# Summation to ∞ and beyond

We know that the summation sequence 1/1+1/2+1/3+1/4… goes to ∞

And the summation sequence 1/1+1/4+1/9+1/16 … = π^2/6

So there is some x beyond which 1^(-x) + 2^(-x) + 3^(-x) … sums to a finite number and beneath which the sum tends to ∞

Does anyone have an idea on what x is, and if possible a proof

Note: According to the Riemann zeta function 1^(-1) + 2^(-1) … goes to -1/12, so my question in that context would be when would the Riemann zeta function give the output as ∞/-∞ , not sure but I think both of these questions are the same ( if not please tell )

Note by Jason Gomez
4 months, 4 weeks ago

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i gave some hints in my latex note, but i can repeat it here ;-)
you are right: i'm used to treat hexadecimals as numbers.
(so i refer to a, b, c, d, e, f as numbers too)
with this you can decode every two "numbers" to represent a char.
so "beating a dead horse" = 79767a67 726d7420 7a207776 7a772073 6c696876
decrypts by: 79='b', 76='e', 7a='a', ...
if you continue that you will figure what 20 means

- 4 months, 2 weeks ago

Lol nice code

- 4 months, 2 weeks ago

you are fast. yeah, you got it.

- 4 months, 2 weeks ago

Intentional, gosh the 7 and 6’s and that 20 kept throwing me off track

- 4 months, 2 weeks ago

Wow it’s a space

- 4 months, 2 weeks ago

Oh no wonder I kept missing the middle word 7a207776 quite deceptive to keep a space for no use and deceptively keep the space named differently

- 4 months, 2 weeks ago

Wait heh

- 4 months, 2 weeks ago

My hypothesis for the Riemann zeta function part was because I thought the function was continuous

- 4 months, 4 weeks ago

Never would have thought that it abruptly jumped from -1/12 to a positive value at 1

- 4 months, 4 weeks ago

1/1 + 1/2 + 1/3 + ... = -1/12

is only true when we want to regularize divergent sums.

- 4 months, 4 weeks ago

What I expected was for it to go towards - ∞ as x>p>1 then at x jump to ∞ and for p>x sum up to smaller and smaller positive values ( that’s what my intuition said to me )

- 4 months, 4 weeks ago

$x>1$ gives a convergent sum. $x\leqslant 1$ gives a divergent sum.

Look up Convergence test for $p$-series.

- 4 months, 4 weeks ago

What was said exactly was that it converges for p>1

- 4 months, 4 weeks ago

Oh whoops, lemme fix my comment. I accidentally wrote "2".

- 4 months, 4 weeks ago

I looked it up right now and it says it’s at 1?!

- 4 months, 4 weeks ago

What is at 1?

- 4 months, 4 weeks ago