2
votes
0answers
123 views

Good book on analytic continuation?

For my Bachelor's thesis, I am investigating divergent series summation methods. One of those is analytic continuation. There are quite a few books on complex analysis that include a chapter or two on ...
0
votes
0answers
117 views

An identity about Dirichlet $\eta$ Function

We know the Dirichlet $\eta$-function is defined as the analytic continuation of $$\eta(s) = \sum_{i=1}^\infty \frac{(-1)^{n-1}}{n^s} \quad \Re(s)>0$$ I find an identity for the values of this ...
4
votes
1answer
304 views

Clarkson's Proof of the Divergence of Reciprocal of Primes

In Tom Apostol's book, he credits the proof of the divergence of the sum of reciprocal of primes to Clarkson. To begin, we assume $\{p_n\}$ is an enumeration of the primes and ...
5
votes
4answers
483 views

Evaluating $\sum\limits_{n=2}^{\infty} \frac{1}{ GPF(n) GPF(n+1)}\,$, where $\operatorname{ GPF}(n)$ is the greatest prime factor

$\operatorname{ GPF}(n)=$Greatest prime factor of $n$, eg. $\operatorname{ GPF}(17)=17$, $\operatorname{ GPF}(18)=3$. $\operatorname{ LPF}(n)=$Least prime factor of $n$, eg. $\operatorname{ ...
8
votes
3answers
192 views

For what $t$ does $\lim\limits_{n \to \infty} \frac{1}{n^t} \sum\limits_{k=1}^n \text{prime}(k)$ converge?

The average of all primes is $$\lim\limits_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \text{prime}(k) ,$$ which diverges. What is the smallest $r$ such that for $t>r$, $$\lim_{n \to \infty} ...
5
votes
1answer
330 views

Are there complex Bernoulli numbers?

I am aware of the generalized Bernoulli numbers, but these are not what I'm looking for. I was wondering if there exists such a thing as fractional, real or even complex Bernoulli numbers ( $B_z$ for ...
9
votes
2answers
797 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 ...