Sums of powers and Bernoulli numbers

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Adding the integers from 1 through $n$ pairs each term with a partner that makes the same total. Square those summands and the partners disagree: $1^{2} + n^{2}$ is one number, $2^{2} + (n - 1)^{2}$ is another. The cubes close again, as the square of the old triangular number, and the fourth powers pick up yet another face. The question is whether one recipe still names the sum for every positive integer exponent.

The squares refuse to pair

On finite geometric and arithmetic series the sum from 1 through $n$ is $n (n + 1) /2$ , because each pair of ends adds to $n + 1$ . That pairing spends two extra freezes: the first term is 1, and the step is 1. Square the summands and a constant pair-sum is gone. The board keeps the exponent frozen at 2 and lets $n$ run through the positive integers 1 through 6. Each left-hand layer is one square. A layer past $n$ has height 0. The right-hand bar is the single height $n (n + 1) (2 n + 1) /6$ .

Drag n. It stays a positive integer from 1 through 6, and the exponent stays 2. Each left bar is one square k^2, and a bar past n has height 0. The right bar is n(n+1)(2n+1)/6. The segment between the tops is level when that formula has paid the stack.

At the start $n$ is 4, so the left stack is $1 + 4 + 9 + 16 = 30$ and the right bar is the same 30. Drag $n$ . The segment between the tops stays level. Both freezes stay in force: $n$ is a positive integer, and the exponent is 2. The identity the tops are meeting is

$\sum_{k = 1}^{n} k^{2} = 1^{2} + 2^{2} + \dots + n^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6} .$

A longer step is still the old sum

The even numbers are the arithmetic series with first term 2 and common difference 2, stopped after $n$ terms, so the sum is $n (n + 1)$ . That row is already on finite geometric and arithmetic series.

The odd numbers are the layers of a square. The $k$ -th layer has length $2 k - 1$ , and $n$ of those layers fill an $n$ by $n$ square.

Drag n. It stays a positive integer from 1 through 6. The k-th layer has length 2k-1 and wraps the square already built. At the start n is 4, so the layers 1, 3, 5, and 7 fill a square of side 4.

$\sum_{k = 1}^{n} (2 k - 1) = 1 + 3 + \dots + (2 n - 1) = n^{2} .$

Square those same odd steps and the total is still a polynomial in this $n$ :

$\sum_{k = 1}^{n} (2 k - 1)^{2} = 1^{2} + 3^{2} + \dots + (2 n - 1)^{2} = \frac{n ( 4 n ^{2} - 1 )}{3} .$

Four exponents, four faces

The first power is the paired sum $n (n + 1) /2$ . The board freezes that number as the side of a square. The area is the sum of the cubes.

Drag n. It stays a positive integer from 1 through 5. The side is 1+2+...+n, the paired sum, so this is a square of that side and not a square of arbitrary length. Its area is the sum of the first n cubes.

$\sum_{k = 1}^{n} k^{3} = \frac{n ^{2} ( n + 1 ) ^{2}}{4} .$

The fourth powers pick up one more factor,

$\sum_{k = 1}^{n} k^{4} = \frac{n ( n + 1 ) ( 2 n + 1 ) ( 3 n ^{2} + 3 n - 1 )}{30} .$

Read as polynomials, the four results wear four faces. The square sum carries a linear term, the cube sum has none, and the fourth-power sum carries a factor the square sum never used. Jacob Bernoulli (1645–1705), looking for an empirical formula for the sum of the powers of the natural numbers, met these special cases. For $n = 1, 2, \dots$ and exponents $p = 1, 2, \dots$ he wrote one expression that specialises to all four.

One polynomial for every exponent

Write

$S_{n}^{p} := 1^{p} + 2^{p} + \dots + n^{p}$

for a positive integer $n$ and an exponent $p = 1, 2, \dots$ . The expression is

$S_{n}^{p} = \frac{1}{p + 1} n^{p + 1} + \frac{1}{2} n^{p} + \frac{B _{2}}{2} (1 p) n^{p - 1} + \frac{B _{3}}{3} (2 p) n^{p - 2} + \dots + \frac{B _{p}}{p} (p - 1 p) n .$

The first term is the highest power, $n^{p + 1} / (p + 1)$ . For the squares, $p = 2$ , that piece is $n^{3} /3$ , and the board's total is larger: the lower powers make up the difference between $n^{3} /3$ and $n (n + 1) (2 n + 1) /6$ . The second term is half of $n^{p}$ . Every later term is a Bernoulli number times a binomial coefficient, times a lower power of $n$ .

The coefficients in this polynomial always add to 1:

$\frac{1}{p + 1} + \frac{1}{2} + \frac{B _{2}}{2} (1 p) + \frac{B _{3}}{3} (2 p) + \dots + \frac{B _{p}}{p} (p - 1 p) = 1.$

Set $n = 1$ . The left side of the polynomial is $S_{1}^{p} = 1$ , and the right side is this sum of coefficients. The identity is the general formula at the first positive integer.

Two values are fixed at the start: $B_{0} := 1$ and $B_{1} := - 1/2$ . For every odd index $n \geq 3$ , $B_{n} = 0$ . The cube sum had no lone factor of $n$ because the term that would have carried $B_{3}$ is zero. The recursion that produces the later numbers is

$\sum_{k = 0}^{n} (k p + 1) B_{k} = 0.$

Expand the left-hand side by multiplying, and wherever a power $B^{n}$ appears write the number $B_{n}$ instead. The same constraint reads

$(1 + B)^{p + 1} - B_{p + 1} = 0.$

For $p = 2, 3, \dots$ this determines $B_{2}, B_{3}, \dots$ . Through $B_{20}$ the values are

$k$	$B_{k}$	$k$	$B_{k}$
$0$	$1$	$10$	$\frac{5}{66}$
$1$	$- \frac{1}{2}$	$12$	$- \frac{691}{2730}$
$2$	$\frac{1}{6}$	$14$	$\frac{7}{6}$
$4$	$- \frac{1}{30}$	$16$	$- \frac{3617}{510}$
$6$	$\frac{1}{42}$	$18$	$\frac{43867}{798}$
$8$	$- \frac{1}{30}$	$20$	$- \frac{174611}{330}$

Odd indices from 3 on stay 0, so the table leaves them out. $B_{4} = - 1/30$ is the last term of the fourth-power sum, $- n /30$ .

The four expansions are one recipe

Substitute the first Bernoulli numbers and the early formulas come back as polynomials:

$S_{n}^{1} S_{n}^{2} S_{n}^{3} S_{n}^{4} = \frac{1}{2} n^{2} + \frac{1}{2} n, = \frac{1}{3} n^{3} + \frac{1}{2} n^{2} + \frac{1}{6} n, = \frac{1}{4} n^{4} + \frac{1}{2} n^{3} + \frac{1}{4} n^{2}, = \frac{1}{5} n^{5} + \frac{1}{2} n^{4} + \frac{1}{3} n^{3} - \frac{1}{30} n .$

The board draws those terms divided by $n^{p + 1}$ , so the picture keeps the shape of the coefficients while $n$ moves. The exponent stays on $1, 2, 3, 4$ . At the start $p = 4$ and $n = 1$ , so each bar is the coefficient itself. The fourth slot is the $B_{3}$ term and sits on the axis. The fifth bar drops by $1/30$ , which is the term $- n /30$ at $n = 1$ . Drag $p$ off 4 and that downward bar leaves, because the formula stops at $B_{p}$ . Drag $n$ upward and each correction is a smaller share of $n^{p + 1}$ .

Drag p or n. p stays on 1 through 4 and n stays a positive integer from 1 through 4. Each bar is that term of the expanded sum, divided by n to the power p+1. At the start p is 4 and n is 1, so the last bar drops by 1/30. The fourth bar is the B_3 term and stays on the axis.

The factorial writing is the same polynomial:

$\frac{S _{n}^{p}}{p !} = \frac{B _{0} ( n + 1 ) ^{p + 1}}{0 ! ( p + 1 )!} + \frac{B _{1} ( n + 1 ) ^{p}}{1 ! p !} + \frac{B _{2} ( n + 1 ) ^{p - 1}}{2 ! ( p - 1 )!} + \dots + \frac{B _{p} ( n + 1 )}{p ! 1 !} .$

$B_{1} = - 1/2$ already sits in the second term. That is the sign the four expansions were built with.

The same numbers inside a series

For every complex $x$ with $0 < ∣ x ∣ < 2 π$ ,

$\frac{x}{e ^{x} - 1} = \frac{B _{0}}{0 !} + \frac{B _{1}}{1 !} x + \frac{B _{2}}{2 !} x^{2} + \dots = \sum_{k = 0}^{\infty} \frac{B _{k}}{k !} x^{k} .$

The board stays on the positive reals inside that disk. The solid curve is $x / (e^{x} - 1)$ . The dashed curve keeps $B_{0}$ , $B_{1}$ , $B_{2}$ , and $B_{4}$ . It omits $B_{3}$ because that number is 0, and it omits $B_{6}$ together with every later term. The segment from $P$ to $Q$ is the height those omitted terms still owe at the sample point. The equality on the page is the full series inside the disk.

Drag t. The sample stays on the positive reals short of 2π. P lies on x/(e^x-1). Q is the partial sum through B_4, and the B_3 term is absent because that number is 0. The segment is the height the omitted terms still owe. The dashed vertical line is the positive edge x=2π.

The same coefficients appear in the power series for

$tan x, cot x, tanh x, coth x, \frac{1}{s i n x}, \frac{1}{s i n h x}, ln ∣ tan x ∣, ln ∣ sin x ∣, ln cos x .$

This page names those functions.

Even powers in the denominator

In 1734 Leonhard Euler found

$1 + \frac{1}{2 ^{2}} + \frac{1}{3 ^{2}} + \dots = \sum_{n = 1}^{\infty} \frac{1}{n ^{2}} = \frac{π ^{2}}{6} .$

Drag n. It counts the first n terms, from 1 through 8. Each layer has height 1/k^2. The horizontal line is π^2/6. The stack of these finitely many layers stays below that line.

For each $k = 1, 2, \dots$ the even power $2 k$ is the same kind of sum, with the Bernoulli number of that even index:

$\sum_{n = 1}^{\infty} \frac{1}{n ^{2 k}} = \frac{( 2 π ) ^{2 k}}{2 ( 2 k )!} ∣ B_{2 k} ∣.$

For $k = 1$ , $B_{2} = 1/6$ , and

$\frac{( 2 π ) ^{2}}{2 \cdot 2 !} \cdot \frac{1}{6} = \frac{π ^{2}}{6} .$

The absolute value carries the sign. $B_{2}$ is positive and $B_{4}$ is negative, while the sum of reciprocals stays positive either way. The formula is stated for the even orders $2 k$ . Johann Bernoulli (1667–1748) and Jacob Bernoulli spent a long time determining the values of these series.

The object is $S_{n}^{p}$ for a positive integer $n$ and a positive integer exponent $p$ : one polynomial in $n$ , with coefficients fixed by $B_{0} = 1$ , $B_{1} = - 1/2$ , and the later $B_{k}$ . The generating series asks the further restriction $0 < ∣ x ∣ < 2 π$ . The reciprocal sums ask the exponent to be a positive even integer $2 k$ , so that $∣ B_{2 k} ∣$ can set $\sum 1/ n^{2 k}$ . Keep only the odd denominators and alternate the sign. That constant is fixed by an Euler number, on Euler numbers.