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Ways to compute $sumlimits_{i=1}^{sqrt n}sqrt i$ [duplicate]
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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
asked Jan 30 at 10:05
Dingo13Dingo13
85113
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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.
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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
asked Jan 30 at 10:05
Dingo13Dingo13
85113
$endgroup$
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
add a comment |
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
asked Jan 30 at 10:05
Dingo13Dingo13
85113
$endgroup$
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
asked Jan 30 at 10:05
Dingo13Dingo13
85113
asked Jan 30 at 10:05
Dingo13Dingo13
85113
edited Jan 30 at 10:11
Dingo13
asked Jan 30 at 10:05
Dingo13Dingo13
85113
asked Jan 30 at 10:05
Dingo13Dingo13
85113
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
Assuming that the upper limit for the sum is $m$, then
$$
sumlimits_{i=1}^{m}sqrt i le int_1^{m+1} ! ! sqrt{x} , dx = frac23 ((m + 1)^{3/2} - 1)
$$
answered Jan 30 at 10:10
lhflhf
167k11172404
$endgroup$
$begingroup$
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
add a comment |
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
$$
sumlimits_{i=1}^{m}sqrt i le int_1^{m+1} ! ! sqrt{x} , dx = frac23 ((m + 1)^{3/2} - 1)
$$
answered Jan 30 at 10:10
lhflhf
167k11172404
$endgroup$
$begingroup$
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
add a comment |
2
$begingroup$
Assuming that the upper limit for the sum is $m$, then
$$
sumlimits_{i=1}^{m}sqrt i le int_1^{m+1} ! ! sqrt{x} , dx = frac23 ((m + 1)^{3/2} - 1)
$$
answered Jan 30 at 10:10
lhflhf
167k11172404
$endgroup$
$begingroup$
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
add a comment |
2
2
2
$begingroup$
Assuming that the upper limit for the sum is $m$, then
$$
sumlimits_{i=1}^{m}sqrt i le int_1^{m+1} ! ! sqrt{x} , dx = frac23 ((m + 1)^{3/2} - 1)
$$
answered Jan 30 at 10:10
lhflhf
167k11172404
$endgroup$
Assuming that the upper limit for the sum is $m$, then
$$
sumlimits_{i=1}^{m}sqrt i le int_1^{m+1} ! ! sqrt{x} , dx = frac23 ((m + 1)^{3/2} - 1)
$$
answered Jan 30 at 10:10
lhflhf
167k11172404
answered Jan 30 at 10:10
lhflhf
167k11172404
answered Jan 30 at 10:10
lhflhf
167k11172404
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
add a comment |
$begingroup$
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
add a comment |
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