Дефинишемо ли суму
S
p
(
n
)
=
∑
k
=
1
n
k
p
.
{\displaystyle S_{p}(n)=\sum _{k=1}^{n}k^{p}.}
Тада је:
S
p
(
1
)
=
1
{\displaystyle S_{p}(1)=1}
S
p
(
n
+
1
)
=
S
p
(
n
)
+
(
n
+
1
)
p
.
{\displaystyle S_{p}(n+1)=S_{p}(n)+(n+1)^{p}.}
Покушајмо сада да
S
p
(
n
)
{\displaystyle S_{p}(n)}
∑
k
=
0
∞
a
p
,
k
n
k
+
1
{\displaystyle \sum _{k=0}^{\infty }a_{p,k}n^{k+1}}
Уврстимо ли то у други израз у овом поглављу добијамо:
∑
k
=
0
∞
a
p
,
k
(
n
+
1
)
k
+
1
=
∑
k
=
0
∞
a
p
,
k
n
k
+
1
+
(
n
+
1
)
p
{\displaystyle \sum _{k=0}^{\infty }a_{p,k}(n+1)^{k+1}=\sum _{k=0}^{\infty }a_{p,k}n^{k+1}+(n+1)^{p}}
Користимо биномну теорему , па следи:
∑
k
=
0
∞
a
p
,
k
[
(
∑
j
=
0
k
+
1
(
k
+
1
j
)
n
j
)
−
n
k
+
1
]
=
(
n
+
1
)
p
{\displaystyle \sum _{k=0}^{\infty }a_{p,k}\left[\left(\sum _{j=0}^{k+1}{\binom {k+1}{j}}n^{j}\right)-n^{k+1}\right]=(n+1)^{p}}
∑
k
=
0
∞
a
p
,
k
∑
j
=
0
k
(
k
+
1
j
)
n
j
=
∑
j
=
0
∞
(
p
j
)
n
j
.
{\displaystyle \sum _{k=0}^{\infty }a_{p,k}\sum _{j=0}^{k}{\binom {k+1}{j}}n^{j}=\sum _{j=0}^{\infty }{\binom {p}{j}}n^{j}.}
Двоструку суму на левој страни преуредимо узимајући у обзир j ≤k
∑
j
=
0
∞
n
j
∑
k
=
j
∞
(
k
+
1
j
)
a
p
,
k
=
∑
j
=
0
∞
(
p
j
)
n
j
.
{\displaystyle \sum _{j=0}^{\infty }n^{j}\sum _{k=j}^{\infty }{\binom {k+1}{j}}a_{p,k}=\sum _{j=0}^{\infty }{\binom {p}{j}}n^{j}.}
и коначно се добија:
∑
k
=
j
∞
(
k
+
1
j
)
a
p
,
k
=
(
p
j
)
.
{\displaystyle \sum _{k=j}^{\infty }{\binom {k+1}{j}}a_{p,k}={\binom {p}{j}}.}
Десна страна је једнака нули за j >p
a
p
,
k
=
0
{\displaystyle a_{p,k}=0}
k >p j !Поххамеровога симбола вреди:
∑
k
=
j
p
(
k
+
1
)
j
a
p
,
k
=
(
p
)
j
{\displaystyle \sum _{k=j}^{p}(k+1)_{j}a_{p,k}=(p)_{j}}
∑
k
=
j
p
(
k
+
1
)
t
(
k
+
1
−
t
)
j
−
t
a
p
,
k
=
(
p
)
t
(
p
−
t
)
j
−
t
{\displaystyle \sum _{k=j}^{p}(k+1)_{t}(k+1-t)_{j-t}a_{p,k}=(p)_{t}(p-t)_{j-t}}
Супституцијом k =k' +t
∑
k
′
=
j
−
t
p
−
t
(
k
′
+
1
)
j
−
t
[
(
k
′
+
t
+
1
)
t
(
p
)
t
a
p
,
k
′
+
t
]
=
(
p
−
t
)
j
−
t
.
{\displaystyle \sum _{k'=j-t}^{p-t}(k'+1)_{j-t}\left[{\frac {(k'+t+1)_{t}}{(p)_{t}}}a_{p,k'+t}\right]=(p-t)_{j-t}.}
односно:
(
k
′
+
t
+
1
)
t
(
p
)
t
a
p
,
k
′
+
t
=
a
p
−
t
,
k
′
.
{\displaystyle {\frac {(k'+t+1)_{t}}{(p)_{t}}}a_{p,k'+t}=a_{p-t,k'}.}
За k' =0
a
p
,
t
=
(
p
)
t
(
t
+
1
)
t
a
p
−
t
,
0
=
(
p
t
)
a
p
−
t
,
0
t
+
1
.
{\displaystyle a_{p,t}={\frac {(p)_{t}}{(t+1)_{t}}}a_{p-t,0}={\binom {p}{t}}{\frac {a_{p-t,0}}{t+1}}.}
а то управо одговара Бернулијевим бројевима , тако да коначно добијамо:
∑
k
=
1
n
k
p
=
∑
j
=
0
p
(
p
j
)
B
p
−
j
j
+
1
n
j
+
1
{\displaystyle \sum _{k=1}^{n}k^{p}=\sum _{j=0}^{p}{\binom {p}{j}}{\frac {B_{p-j}}{j+1}}n^{j+1}}
![{\displaystyle \sum _{k=0}^{\infty }a_{p,k}\left[\left(\sum _{j=0}^{k+1}{\binom {k+1}{j}}n^{j}\right)-n^{k+1}\right]=(n+1)^{p}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52ae5d9e150bd8bceb33c20ef3adb615896a0520)
![{\displaystyle \sum _{k'=j-t}^{p-t}(k'+1)_{j-t}\left[{\frac {(k'+t+1)_{t}}{(p)_{t}}}a_{p,k'+t}\right]=(p-t)_{j-t}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f07d4eaa2d40947109ceb0f88f644fd4814d2d85)
