One-liner mini-challenge [HP-71B]

[Gerson W. Barbosa]

(04-13-2015 11:40 PM)Paul Dale Wrote:  Your reference lists an expansion of Hurwitz zeta in terms of the Bernoulli polynomials valid for integer arguments such as here.

In fact one of the arguments I use is not integer, but the reason I wasn't able to use the integral representation of Hurwitz zeta on the HP-71B is that it requires Re(s) to be greater than 1. Anyway, I'm using an empirical approximation I'd found prior to checking the Wikipedia article. So far I have only the first few terms of the series, but they suffice for this application.

If Hurwitz zeta were available, then an exact solution would be possible:

\[\zeta\left ( \frac{1}{2} \right )= \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+
\frac{1}{\sqrt{5}}+\cdots+\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt​{n+2}}+\cdots\]

\[\zeta\left ( \frac{1}{2},n+1 \right )=\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+\frac{1}{\sqrt{n+3}}+\frac{1}{\sqrt{​n+4}}+\cdots\]

\[\therefore \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+
\frac{1}{\sqrt{5}}+\cdots+\frac{1}{\sqrt{n}}= \zeta \left ( \frac{1}{2} \right )- \zeta \left ( \frac{1}{2},n+1 \right )\]

We can use W|A to check our previous result for n = 34:

zeta(1/2) - zeta(1/2, 35)

Gerson.