The Bernoulli numbers play a key role in finding the sum of natural numbers (up to n) all raised to a certain power m.

• When Jakob Bernoulli first discovered this, he solved 1^(10) + 2^(10) + ... + 1000^(10) in just seven minutes.

• Even though he did not provide a proof for his method, he did discover the numbers which appear frequently in number theory.

• Note that this sum closely resembles the Reimann Zeta function of m (but it does not go on forever).

  • This will build the foundation for the Ramanujan Summation later in time.

These numbers also show up in the Taylor series of trigonometric and hyperbolic trigonometric functions, like tan(x) and coth(x)

We prove that 1^2 + 2^2 + 3^2 = 14 using the Bernoulli numbers.

These Bernoulli numbers, along with the coefficients of binomial theorem, make up the Bernoulli polynomials.

• On the right, are some generating function for the Bernoulli polynomials, along with some preliminary Bernoulli functions.

• Similar to the last example, using these polynomials specifically, we prove that 1^2 + 2^2 + 3^2 = 14.

This formula dictates the difference between an integral and the approximation of the area under the curve using the rectangle method.

Proof: 

• Taking any function f(x), by conveniently defining f(x) in terms of Bernoulli Polynomials, we can integrate by parts 

• After some algebraic manipulation (moving terms over), we simplify our expression to the Euler-Maclaurin series.

• Note, I first derive the formula for the case of p = 1. However, I list (right below) the formula when it is extended to higher p values. 

This formula has tremendous applications, including the Basel Problem. 

• Based on this formula, Euler approximated the problem to 20 decimal places and guessed the answer was pi^2/6, which he later proved. 

The Ramanujan Summation assigns a numerical value to divergent infinite series. 

This formula actually does not indicate the sum in a traditional sense, as indicated by the imaginary numbers in the formula.

• Just like how 'i = sqrt(-1)' does not exist in real life, this value does not equal the sum of various series.

This formula is derived from the Euler-Maclaurin formula, as one can see a close similarity between the two.

• Specifically, the Ramanujan Summation deals with when p approaches infinity. 

This formula agrees with the mathematician Ramanujan's derivation that the sum of all natural numbers is equal to -1/12, a very counterintuitive idea.

• This can be seen on the right when we make f(x) = x and use the Bose-Einstein integral to reach -1/12.

• It's also very interesting to see how it uses pi^2/6 in it's solution, due to the Reimann Zeta function(2). 

Even though this number is not its true value, it plays a pivot role in the summation itself. 

• For example, the sum of all natural numbers from 1 to n is n(n+1)/2. 

  • When plotted, this is a parabola that has n-intercepts at (-1,0) and (0,0). 

  • The integral of this function from -1 to 0 is equal to -1/12.

• We can extend it even further: the sum of the squares of natural numbers from 1 to n is n(n+1)(n+2)/6.

  • When graphed, the n-intercepts lie at (-2,0), (-1,0), and (0,0). 

  • Take the integral of our function from the first and last intercept and we get 0. 

• These values coincide with the Reimann Zeta Functions at -1 and -2 (see right for why they are negative numbers and not positive).

The Basel Problem in general is stated at the top of the sheet on the right.

Euler unintuitively started with sin(x) and expanded it to its Taylor series.  

By the Weierstrass Factorization Theorem in complex analysis, Euler wrote sin(x)/x as a product of its roots. 

By noticing patterns and manipulating algebra, Euler figured out the coefficient of x^2 in the new expression for sin(x)/x.

Since coefficients have to match, Euler set the new coefficient and the previous one from the Taylor series equal to each other and found that a series of rational numbers, as it goes on to infinity, converges to (pi^2)/6