RAMANUJAN’S EASIEST FORMULA

RAMANUJAN’S EASIEST FORMULA

John Baez

November 20, 2020Whittier College Math Club

π is a bit bigger than 3; e is a bit less. You should be curious abouttheir average.

Is there anything interesting about their geometric mean?

√πe = 2.92228236532 . . .

In 1914, Ramanujan posed this puzzle in The Journal of the IndianMathematical Society:

Prove that

(11+

11 · 3

+1

1 · 3 · 5+ · · ·

)+

1

1 + 11+ 2

1+ 31+ 4

1+ 5

. . .

=

√πe2

This is an infinite series:

11+

11 · 3

+1

1 · 3 · 5+ · · ·

This is a continued fraction:

1

1 + 11+ 2

1+ 31+ 4

1+ 5

...

Infinite series are easier, so let’s start with that!

If we knew which function had this Taylor series:

f(x) =x1+

x3

1 · 3+

x5

1 · 3 · 5+ · · ·

then we’d know

f(1) =11+

11 · 3

+1

1 · 3 · 5+ · · ·

We could try to guess f(x), but let’s figure out a differentialequation it satisfies, and solve that.

If we differentiate

f(x) =x1+

x3

1 · 3+

x5

1 · 3 · 5+ · · ·

we get

f ′(x) = 1 +x2

1+

x4

1 · 3+ · · ·

This looks a lot like

xf(x) =x2

1+

x4

1 · 3+ · · ·

Indeed, we have

f ′(x) = xf(x) + 1

So, let’s solve f ′(x) = xf(x) + 1 or

f ′(x) − xf(x) = 1

This is a first-order linear ODE. The trick is to use an “integratingfactor”. Multiply both sides by e−

∫x dx = e−x2/2 :

e−x2/2f ′(x) − xe−x2/2f(x) = e−x2/2

ddx

(e−x2/2f(x)

)= e−x2/2

e−x2/2f(x) =∫

e−x2/2 dx

f(x) = ex2/2∫

e−x2/2 dx

So we’re getting

x1+

x3

1 · 3+

x5

1 · 3 · 5+ · · · = ex2/2

∫e−x2/2 dx

But there’s a constant of integration! Really we have

x1+

x3

1 · 3+

x5

1 · 3 · 5+ · · · = ex2/2

(∫ x

0e−t2/2 dt + C

)But the left side is zero when x = 0, so C = 0:

x1+

x3

1 · 3+

x5

1 · 3 · 5+ · · · = ex2/2

∫ x

0e−t2/2 dt

Remember, we really care about this when x = 1:

1 +1

1 · 3+

11 · 3 · 5

+ · · · =√

e∫ 1

0e−t2/2 dt

We wanted to show

(11+

11 · 3

+1

1 · 3 · 5+ · · ·

)+

1

1 + 11+ 2

1+ 31+ 4

1+ 5

...

=

√πe2

Now we know

1 +1

1 · 3+

11 · 3 · 5

+ · · · =√

e∫ 1

0e−x2/2 dx

so we “just” need to show

1

1 + 11+ 2

1+ 31+ 4

1+ 5

...

=

√πe2−√

e∫ 1

0e−t2/2 dt

This still looks hard:

1

1 + 11+ 2

1+ 31+ 4

1+ 5

...

=

√πe2−√

e∫ 1

0e−t2/2 dt

This famous integral is a clue:∫ ∞

−∞

e−t2/2 dt =√

2π

It’s the area under a Gaussian:

Take half the area under the Gaussian:∫ ∞

0e−t2/2 dt =

√π

2

This is starting to look a bit like what we want to prove:

1

1 + 11+ 2

1+ 31+ 4

1+ 5

...

=

√πe2−√

e∫ 1

0e−t2/2 dt

so multiply it by√

e:

√e∫ ∞

0e−t2/2 dt =

√πe2

and plug this into the equation we want to prove!

We get

1

1 + 11+ 2

1+ 31+ 4

1+ 5

...

=√

e∫ ∞

0e−t2/2 dt −

√e∫ 1

0e−t2/2 dt

or just:

1

1 + 11+ 2

1+ 31+ 4

1+ 5

. . .

=√

e∫ ∞

1e−t2/2 dt

If we can prove this, we’re done! This is Ramanujan’s “puzzlewithin a puzzle”.

In the first part of his puzzle, he was secretly asking us to show

f(x) =x1+

x3

1 · 3+

x5

1 · 3 · 5+ · · ·

obeys the differential equation f ′(x) = xf(x) + 1 ,and then solvethis getting

x1+

x3

1 · 3+

x5

1 · 3 · 5+ · · · = ex2/2

∫ x

0e−t2/2 dt

In his “puzzle within a puzzle”, he’s secretly asking us to show

g(x) =1

x + 1x+ 2

x+ 3x+ 4

...

obeys the differential equation g′(x) = xg(x) − 1, and then solvethis getting

1

x + 1x+ 2

x+ 3x+ 4

. . .

= ex2/2∫ ∞

xe−t2/2 dt

This part is harder.

Adding them together we get(x1+

x3

1 · 3+

x5

1 · 3 · 5+ · · ·

)+

1

x + 1x+ 2

x+ 3x+ 4

...

=

ex2/2∫ x

0e−x2/2 dx + ex2/2

∫ ∞

xe−t2/2 dt =

ex2/2∫ ∞

0e−t2/2 dt =√π

2ex2/2

and setting x = 1 we’re done:(11+

11 · 3

+1

1 · 3 · 5+ · · ·

)+

1

1 + 11+ 2

1+ 31+ 4

1+ 5

. . .

=

√πe2

In 1913, Ramanujan wrote a letter to Hardy introducing himself. Itcontained many formulas he’d proved. His formula (1.8) wasequivalent to this:

1

x + 1x+ 2

x+ 3x+ 4

...

= ex2/2∫ ∞

xe−t2/2 dt

Later Hardy wrote:

The first question was whether I could recogniseanything. I had proved things rather like (1.7) myself, andseemed vaguely familiar with (1.8). Actually (1.8) isclassical; it is a formula of Laplace first proved properlyby Jacobi. [....] On the whole the integral formulaeseemed the least impressive.

Laplace’s derivation of the formula

1

x + 1x+ 2

x+ 3x+ 4

...

= ex2/2∫ ∞

xe−t2/2 dt

made no sense to me.

Jacobi wrote a 2-page paper about it in 1834. Even though it wasin Latin and full of errors, it was pretty easy to understand, so Iblogged about it.

On Twitter my friend Leo Stein found a simpler proof, which youcan see here:

John Baez, Chasing the Tail of the Gaussian (Part 2), Then-Category Cafe.

Let’s write:

g(x) = ex2/2∫ ∞

xe−t2/2 dt

as a continued fraction. If we differentiate g we can see that

g′(x) = xg(x) − 1

But keep on differentiating g, and see what happens:

g′′(x) = xg′(x) + g(x)

g′′′(x) = xg′′(x) + 2g′(x)

g(4)(x) = xg′′′(x) + 3g′′(x)

We notice a pattern which we can prove inductively:

g(n+2) = xg(n+1) + (n + 1)g(n) for n ≥ 0

To get a continued fraction, let’s take our formula

g(n+2) = xg(n+1) + (n + 1)g(n) for n ≥ 0

and divide it by g(n+1):

g(n+2)

g(n+1)= x + (n + 1)

g(n)

g(n+1)for n ≥ 0

This looks simpler in terms of the ratios rn = g(n+1)/g(n):

rn+1 = x +(n + 1)

rn

or solving for rn:

rn =n + 1−x + rn+1

for n ≥ 0 ♠

rn =n + 1−x + rn+1

for n ≥ 0 ♠

In terms of r0 = g′/g, our original equation g′ = xg − 1 gets asimilar look:

g =1

x − r0

Starting from here, and repeatedly using ♠, we get

g =1

x − r0=

1

x − 1−x+r1

=1

x − 1−x+ 2

−x+r2

=1

x − 1−x+ 2

−x+ 3−x+r3

= · · ·

and so on.

If we go on forever, we get

g =1

x − 1−x+ 2

−x+ 3−x+ 4

...

A bit of algebra gives

g =1

x + 1x+ 2

x+ 3x+ 4

...

or in other words:

ex2/2∫ ∞

xe−t2/2 dt =

1

x + 1x+ 2

x+ 3x+ 4

. . .

So we’re done!

(11+

11 · 3

+1

1 · 3 · 5+ · · ·

)+

1

1 + 11+ 2

1+ 31+ 4

1+ 5

. . .

=

√πe2