# 2(5!) − cot [4 arctan 0.2 + (i/2) ln i] − 1

1. Jan 20, 2016

### Jenab2

Don't ever divide anything by the quantity in the title.

Post your favorite "fancy zeros" here.

2. Jan 21, 2016

### axmls

$$\frac{1}{12} + \sum _{n = 1} ^\infty n$$

3. Jan 23, 2016

### Jenab2

That should probably be

−1/12 + Σ(2,∞) 1/n⁴

Edit: whoops, no. That doesn't seem quite right, either. I evaluated ten million terms of the sum and came up with −0.0010100996222299347, so

−1/12 + 1/999 + Σ(2,∞) 1/n⁴

seems to be nearer to zero.

Last edited: Jan 23, 2016
4. Jan 23, 2016

### axmls

Nope, it's written as I intended.

5. Jan 23, 2016

### Staff: Mentor

Unless you specify how divergent sums are to be evaluated, the formula is not well-defined.
Yes there is a specific way that leads to -1/12, but this is by far not the only way to assign finite values to divergent sums.

6. Jan 23, 2016

### axmls

It was a somewhat tongue-in-cheek answer, if that wasn't clear.

7. Jan 23, 2016

### HomogenousCow

I tried something like this to mess with my maths teacher in senior year once, replaced pi with some weird sums.

8. Jan 23, 2016

### Jenab2

1 / Σ(1,∞) n = 0
1 / { 1/a + 1 / Σ(1,∞) n } = a, a≠0.

9. Jan 23, 2016

### axmls

10. Jan 23, 2016

### mfig

$\begin{Vmatrix} \vec\nabla \times \vec\nabla f \end{Vmatrix}$

Last edited: Jan 23, 2016
11. Jan 24, 2016

### Jenab2

Ah. My difficulty in appreciating the assignment was caused by my thinking of scalars in vector terms. Consider velocities in the same direction, classically being added, tail to head:

v₁ + v₂ + v₃ + v₄ + ...

where each velocity is in the direction of the +x axis and the magnitude of the velocities is proportional to the subscript.

How is it that an object, moving through an infinite succession of changes-of-velocity, all of them forward, might end up moving BACKWARD at a speed of 1/12 velocity units?

I'd figured that this was a case of getting a strange result out of an indeterminate form.