Mixing
Is it possible a closed expression for $1^k + 2^k + … + n^k$? If so, am I on the way?
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I was asked to calculate the sequence $1^k + 2^k + ... + n^k$ and if it is possible to find a closed expression for it for every $k in mathbb{N}$.
I could find that I can rewrite $sum_{i=1}^ni^k$ into $1^k + sum_{i=1}^{n-1} (1+i)^k$. Also it seemed to me that I could without prejudice use the superior summation limit as $n$, so that I would have a sum of binomial expressions (using $(n-1)$ the last term would be $n^k$ and for that I was unsure if I could use the binomial theorem within the sum).
Rewriting $1^k + sum_{i=1}^{n} (1+i)^k$ again (using the mentioned theorem) I got:
$1 + sum_{i=1}^{n} sum_{j=0}^k {k choose j}i^j$, but now I'm stuck. Any tips? Am I on a reasonable course?
summation induction binomial-coefficients binomial-theorem
asked Aug 6 '16 at 8:16
izzortsizzorts
13618
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add a comment |
$begingroup$
I was asked to calculate the sequence $1^k + 2^k + ... + n^k$ and if it is possible to find a closed expression for it for every $k in mathbb{N}$.
I could find that I can rewrite $sum_{i=1}^ni^k$ into $1^k + sum_{i=1}^{n-1} (1+i)^k$. Also it seemed to me that I could without prejudice use the superior summation limit as $n$, so that I would have a sum of binomial expressions (using $(n-1)$ the last term would be $n^k$ and for that I was unsure if I could use the binomial theorem within the sum).
Rewriting $1^k + sum_{i=1}^{n} (1+i)^k$ again (using the mentioned theorem) I got:
$1 + sum_{i=1}^{n} sum_{j=0}^k {k choose j}i^j$, but now I'm stuck. Any tips? Am I on a reasonable course?
summation induction binomial-coefficients binomial-theorem
asked Aug 6 '16 at 8:16
izzortsizzorts
13618
$endgroup$
1
$begingroup$
take a look at this en.wikipedia.org/wiki/Faulhaber's_formula
$endgroup$
– msm
Aug 6 '16 at 8:19
4
$begingroup$
And search the site. This has been asked so many times on our site that I have lost count.
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– Jyrki Lahtonen
Aug 6 '16 at 8:23
$begingroup$
math.stackexchange.com/questions/1878810/…
$endgroup$
– Ahmed S. Attaalla
Aug 6 '16 at 12:59
add a comment |
$begingroup$
I was asked to calculate the sequence $1^k + 2^k + ... + n^k$ and if it is possible to find a closed expression for it for every $k in mathbb{N}$.
I could find that I can rewrite $sum_{i=1}^ni^k$ into $1^k + sum_{i=1}^{n-1} (1+i)^k$. Also it seemed to me that I could without prejudice use the superior summation limit as $n$, so that I would have a sum of binomial expressions (using $(n-1)$ the last term would be $n^k$ and for that I was unsure if I could use the binomial theorem within the sum).
Rewriting $1^k + sum_{i=1}^{n} (1+i)^k$ again (using the mentioned theorem) I got:
$1 + sum_{i=1}^{n} sum_{j=0}^k {k choose j}i^j$, but now I'm stuck. Any tips? Am I on a reasonable course?
summation induction binomial-coefficients binomial-theorem
asked Aug 6 '16 at 8:16
izzortsizzorts
13618
$endgroup$
I was asked to calculate the sequence $1^k + 2^k + ... + n^k$ and if it is possible to find a closed expression for it for every $k in mathbb{N}$.
I could find that I can rewrite $sum_{i=1}^ni^k$ into $1^k + sum_{i=1}^{n-1} (1+i)^k$. Also it seemed to me that I could without prejudice use the superior summation limit as $n$, so that I would have a sum of binomial expressions (using $(n-1)$ the last term would be $n^k$ and for that I was unsure if I could use the binomial theorem within the sum).
Rewriting $1^k + sum_{i=1}^{n} (1+i)^k$ again (using the mentioned theorem) I got:
$1 + sum_{i=1}^{n} sum_{j=0}^k {k choose j}i^j$, but now I'm stuck. Any tips? Am I on a reasonable course?
summation induction binomial-coefficients binomial-theorem
summation induction binomial-coefficients binomial-theorem
asked Aug 6 '16 at 8:16
izzortsizzorts
13618
asked Aug 6 '16 at 8:16
izzortsizzorts
13618
asked Aug 6 '16 at 8:16
izzortsizzorts
13618
asked Aug 6 '16 at 8:16
izzortsizzorts
13618
asked Aug 6 '16 at 8:16
izzortsizzorts
13618
13618
1
$begingroup$
take a look at this en.wikipedia.org/wiki/Faulhaber's_formula
$endgroup$
– msm
Aug 6 '16 at 8:19
4
$begingroup$
And search the site. This has been asked so many times on our site that I have lost count.
$endgroup$
– Jyrki Lahtonen
Aug 6 '16 at 8:23
$begingroup$
math.stackexchange.com/questions/1878810/…
$endgroup$
– Ahmed S. Attaalla
Aug 6 '16 at 12:59
add a comment |
1
$begingroup$
take a look at this en.wikipedia.org/wiki/Faulhaber's_formula
$endgroup$
– msm
Aug 6 '16 at 8:19
4
$begingroup$
And search the site. This has been asked so many times on our site that I have lost count.
$endgroup$
– Jyrki Lahtonen
Aug 6 '16 at 8:23
$begingroup$
math.stackexchange.com/questions/1878810/…
$endgroup$
– Ahmed S. Attaalla
Aug 6 '16 at 12:59
1
1
$begingroup$
take a look at this en.wikipedia.org/wiki/Faulhaber's_formula
$endgroup$
– msm
Aug 6 '16 at 8:19
$begingroup$
take a look at this en.wikipedia.org/wiki/Faulhaber's_formula
$endgroup$
– msm
Aug 6 '16 at 8:19
4
4
$begingroup$
And search the site. This has been asked so many times on our site that I have lost count.
$endgroup$
– Jyrki Lahtonen
Aug 6 '16 at 8:23
$begingroup$
And search the site. This has been asked so many times on our site that I have lost count.
$endgroup$
– Jyrki Lahtonen
Aug 6 '16 at 8:23
$begingroup$
math.stackexchange.com/questions/1878810/…
$endgroup$
– Ahmed S. Attaalla
Aug 6 '16 at 12:59
$begingroup$
math.stackexchange.com/questions/1878810/…
$endgroup$
– Ahmed S. Attaalla
Aug 6 '16 at 12:59
add a comment |
1 Answer
1
active
oldest
votes
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Hint: You could find in this answer formulas for the sum of $k$-th powers of the first $n$ natural numbers.
begin{align*}
sum_{j=1}^nj^k&=[z^n]frac{1}{1-z}(zD_z)^kfrac{1}{1-z}\
&=sum_{j=1}^k{kbrace j}frac{(n+1)^{underline{j+1}}}{j+1}
end{align*}
with $D_z:=frac{d}{dz}$ the differential operator and ${kbrace j}$ Stirling numbers of the second kind.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 6 '16 at 14:45
Markus ScheuerMarkus Scheuer
61.2k456145
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Hint: You could find in this answer formulas for the sum of $k$-th powers of the first $n$ natural numbers.
begin{align*}
sum_{j=1}^nj^k&=[z^n]frac{1}{1-z}(zD_z)^kfrac{1}{1-z}\
&=sum_{j=1}^k{kbrace j}frac{(n+1)^{underline{j+1}}}{j+1}
end{align*}
with $D_z:=frac{d}{dz}$ the differential operator and ${kbrace j}$ Stirling numbers of the second kind.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 6 '16 at 14:45
Markus ScheuerMarkus Scheuer
61.2k456145
$endgroup$
add a comment |
$begingroup$
Hint: You could find in this answer formulas for the sum of $k$-th powers of the first $n$ natural numbers.
begin{align*}
sum_{j=1}^nj^k&=[z^n]frac{1}{1-z}(zD_z)^kfrac{1}{1-z}\
&=sum_{j=1}^k{kbrace j}frac{(n+1)^{underline{j+1}}}{j+1}
end{align*}
with $D_z:=frac{d}{dz}$ the differential operator and ${kbrace j}$ Stirling numbers of the second kind.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 6 '16 at 14:45
Markus ScheuerMarkus Scheuer
61.2k456145
$endgroup$
add a comment |
$begingroup$
Hint: You could find in this answer formulas for the sum of $k$-th powers of the first $n$ natural numbers.
begin{align*}
sum_{j=1}^nj^k&=[z^n]frac{1}{1-z}(zD_z)^kfrac{1}{1-z}\
&=sum_{j=1}^k{kbrace j}frac{(n+1)^{underline{j+1}}}{j+1}
end{align*}
with $D_z:=frac{d}{dz}$ the differential operator and ${kbrace j}$ Stirling numbers of the second kind.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 6 '16 at 14:45
Markus ScheuerMarkus Scheuer
61.2k456145
$endgroup$
Hint: You could find in this answer formulas for the sum of $k$-th powers of the first $n$ natural numbers.
begin{align*}
sum_{j=1}^nj^k&=[z^n]frac{1}{1-z}(zD_z)^kfrac{1}{1-z}\
&=sum_{j=1}^k{kbrace j}frac{(n+1)^{underline{j+1}}}{j+1}
end{align*}
with $D_z:=frac{d}{dz}$ the differential operator and ${kbrace j}$ Stirling numbers of the second kind.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 6 '16 at 14:45
Markus ScheuerMarkus Scheuer
61.2k456145
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Aug 6 '16 at 14:45
Markus ScheuerMarkus Scheuer
61.2k456145
answered Aug 6 '16 at 14:45
Markus ScheuerMarkus Scheuer
61.2k456145
answered Aug 6 '16 at 14:45
Markus ScheuerMarkus Scheuer
61.2k456145
61.2k456145
add a comment |
add a comment |
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1
$begingroup$
take a look at this en.wikipedia.org/wiki/Faulhaber's_formula
$endgroup$
– msm
Aug 6 '16 at 8:19
4
$begingroup$
And search the site. This has been asked so many times on our site that I have lost count.
$endgroup$
– Jyrki Lahtonen
Aug 6 '16 at 8:23
$begingroup$
math.stackexchange.com/questions/1878810/…
$endgroup$
– Ahmed S. Attaalla
Aug 6 '16 at 12:59