POWER SUMS, BERNOULLI NUMBERS, AND RIEMANN’S

ζ-FUNCTION.

1. Power sums

We begin with a definition of power sums, Sm(n). This quantity is defined forpositive integers m > 0 and n > 1 as the sum of m-th powers of the first n − 1integers:

Sm(n) =n−1∑

i=1

im = 1m + . . . + (n − 1)m.

Our target is to find an explicite closed formula for Sm(n) in terms of m and n.We already know some special cases of the formula in question. In particular, it iseasy to see that

Sm(2) = 1,

and it is not difficult to verify the following formulas using mathematical induction

S1(n) =n(n − 1)

2,

S2(n) =n(n − 1)(2n − 1)

6,

S3(n) =n2(n − 1)2

4.

Although it is not difficult to prove the above formulas if they are stated, it doesnot seem easy to guess the exact shape of them. One can, however, make a generalguess, which we formulate and prove as a proposition.

Proposition 1. Sm(n) may be represented as a polynomial in n of degree m + 1.The constant term of this polynomial is zero, and the leading term of this polynomial

is nm+1/(m + 1)

Proof. Recall the standard binomial formula: for a positive integer d, one has

(l + k)d = ld +

(

d

1

)

ld−1k +

(

d

2

)

ld−2k2 + . . . + kd.

We will need a special case of this formula. Namely, we put d = m + 1, l = 1,and substruct the term km+1 from both sides to obtain

(1 + k)m+1− km+1 = 1 +

(

m + 1

1

)

k +

(

m + 1

2

)

k2 + . . . +

(

m + 1

m

)

km.

Let us now plug in the values k = 0, 1, 2, . . . , n − 1 into the latter identity, andwrite down the n formulas obtained in this way:

1

2

1m+1 − 0m+1 = 1

2m+1 − 1m+1 = 1+(

m+11

)

1+(

m+12

)

12+ . . . +(

m+1m

)

1m

3m+1 − 2m+1 = 1+(

m+11

)

2+(

m+12

)

22+ . . . +(

m+1m

)

2m

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

nm+1 − (n − 1)m+1 = 1+(

m+11

)

(n − 1)+(

m+12

)

(n − 1)2+ . . . +(

m+1m

)

(n − 1)m.

Let us now sum these n identities together, observing the cancellations in theleft-hand side and the appearence of power sums in the right:

nm+1 = n +

(

m + 1

1

)

S1(n) +

(

m + 1

2

)

S2(n) + . . . +

(

m + 1

m

)

Sm(n).

Taking into the account that(

m + 1

m

)

= m + 1

we obtain finally

(m+1)Sm(n) = nm+1−

(

m + 1

1

)

S1(n)−

(

m + 1

2

)

S2(n)− . . .−

(

m + 1

m − 1

)

Sm−1(n)

Exercise 1. Finish the proof of Proposition 1 using the latter identity and a math-

ematical induction argument.

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Exercise 2. Use the proof of Proposition 1 in order to find the formula for S5(n).

2. Bernoulli numbers

Althuogh Proposition 1 provides valuable qualitative information about the for-mulas for Sm(n), and the proof of this proposition actually allows to find theseformulas, we will now present another way to write these formulas down. In orderto do so we define the sequence B0, B1, B2 . . . of Bernoulli numbers as the coeffi-cients of the Maclaurin series for the function t/(et − 1):

t

et − 1=

∞∑

m=0

Bm

tm

m!.

One can reformulate the above formula by saying that the function t/(et − 1) isthe generating function for the sequence of rational numbers Bm/m!. A straight-forward but tedious calculation of the Maclaurin series allows us to find the firstfew Bernoulli numbers:

B0 = 1, B1 = −1

2, B2 =

1

6, B3 = 0, B4 = −

1

30, B5 = 0. B6 =

1

42, . . .

We now make Proposition 1 more precise by establishing a connection betweenpower sums and Bernoulli numbers.

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Theorem. (Bernoulli) For an integers n > 1 and m > 0

(m + 1)Sm(n) =

m∑

j=0

(

m + 1

j

)

Bjnm+1−j .

Proof. We begin with a standard Maclaurin series for the exponential function:

ex =

∞∑

m=0

xm

m!.

We put x = kt in the above formula, an obtain

ekt =

∞∑

m=0

km tm

m!.

We now write the latter formula down for k = 0, 1, 2, . . . , n − 1:

1 = 1

e1t =

∞∑

m=0

1m tm

m!

e2t =

∞∑

m=0

2m tm

m!

. . . = . . .

. . . = . . .

. . . = . . .

e(n−1)t =

∞∑

m=0

(n − 1)m tm

m!.

We add these n identities, and observe the appearance of the power sums in theright-hand side:

(1) 1 + e1t + e2t + . . . + e(n−1)t = 1 +

∞∑

m=0

Sm(n)tm

m!

Transform the left-hand side using the geometric series formula

1 + e1t + e2t + . . . + e(n−1)t =ent − 1

et − 1=

ent − 1

t

t

et − 1,

notice that both factors have Maclaurin series

ent − 1

t=

∞∑

k=1

nk tk−1

k!

t

et − 1=

∞∑

j=0

Bj

tj

j!,

and multiply these two series

ent − 1

t

t

et − 1=

∞∑

m=0

∑

m=k+j−1

k≥1 j≥0

nkBj

tm

k!j!= 1 +

∞∑

m=0

Sm(n)tm

m!.

4

Equate like powers of t in the above identity and obtain

Sm(n)

m!=

∑

m=k+j−1

k≥1 j≥0

nkBj

1

k!j!.

In order to finish the proof we multiply both sides of this identity by (m + 1)!, anduse the obvious fact that

(m + 1)!

(m + 1 − j)!j!=

(

m + 1

j

)

.

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Exercise 3. Prove that B2n+1 = 0 for n ≥ 1.Hint. The function t/(et − 1) + t/2 is even.

Exercise 4. Define the sequence of rational numbers b0, b1, b2, . . . as follows. Put

b0 = 0, and for m ≥ 1

(m + 1)bm = −

m−1∑

k=0

(

m + 1

k

)

bk.

Prove that bm = Bm.

Hint. In the definition of Bernoulli numbers, multiply both sides by et − 1, and

write the Maclourin series in t for this function. Equate like coefficients of like

powers of t, and show that Bernoulli numbers satisfy the above identity. Explain,

why this fact implies bm = Bm.

Show that it suffices to prove that Bernoulli numbers Bm satisfy the above iden-

tity.

This exercise presents an alternative definition for Bernoulli numbers.

3. Riemann’s ζ-function

Riemann’s ζ-function is a function on s defined by the series

ζ(s) =

∞∑

n=1

1

ns.

Honestly, this series is considered for a complex variable s. We do not do that here.Instead, for a positive integer m, we consider the values

ζ(2m) =

∞∑

n=1

1

n2m=

1

12m+

1

22m+

1

32m+

1

42m+ . . . .

This series obviously converges (why?), and may be taken as an analogue of thepower sums. The sums, however are now infinite, and the exponents are negative,thus the analogy is not close. This makes even more impressive the followingclassical result which connects these infinite sums to Bernoulli numbers

Theorem. (Euler)

2ζ(2m) = (−1)m+1 (2π)2m

(2m)!B2m.

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Proof. The proof requires the following identity from real analysis

cotx =1

x− 2

∞∑

n=1

x

n2π2 − x2.

We take this identity for granted.We will now make use of the infinite geometric series formula in order to trans-

form the summands in the right-hand side:

x

n2π2 − x2=

x

n2π2

1

1 − (x/nπ)2=

x

n2π2

∞∑

l=0

( x

nπ

)2l

.

We thus obtain

x cotx = 1 − 2

∞∑

n=1

∞∑

l=1

( x

nπ

)2l

= 1 − 2

∞∑

l=1

x2l

π2l

∞∑

n=1

1

n2l= 1 − 2

∞∑

l=1

x2l

π2lζ(2l).

Now let us transform x cotx. We need some basic facts from complex analysis,namely

cosx =eix + e−ix

2and , sin x =

eix − e−ix

2iwhich we also take for granted. Thus

x cotx = ixeix + e−ix

eix − e−ix= ix

e2ix + 1

e2ix − 1= ix

e2ix − 1 + 2

e2ix − 1= ix+

2ix

e2ix − 1= ix+

∞∑

m=0

Bm

(2ix)m

m!

We thus have two power series (in x) representations for the function x cotx,and can equate them. After that we equate like powers of x, and finish the proofof the theorem.

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