THE FUCK?

[Anonymous No.10691457]

[eqn] \sum_{m=2}^{\infty} \sum_{n=2}^{\infty} n^{-m} = 1 [/eqn]

[Anonymous No.10691492]

By the geometric series for |x|<1,

[math] \sum_{m=0}^\infty x^m = \frac{1}{1-x} [/math]

so

[math] \sum_{m=2}^\infty x^m = \frac{1}{1-x} - (1 + x) [/math]

so

[math] \sum_{m=2}^\infty (1/n)^m = \frac{1}{1-1/n} - (1 + 1/n) = \frac{1}{1-1/n} - \frac{1}{1-1/(n+1)} [/math]

[math] \sum_{n=2}^\infty \left(\frac{1}{1-1/n} - \frac{1}{1-1/(n+1)} \right) = 1 + (-a+a) + (-b+b) + (-c+c) + ... [/math]

[Anonymous No.10691509]

>>10691492
By this logic [math]0 = (1 - 1) + (1 - 1) + \cdots = 1 + (-1+1)+(-1+1)+\cdots[/math].

[Anonymous No.10691514]

>>10691509
=1

[Anonymous No.10691516]

>>10691492
Thank you mr. smart man!
I'll pretend I understand this!

[Anonymous No.10692731]

>>10691509
No, because the difference between the partial sums in the telescoping series, and the number 1, gets smaller for each new pair of elements

[Anonymous No.10692762]

>>10691509
Let me give a formal argument for the last line: Looking at the parial sums n=2 to N of the last infinite sum, we see that the first term is always 1/(1-1/2), then follows a sequence of clearly canceling terms, and eventually one term of -1/(1-1/(N+1)). In other words, take together, the value of the partial sums is always 1-1/N.

[Anonymous No.10692788]

>>10691492
I think the argument would be more compelling if in the last line you show the partial sums converge.

[Anonymous No.10693041]

>>10691457
If you notice, the n-sum is idendical to the definition of the riemann zeta function, so we could write this as \sum_{m=2}^{\infty} \zeta (m) = 1

This is strange, as the limit as \zeta (n) where n approaches infinity approaches 1.

Rest assured, I ran this through a python script and got indeed that op's equation is indeed valid.