceil}sqrt i$.



This sum is $leq n$, but what is a good bound and what is the method to compute this type of sums?






sequences-and-series summation







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • How closely can we estimate $sum_{i=0}^n sqrt{i}$

    7 answers




I have the following sum $sumlimits_{i=1}^{lceil sqrt n rceil}sqrt i$.



This sum is $leq n$, but what is a good bound and what is the method to compute this type of sums?





This question already has an answer here:




  • How closely can we estimate $sum_{i=0}^n sqrt{i}$

    7 answers







sequences-and-series summation




sequences-and-series summation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 30 at 10:11







Dingo13

















asked Jan 30 at 10:05









Dingo13Dingo13

85113




85113




marked as duplicate by Aryabhata, Cesareo, José Carlos Santos sequences-and-series
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 Feb 1 at 9:26


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Aryabhata, Cesareo, José Carlos Santos sequences-and-series
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 Feb 1 at 9:26


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 2




    $begingroup$
    There is no general closed formula for these sums. (But what is a sum up to $sqrt n$ when $sqrt n$ is not an integer?)
    $endgroup$
    – Did
    Jan 30 at 10:08










  • $begingroup$
    @Did Thank you Did. I put the ceiling function.
    $endgroup$
    – Dingo13
    Jan 30 at 10:10








  • 1




    $begingroup$
    "Good bounds": $$int_0^{lceilsqrt nrceil}sqrt xdxleqslant S_nleqslantint_1^{lceilsqrt nrceil+1}sqrt xdx$$ hence $$frac23lceilsqrt nrceilsqrt{lceilsqrt nrceil}leqslant S_nleqslantfrac23(lceilsqrt nrceil+1)left(sqrt{lceilsqrt nrceil+1}-1right)$$ which can be simplified into $$frac23n^{3/4}leqslant S_nleqslantfrac23(sqrt n+2)^{3/2}$$ in particular, $$S_n=frac23n^{3/4}+O(n^{1/4})$$
    $endgroup$
    – Did
    Jan 30 at 10:19












  • $begingroup$
    @Did Thank you very much for your detailed comment.
    $endgroup$
    – Dingo13
    Jan 30 at 10:21










  • $begingroup$
    @Did, please make that comment an answer.
    $endgroup$
    – lhf
    Jan 30 at 10:53














  • 2




    $begingroup$
    There is no general closed formula for these sums. (But what is a sum up to $sqrt n$ when $sqrt n$ is not an integer?)
    $endgroup$
    – Did
    Jan 30 at 10:08










  • $begingroup$
    @Did Thank you Did. I put the ceiling function.
    $endgroup$
    – Dingo13
    Jan 30 at 10:10








  • 1




    $begingroup$
    "Good bounds": $$int_0^{lceilsqrt nrceil}sqrt xdxleqslant S_nleqslantint_1^{lceilsqrt nrceil+1}sqrt xdx$$ hence $$frac23lceilsqrt nrceilsqrt{lceilsqrt nrceil}leqslant S_nleqslantfrac23(lceilsqrt nrceil+1)left(sqrt{lceilsqrt nrceil+1}-1right)$$ which can be simplified into $$frac23n^{3/4}leqslant S_nleqslantfrac23(sqrt n+2)^{3/2}$$ in particular, $$S_n=frac23n^{3/4}+O(n^{1/4})$$
    $endgroup$
    – Did
    Jan 30 at 10:19












  • $begingroup$
    @Did Thank you very much for your detailed comment.
    $endgroup$
    – Dingo13
    Jan 30 at 10:21










  • $begingroup$
    @Did, please make that comment an answer.
    $endgroup$
    – lhf
    Jan 30 at 10:53








2




2




$begingroup$
There is no general closed formula for these sums. (But what is a sum up to $sqrt n$ when $sqrt n$ is not an integer?)
$endgroup$
– Did
Jan 30 at 10:08




$begingroup$
There is no general closed formula for these sums. (But what is a sum up to $sqrt n$ when $sqrt n$ is not an integer?)
$endgroup$
– Did
Jan 30 at 10:08












$begingroup$
@Did Thank you Did. I put the ceiling function.
$endgroup$
– Dingo13
Jan 30 at 10:10






$begingroup$
@Did Thank you Did. I put the ceiling function.
$endgroup$
– Dingo13
Jan 30 at 10:10






1




1




$begingroup$
"Good bounds": $$int_0^{lceilsqrt nrceil}sqrt xdxleqslant S_nleqslantint_1^{lceilsqrt nrceil+1}sqrt xdx$$ hence $$frac23lceilsqrt nrceilsqrt{lceilsqrt nrceil}leqslant S_nleqslantfrac23(lceilsqrt nrceil+1)left(sqrt{lceilsqrt nrceil+1}-1right)$$ which can be simplified into $$frac23n^{3/4}leqslant S_nleqslantfrac23(sqrt n+2)^{3/2}$$ in particular, $$S_n=frac23n^{3/4}+O(n^{1/4})$$
$endgroup$
– Did
Jan 30 at 10:19






$begingroup$
"Good bounds": $$int_0^{lceilsqrt nrceil}sqrt xdxleqslant S_nleqslantint_1^{lceilsqrt nrceil+1}sqrt xdx$$ hence $$frac23lceilsqrt nrceilsqrt{lceilsqrt nrceil}leqslant S_nleqslantfrac23(lceilsqrt nrceil+1)left(sqrt{lceilsqrt nrceil+1}-1right)$$ which can be simplified into $$frac23n^{3/4}leqslant S_nleqslantfrac23(sqrt n+2)^{3/2}$$ in particular, $$S_n=frac23n^{3/4}+O(n^{1/4})$$
$endgroup$
– Did
Jan 30 at 10:19














$begingroup$
@Did Thank you very much for your detailed comment.
$endgroup$
– Dingo13
Jan 30 at 10:21




$begingroup$
@Did Thank you very much for your detailed comment.
$endgroup$
– Dingo13
Jan 30 at 10:21












$begingroup$
@Did, please make that comment an answer.
$endgroup$
– lhf
Jan 30 at 10:53




$begingroup$
@Did, please make that comment an answer.
$endgroup$
– lhf
Jan 30 at 10:53










1 Answer
1






active

oldest

votes


















2












$begingroup$

Assuming that the upper limit for the sum is $m$, then
$$
sumlimits_{i=1}^{m}sqrt i le int_1^{m+1} ! ! sqrt{x} , dx = frac23 ((m + 1)^{3/2} - 1)
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you lhf. So a bound for the sum when $m=sqrt{n}$ is $c cdot n^{3/4}$ for a constant $c$?
    $endgroup$
    – Dingo13
    Jan 30 at 10:16




















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Assuming that the upper limit for the sum is $m$, then
$$
sumlimits_{i=1}^{m}sqrt i le int_1^{m+1} ! ! sqrt{x} , dx = frac23 ((m + 1)^{3/2} - 1)
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you lhf. So a bound for the sum when $m=sqrt{n}$ is $c cdot n^{3/4}$ for a constant $c$?
    $endgroup$
    – Dingo13
    Jan 30 at 10:16


















2












$begingroup$

Assuming that the upper limit for the sum is $m$, then
$$
sumlimits_{i=1}^{m}sqrt i le int_1^{m+1} ! ! sqrt{x} , dx = frac23 ((m + 1)^{3/2} - 1)
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you lhf. So a bound for the sum when $m=sqrt{n}$ is $c cdot n^{3/4}$ for a constant $c$?
    $endgroup$
    – Dingo13
    Jan 30 at 10:16
















2












2








2





$begingroup$

Assuming that the upper limit for the sum is $m$, then
$$
sumlimits_{i=1}^{m}sqrt i le int_1^{m+1} ! ! sqrt{x} , dx = frac23 ((m + 1)^{3/2} - 1)
$$






share|cite|improve this answer









$endgroup$



Assuming that the upper limit for the sum is $m$, then
$$
sumlimits_{i=1}^{m}sqrt i le int_1^{m+1} ! ! sqrt{x} , dx = frac23 ((m + 1)^{3/2} - 1)
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 30 at 10:10









lhflhf

167k11172404




167k11172404












  • $begingroup$
    Thank you lhf. So a bound for the sum when $m=sqrt{n}$ is $c cdot n^{3/4}$ for a constant $c$?
    $endgroup$
    – Dingo13
    Jan 30 at 10:16




















  • $begingroup$
    Thank you lhf. So a bound for the sum when $m=sqrt{n}$ is $c cdot n^{3/4}$ for a constant $c$?
    $endgroup$
    – Dingo13
    Jan 30 at 10:16


















$begingroup$
Thank you lhf. So a bound for the sum when $m=sqrt{n}$ is $c cdot n^{3/4}$ for a constant $c$?
$endgroup$
– Dingo13
Jan 30 at 10:16






$begingroup$
Thank you lhf. So a bound for the sum when $m=sqrt{n}$ is $c cdot n^{3/4}$ for a constant $c$?
$endgroup$
– Dingo13
Jan 30 at 10:16





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