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Is there a way to find a formulation of $sum_{n=1}^N sqrt{n}$ [closed]
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I'm searching a formulation of $displaystylesum_{n=1}^N sqrt{n}$ with a plinomial form or a form of serie. The formulation must be, with $Ninmathbb{N}$, like:
$$ sum_{n=1}^N n = frac{N(N+1)}{2}$$
$$ sum_{n=1}^N n^p = sum_{n=1}^p a_nN^n $$
where $a_n$ is a known sucession of numbers relationated with the Bernoulli numbers, for example. The formulation must not include $N$ in the superior or inferior part of the sum $sum$. Any ideas?
sequences-and-series summation
asked Jan 13 at 21:08
El boritoEl borito
674216
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closed as unclear what you're asking by Simply Beautiful Art, Did, Xander Henderson, RRL, Jack D'Aurizio Jan 14 at 15:48
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I'm searching a formulation of $displaystylesum_{n=1}^N sqrt{n}$ with a plinomial form or a form of serie. The formulation must be, with $Ninmathbb{N}$, like:
$$ sum_{n=1}^N n = frac{N(N+1)}{2}$$
$$ sum_{n=1}^N n^p = sum_{n=1}^p a_nN^n $$
where $a_n$ is a known sucession of numbers relationated with the Bernoulli numbers, for example. The formulation must not include $N$ in the superior or inferior part of the sum $sum$. Any ideas?
sequences-and-series summation
asked Jan 13 at 21:08
El boritoEl borito
674216
$endgroup$
closed as unclear what you're asking by Simply Beautiful Art, Did, Xander Henderson, RRL, Jack D'Aurizio Jan 14 at 15:48
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
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No, but your sum subtracted from $frac{3}{2} ; N^{3/2}$ approaches a constant limit, call it $C,$ so you get a fairly good estimate that way
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– Will Jagy
Jan 13 at 21:36
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Did you mean $frac{2}{3}$ $N^{3/2}$ ?
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– J. W. Tanner
Jan 13 at 23:07
$begingroup$
@J.W.Tanner yes.
$endgroup$
– Will Jagy
Jan 13 at 23:43
add a comment |
$begingroup$
I'm searching a formulation of $displaystylesum_{n=1}^N sqrt{n}$ with a plinomial form or a form of serie. The formulation must be, with $Ninmathbb{N}$, like:
$$ sum_{n=1}^N n = frac{N(N+1)}{2}$$
$$ sum_{n=1}^N n^p = sum_{n=1}^p a_nN^n $$
where $a_n$ is a known sucession of numbers relationated with the Bernoulli numbers, for example. The formulation must not include $N$ in the superior or inferior part of the sum $sum$. Any ideas?
sequences-and-series summation
asked Jan 13 at 21:08
El boritoEl borito
674216
$endgroup$
I'm searching a formulation of $displaystylesum_{n=1}^N sqrt{n}$ with a plinomial form or a form of serie. The formulation must be, with $Ninmathbb{N}$, like:
$$ sum_{n=1}^N n = frac{N(N+1)}{2}$$
$$ sum_{n=1}^N n^p = sum_{n=1}^p a_nN^n $$
where $a_n$ is a known sucession of numbers relationated with the Bernoulli numbers, for example. The formulation must not include $N$ in the superior or inferior part of the sum $sum$. Any ideas?
sequences-and-series summation
sequences-and-series summation
asked Jan 13 at 21:08
El boritoEl borito
674216
asked Jan 13 at 21:08
El boritoEl borito
674216
edited Feb 1 at 16:02
El borito
asked Jan 13 at 21:08
El boritoEl borito
674216
asked Jan 13 at 21:08
El boritoEl borito
674216
asked Jan 13 at 21:08
El boritoEl borito
674216
674216
closed as unclear what you're asking by Simply Beautiful Art, Did, Xander Henderson, RRL, Jack D'Aurizio Jan 14 at 15:48
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Simply Beautiful Art, Did, Xander Henderson, RRL, Jack D'Aurizio Jan 14 at 15:48
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
No, but your sum subtracted from $frac{3}{2} ; N^{3/2}$ approaches a constant limit, call it $C,$ so you get a fairly good estimate that way
$endgroup$
– Will Jagy
Jan 13 at 21:36
$begingroup$
Did you mean $frac{2}{3}$ $N^{3/2}$ ?
$endgroup$
– J. W. Tanner
Jan 13 at 23:07
$begingroup$
@J.W.Tanner yes.
$endgroup$
– Will Jagy
Jan 13 at 23:43
add a comment |
2
$begingroup$
No, but your sum subtracted from $frac{3}{2} ; N^{3/2}$ approaches a constant limit, call it $C,$ so you get a fairly good estimate that way
$endgroup$
– Will Jagy
Jan 13 at 21:36
$begingroup$
Did you mean $frac{2}{3}$ $N^{3/2}$ ?
$endgroup$
– J. W. Tanner
Jan 13 at 23:07
$begingroup$
@J.W.Tanner yes.
$endgroup$
– Will Jagy
Jan 13 at 23:43
2
2
$begingroup$
No, but your sum subtracted from $frac{3}{2} ; N^{3/2}$ approaches a constant limit, call it $C,$ so you get a fairly good estimate that way
$endgroup$
– Will Jagy
Jan 13 at 21:36
$begingroup$
No, but your sum subtracted from $frac{3}{2} ; N^{3/2}$ approaches a constant limit, call it $C,$ so you get a fairly good estimate that way
$endgroup$
– Will Jagy
Jan 13 at 21:36
$begingroup$
Did you mean $frac{2}{3}$ $N^{3/2}$ ?
$endgroup$
– J. W. Tanner
Jan 13 at 23:07
$begingroup$
Did you mean $frac{2}{3}$ $N^{3/2}$ ?
$endgroup$
– J. W. Tanner
Jan 13 at 23:07
$begingroup$
@J.W.Tanner yes.
$endgroup$
– Will Jagy
Jan 13 at 23:43
$begingroup$
@J.W.Tanner yes.
$endgroup$
– Will Jagy
Jan 13 at 23:43
add a comment |
3 Answers
3
active
oldest
votes
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There is no closed form for this summation, but for $N$ being large enough we can write$$sum_{n=1}^N sqrt{n}{=Nsqrt{N}cdot {1over N}sum_{n=1}^N sqrt{nover N}\approx Nsqrt N cdot int _{0}^1sqrt x{dx}\={2over 3}Nsqrt N}$$another way of expressing that is $$sum_{n=1}^N sqrt{n}=OBig(Nsqrt NBig)$$using Big O Notation.
answered Jan 13 at 22:39
Mostafa AyazMostafa Ayaz
15.6k3939
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add a comment |
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Another way is
$$
eqalign{
& sumlimits_{n = 1}^N {sqrt n } = sumlimits_{n = 1}^N {{1 over {n^{, - 1/2} }}} = H_{,N} ^{( - 1/2)} = cr
& = sumlimits_{n = 1}^infty {{1 over {n^{, - 1/2} }}} - sumlimits_{n = N + 1}^infty {{1 over {n^{, - 1/2} }}} = cr
& = zeta left( { - 1/2,;1} right) - zeta left( { - 1/2,;N + 1} right) cr}
$$
in terms of the Generalized Harmonic Number and Hurwitz zeta function
answered Jan 14 at 0:05
G CabG Cab
19.5k31238
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add a comment |
$begingroup$
Using the same approach as G Cab
$$ S_N=sumlimits_{n = 1}^N {sqrt n } = H_N^{left(-frac{1}{2}right)} $$ and, for large values of $N$, the asymptotics would be
$$S_N=frac{2 Nsqrt{N}}{3}+frac{sqrt{N}}{2}+zeta
left(-frac{1}{2}right)+frac1{24sqrt{N}}-frac1{1920 N^2sqrt{N}}+Oleft(frac{1}{N^{7/2}}right)$$
Trying for $N=100$, the "exact" value would be $671.46294710314775$ while the above expansion gives $671.46294710314765$
answered Jan 14 at 8:13
Claude LeiboviciClaude Leibovici
121k1157133
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is no closed form for this summation, but for $N$ being large enough we can write$$sum_{n=1}^N sqrt{n}{=Nsqrt{N}cdot {1over N}sum_{n=1}^N sqrt{nover N}\approx Nsqrt N cdot int _{0}^1sqrt x{dx}\={2over 3}Nsqrt N}$$another way of expressing that is $$sum_{n=1}^N sqrt{n}=OBig(Nsqrt NBig)$$using Big O Notation.
answered Jan 13 at 22:39
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
There is no closed form for this summation, but for $N$ being large enough we can write$$sum_{n=1}^N sqrt{n}{=Nsqrt{N}cdot {1over N}sum_{n=1}^N sqrt{nover N}\approx Nsqrt N cdot int _{0}^1sqrt x{dx}\={2over 3}Nsqrt N}$$another way of expressing that is $$sum_{n=1}^N sqrt{n}=OBig(Nsqrt NBig)$$using Big O Notation.
answered Jan 13 at 22:39
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
There is no closed form for this summation, but for $N$ being large enough we can write$$sum_{n=1}^N sqrt{n}{=Nsqrt{N}cdot {1over N}sum_{n=1}^N sqrt{nover N}\approx Nsqrt N cdot int _{0}^1sqrt x{dx}\={2over 3}Nsqrt N}$$another way of expressing that is $$sum_{n=1}^N sqrt{n}=OBig(Nsqrt NBig)$$using Big O Notation.
answered Jan 13 at 22:39
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
There is no closed form for this summation, but for $N$ being large enough we can write$$sum_{n=1}^N sqrt{n}{=Nsqrt{N}cdot {1over N}sum_{n=1}^N sqrt{nover N}\approx Nsqrt N cdot int _{0}^1sqrt x{dx}\={2over 3}Nsqrt N}$$another way of expressing that is $$sum_{n=1}^N sqrt{n}=OBig(Nsqrt NBig)$$using Big O Notation.
answered Jan 13 at 22:39
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 13 at 22:39
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 13 at 22:39
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 13 at 22:39
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
add a comment |
add a comment |
$begingroup$
Another way is
$$
eqalign{
& sumlimits_{n = 1}^N {sqrt n } = sumlimits_{n = 1}^N {{1 over {n^{, - 1/2} }}} = H_{,N} ^{( - 1/2)} = cr
& = sumlimits_{n = 1}^infty {{1 over {n^{, - 1/2} }}} - sumlimits_{n = N + 1}^infty {{1 over {n^{, - 1/2} }}} = cr
& = zeta left( { - 1/2,;1} right) - zeta left( { - 1/2,;N + 1} right) cr}
$$
in terms of the Generalized Harmonic Number and Hurwitz zeta function
answered Jan 14 at 0:05
G CabG Cab
19.5k31238
$endgroup$
add a comment |
$begingroup$
Another way is
$$
eqalign{
& sumlimits_{n = 1}^N {sqrt n } = sumlimits_{n = 1}^N {{1 over {n^{, - 1/2} }}} = H_{,N} ^{( - 1/2)} = cr
& = sumlimits_{n = 1}^infty {{1 over {n^{, - 1/2} }}} - sumlimits_{n = N + 1}^infty {{1 over {n^{, - 1/2} }}} = cr
& = zeta left( { - 1/2,;1} right) - zeta left( { - 1/2,;N + 1} right) cr}
$$
in terms of the Generalized Harmonic Number and Hurwitz zeta function
answered Jan 14 at 0:05
G CabG Cab
19.5k31238
$endgroup$
add a comment |
$begingroup$
Another way is
$$
eqalign{
& sumlimits_{n = 1}^N {sqrt n } = sumlimits_{n = 1}^N {{1 over {n^{, - 1/2} }}} = H_{,N} ^{( - 1/2)} = cr
& = sumlimits_{n = 1}^infty {{1 over {n^{, - 1/2} }}} - sumlimits_{n = N + 1}^infty {{1 over {n^{, - 1/2} }}} = cr
& = zeta left( { - 1/2,;1} right) - zeta left( { - 1/2,;N + 1} right) cr}
$$
in terms of the Generalized Harmonic Number and Hurwitz zeta function
answered Jan 14 at 0:05
G CabG Cab
19.5k31238
$endgroup$
Another way is
$$
eqalign{
& sumlimits_{n = 1}^N {sqrt n } = sumlimits_{n = 1}^N {{1 over {n^{, - 1/2} }}} = H_{,N} ^{( - 1/2)} = cr
& = sumlimits_{n = 1}^infty {{1 over {n^{, - 1/2} }}} - sumlimits_{n = N + 1}^infty {{1 over {n^{, - 1/2} }}} = cr
& = zeta left( { - 1/2,;1} right) - zeta left( { - 1/2,;N + 1} right) cr}
$$
in terms of the Generalized Harmonic Number and Hurwitz zeta function
answered Jan 14 at 0:05
G CabG Cab
19.5k31238
answered Jan 14 at 0:05
G CabG Cab
19.5k31238
answered Jan 14 at 0:05
G CabG Cab
19.5k31238
answered Jan 14 at 0:05
G CabG Cab
19.5k31238
19.5k31238
add a comment |
add a comment |
$begingroup$
Using the same approach as G Cab
$$ S_N=sumlimits_{n = 1}^N {sqrt n } = H_N^{left(-frac{1}{2}right)} $$ and, for large values of $N$, the asymptotics would be
$$S_N=frac{2 Nsqrt{N}}{3}+frac{sqrt{N}}{2}+zeta
left(-frac{1}{2}right)+frac1{24sqrt{N}}-frac1{1920 N^2sqrt{N}}+Oleft(frac{1}{N^{7/2}}right)$$
Trying for $N=100$, the "exact" value would be $671.46294710314775$ while the above expansion gives $671.46294710314765$
answered Jan 14 at 8:13
Claude LeiboviciClaude Leibovici
121k1157133
$endgroup$
add a comment |
$begingroup$
Using the same approach as G Cab
$$ S_N=sumlimits_{n = 1}^N {sqrt n } = H_N^{left(-frac{1}{2}right)} $$ and, for large values of $N$, the asymptotics would be
$$S_N=frac{2 Nsqrt{N}}{3}+frac{sqrt{N}}{2}+zeta
left(-frac{1}{2}right)+frac1{24sqrt{N}}-frac1{1920 N^2sqrt{N}}+Oleft(frac{1}{N^{7/2}}right)$$
Trying for $N=100$, the "exact" value would be $671.46294710314775$ while the above expansion gives $671.46294710314765$
answered Jan 14 at 8:13
Claude LeiboviciClaude Leibovici
121k1157133
$endgroup$
add a comment |
$begingroup$
Using the same approach as G Cab
$$ S_N=sumlimits_{n = 1}^N {sqrt n } = H_N^{left(-frac{1}{2}right)} $$ and, for large values of $N$, the asymptotics would be
$$S_N=frac{2 Nsqrt{N}}{3}+frac{sqrt{N}}{2}+zeta
left(-frac{1}{2}right)+frac1{24sqrt{N}}-frac1{1920 N^2sqrt{N}}+Oleft(frac{1}{N^{7/2}}right)$$
Trying for $N=100$, the "exact" value would be $671.46294710314775$ while the above expansion gives $671.46294710314765$
answered Jan 14 at 8:13
Claude LeiboviciClaude Leibovici
121k1157133
$endgroup$
Using the same approach as G Cab
$$ S_N=sumlimits_{n = 1}^N {sqrt n } = H_N^{left(-frac{1}{2}right)} $$ and, for large values of $N$, the asymptotics would be
$$S_N=frac{2 Nsqrt{N}}{3}+frac{sqrt{N}}{2}+zeta
left(-frac{1}{2}right)+frac1{24sqrt{N}}-frac1{1920 N^2sqrt{N}}+Oleft(frac{1}{N^{7/2}}right)$$
Trying for $N=100$, the "exact" value would be $671.46294710314775$ while the above expansion gives $671.46294710314765$
answered Jan 14 at 8:13
Claude LeiboviciClaude Leibovici
121k1157133
answered Jan 14 at 8:13
Claude LeiboviciClaude Leibovici
121k1157133
answered Jan 14 at 8:13
Claude LeiboviciClaude Leibovici
121k1157133
answered Jan 14 at 8:13
Claude LeiboviciClaude Leibovici
121k1157133
121k1157133
add a comment |
add a comment |
2
$begingroup$
No, but your sum subtracted from $frac{3}{2} ; N^{3/2}$ approaches a constant limit, call it $C,$ so you get a fairly good estimate that way
$endgroup$
– Will Jagy
Jan 13 at 21:36
$begingroup$
Did you mean $frac{2}{3}$ $N^{3/2}$ ?
$endgroup$
– J. W. Tanner
Jan 13 at 23:07
$begingroup$
@J.W.Tanner yes.
$endgroup$
– Will Jagy
Jan 13 at 23:43