1^{k}+2^{k}+3^{k}+4^{k}+...+n^{k} = n^{k+1}/(k+1) + n^{k}/2 + ∑_{i=2}^{k} C_{k+1}^{i} B_{i} n^{k+1-i}
In which the B_{i} are those Bernoulli numbers (although Bernoulli mentiens them only in 1713 !).
This relation may also be written in a symbolic way, with "exponents" of B being indexes :
∑_{i=1}^{n} (i^{k}) = ["(n+B)^{k+1} - B^{k+1}"]/(k+1) |
The B_{i} may be calculated, from B_{0} = 1, by recurrence relations :
B_{2} - 2 B_{1} + 1 = B_{2} that is B_{1} = 1/2
B_{3} - 3 B_{2} + 3 B_{1} - 1 = B_{3} that is B_{2} = 1/6
B_{4} - 4 B_{3} + 6 B_{2} - 4 B_{1} + 1 = B_{4} that is B_{3} = 0
And generally from the symbolic notation (indexes as exponents of B) :
" (B - 1)^{k} = B^{k} " |
And the sum of powers :
1^{1}+2^{1}+3^{1}+4^{1}+...+n^{1} = n²/2 + n/2 = n(n+1)/2 (somme des entiers)
1^{2}+2^{2}+3^{2}+4^{2}+...+n^{2} = n³/3 + n²/2 + n/6 = n(n+1)(2n+1)/6 (somme des carrés)
1^{3}+2^{3}+3^{3}+4^{3}+...+n^{3} = n^{4}/4 + n^{3}/2 + n^{2}/4 = n²(n+1)²/4 (somme des cubes)
1^{4}+2^{4}+3^{4}+4^{4}+...+n^{4} = n^{5}/5 + n^{4}/2 + n^{3}/3 - n/30
1^{5}+2^{5}+3^{5}+4^{5}+...+n^{5} = n^{6}/6 + n^{5}/2 + 5n^{4}/12 - n^{2}/12
...
And the sum of 1000 first 10^{th} powers in a blink
("intra semiquadrantem horae" wrote Bernoulli) :
∑_{i=1}^{n} i^{10} =(n^{11} + 11n^{10}/2 + 55n^{9}/6 - 330n^{7}/30 + 462n^{5}/42 - 165n^{3}/30 + 11n x 5/66)/11
that is for n=1000 : 91 409 924 241 424 243 424 241 924 242 500.
Whoosh ! just for copying that, it takes more that a semiquadrantem horae (7mn30) !
We find also Bernoulli numbers in many realations :
For instance the Taylor series coeficients B_{n} in series X/(e^{X}-1)
that is X/(e^{X}-1) = ∑ B_{n}/n! X^{n}
Or in series for cotg(X)=1/X+∑_{1}^{∞} (-1)^{n} 2^{2n} B_{2n} X^{2n-1}/(2n)!
Finally the Riemann zeta function
ζ(x) = ∑1/n^{x}
gives :
ζ(2n) = (-1)^{n-1} (2π)^{2n} B_{2n}/2(2n)!
Specifically the sum of inverse of squares ζ(2) = 1+1/4+1/9+1/16+...= π²/6
and the sum of inverse fourth powers
ζ(4) = 1+1/16+1/81+1/256+...= π^{4}/90