Euler numbers

The sum of reciprocal even powers over every positive integer closes with a Bernoulli number. Drop the even denominators and alternate the sign. The partial sums of

$1-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\cdots$

still settle. Sums of powers and Bernoulli numbers names $\sum 1/n^{2k}$ by $|B_{2k}|$. That table does not name the constant these odd denominators settle on.

Freeze the exponent at 3

The first case keeps the exponent at 3, so each term is the cube of an odd reciprocal, with alternating sign. Every partial sum here sits between $0.96$ and $1$, so the board draws $100$ times the gap above ${\pi }^3/32$. On that scale the dashed line, the constant itself, lies at height $0$.

At the start a single term, $1$, stands above the line. The next term is $-1/27$, and the partial sum drops below the line. Each later term is smaller and flips the side. You can drag only as far as 8 terms. The infinite sum is the dashed line; the segment is the height the omitted terms still owe. The factor in front of ${\pi }^3/32$ for this one exponent is $|E_2|$, and that factor is $1$, so the line can be drawn before the sequence $E_k$ has a definition. The exponent is frozen. Higher odd powers need the whole sequence.

Coefficients of $1/\cosh x$

For every complex $x$ with $|x|<\pi /2$,

$\frac{1}{\cosh x}=1+\frac{E_1}{1!}x+\frac{E_2}{2!}{x}^2+\cdots ={\sum }_{k=0}^{\infty }\frac{E_k}{k!}{x}^k.$

These coefficients $E_k$ are the Euler numbers. $E_0=1$. Every odd index is zero, so only even powers are present. On the real axis, $1/\cosh x$ stays smooth past $x=\pi /2$. The bound is a disk in the complex plane: that is the range where the series is claimed.

The board draws the real function against the partial sum that keeps $E_0$, $E_2=-1$, and $E_4=5$:

$1-\frac{x^2}{2}+\frac{5x^4}{24}.$

It omits $E_6=-61$ and every later term. Drag the sample past either dashed line. The polynomial leaves the curve. The curve itself stays smooth. Inside the disk the segment is only the tail of the series. Outside the disk the series is no longer the claim on the page.

The next even value is forced

Multiply out $(E+1{)}^n+(E-1{)}^n$ and, wherever a power $E^n$ appears, write the number $E_n$ instead. For $n=1,2,\dots$,

$(E+1{)}^n+(E-1{)}^n=0.$

For $n=2$ the even part is $E_0+E_2=0$, so $E_2=-1$. For $n=4$,

$E_0+6E_2+E_4=0,$

and $1+6(-1)+E_4=0$, so $E_4=5$. For $n=6$ the same reading gives $E_6=-61$. An odd $n$ forces the odd Euler numbers to stay $0$, which is why the series above never grows an odd power. Leonhard Euler obtained the relation from the product

$\sin \pi x=\pi x{\prod }_{m=1}^{\infty }\left(1-\frac{x^2}{m^2}\right),$

valid for every complex $x$. This page uses the relation as the recursion that fills the table. It does not derive the product.

The recursion produces the following values. Odd indices stay $0$, so the table leaves them out.

Read against the Bernoulli numbers

The Euler numbers and the Bernoulli numbers are also linked by a second symbolic line. Expand, and replace each power $B^m$ by the Bernoulli number $B_m$. There is no subscript on the $B$ inside the parentheses.

$E_{2n}=\frac{4^{2n+1}}{2n+1}{\left(B-\frac{1}{4}\right)}^{2n+1},n=1,2,\dots .$

The power-sum page fixed $B_1=-1/2$, together with $B_2=1/6$ and $B_3=0$. For $n=1$ the line is $(64/3)$ times $(B-1/4{)}^3$. That expansion equals

$-\frac{3}{4}\cdot \frac{1}{6}+\frac{3}{16}\cdot B_1-\frac{1}{64}=-\frac{1}{8}-\frac{3}{32}-\frac{1}{64}=-\frac{15}{64}$

after $B_1=-1/2$ is substituted, and $(64/3)\cdot (-15/64)=-5$. The table lists $E_2=-1$. The symbolic line does not reproduce that entry. The numbers in the table are the ones computed from $(E+1{)}^n+(E-1{)}^n=0$ and from the series for $1/\cosh x$.

The same table, signs flipped, for secant

The same coefficients appear in the power series for $1/\cos x$. Because $\cos x=\cosh (ix)$, the even term $E_{2m}{x}^{2m}/(2m)!$ in the series for $1/\cosh x$ becomes $E_{2m}(-1{)}^m{x}^{2m}/(2m)!$. Through $E_4$,

$1+\frac{x^2}{2}+\frac{5x^4}{24}.$

Secant has vertical asymptotes on the real axis at the odd multiples of $\pi /2$. The board keeps the sample strictly between the walls at $x=\pm \pi /2$. The dashed curve is that partial sum. Further series that carry these coefficients are not written on this page.

Every higher odd power

For each $k=1,2,\dots$,

$1-\frac{1}{3^{2k+1}}+\frac{1}{5^{2k+1}}-\cdots ={\sum }_{n=0}^{\infty }\frac{(-1{)}^n}{(2n+1{)}^{2k+1}}=\frac{{\pi }^{2k+1}}{2^{2k+2}(2k)!}|E_{2k}|.$

The first board was $k=1$: $|E_2|=1$ and the constant is ${\pi }^3/32$. For $k=2$, $|E_4|=5$ and the constant is $5{\pi }^5/1536$. The first omitted term is already $1/3^5=1/243$, so those partial sums sit on the constant at once; a second board would show a single bar. The absolute value carries the sign. $E_2$ is negative and $E_4$ is positive, while the alternating sum over odd denominators stays positive either way.

The object is the sequence $E_k$ of Taylor coefficients of $1/\cosh x$ for complex $x$ with $|x|<\pi /2$. Odd indices are $0$. The recursion $(E+1{)}^n+(E-1{)}^n=0$ and the table fix the later values. The reciprocal sums ask the exponent $2k+1$ on the odd denominators, so that $|E_{2k}|$ can set the constant. The sum over every positive integer, even denominators included, is sums of powers and Bernoulli numbers.