## Infinity Is Worth No More Than -1/12

On 16 January 1913, a confused manuscript reached the famous mathematician G. H. Hardy in Cambridge. Other researchers have received similar letters before, and rejected it due to the seemingly incoherent formulae mixed with trivial mathematical results. Professional mathematicians are used to receiving manuscripts by amateurs who believe to have solved famous problems, but this particularly odd scribble caught the eye: $1+2+3+4+5+6+\ldots+\infty=-\frac{1}{12}$ Did this amateur mathematician really think that the sum of all natural numbers, a value that will exceed any given boundary at some point, will wind up being a negative fraction? [Read More]

## Does the Euler Product Converge?

Usually, I don’t care too much about convergence as a general overview of the argument is what I aim at here, and otherwise I’ll just trust that things “behave well”. But some words concerning convergence won’t harm. It’s a well known fact that the harmonic series (which we shortly touched in the previous post) $$\sum1/n$$ diverges. I think the best (though not easiest) proof of this to compare it with the corresponding integral: [Read More]

## Euler Product Revisited

From the previous post we know that the harmless looking series $$\sum n^{-s}$$ can be extended to the product $$\prod (1-p^{-s})^{-1}$$. At first sight, this does not seem terribly helpful, and it actually makes the rather easy series more complicated. So what’s the big deal? It’s what has actually been suppressed in the above notation: The sequences we run through. The series runs over all natural numbers (or positive integers, if you prefer), the product runs through all prime numbers. [Read More]

## In the Beginning, There Was... Euler's Formula!

I will start this blog the way Bernhard Riemann started his paper: with Euler’s product formula that John Derbyshire called the golden key: $\zeta(s)=\sum_{n\ge1}n^{-s}=\prod_{p}(1-p^{-s})^{-1}$ This holds for any complex number $$s$$ with $$\Re s > 1$$. If you look up a proof in any modern textbook, you will find a number technical rearrangements that end up in an examination of the absolute convergence on both sides. But actually, the formula is nothing but a fancy way of writing out the Sieve of Eratosthenes. [Read More]