4
votes
4answers
155 views

Find the regularized sum $\frac{1}{\ln(2)}+\frac{1}{\ln(3)}+\frac{1}{\ln(4)}+…$

By considering the integral Zeta function $$F(s)=s+\frac{1}{2^s\ln(2)}+\frac{1}{3^s\ln(3)}+\frac{1}{4^s\ln(4)}+...$$ Evaluate $$\frac{1}{\ln(2)}+\frac{1}{\ln(3)}+\frac{1}{\ln(4)}+...$$ EDIT: ...
2
votes
1answer
154 views

Meaning of equality in zeta regularization

It is known that $$\sum\limits_{n = 1}^\infty{n = 1 + 2 + 3 + \cdots} = \infty$$ but it is also known that $$\sum\limits_{n = 1}^\infty{n = 1 + 2 + 3 + \cdots} = -\frac{1}{{12}}$$ which can obtained ...
9
votes
2answers
718 views

Values of the Riemann Zeta function and the Ramanujan Summation - How strong is the connection?

The Ramanujan Summation of some infinite sums is consistent with a couple of sets of values of the Riemann zeta function. We have, for instance, $$\zeta(-2n)=\sum_{n=1}^{\infty} n^{2k} = 0 ...