On the Riemann-Zeta Function!

Some extra-explanation to @Michael Mendrin 's solution:

First of all, note that: $\zeta(2) + \zeta(3) + \ldots + \zeta(n+1)$ $= \sum_{p=1}^\infty \left( \dfrac{1}{p^2} + \dfrac{1}{p^3} + \ldots + \dfrac{1}{p^{n+1}} \right) = \sum_{p=1}^\infty \dfrac{1}{p^2} \left( 1 + \dfrac{1}{p} + \dfrac{1}{p^2} + \ldots + \dfrac{1}{p^{n-1}} \right)$ $= n + \sum_{p=2}^\infty \dfrac{1}{p^2} \cdot \dfrac{1 - \frac{1}{p^n}}{1 - \frac{1}{p}} = n + \sum_{p=2}^\infty \dfrac{p^n - 1}{p^{n+1}(p-1)} \quad ...(1)$

Since $\dfrac{p^n - 1}{p^{n+1}(p-1)} = \dfrac{1}{p(p-1)} - \dfrac{1}{p^{n+1}(p-1)}$

and $\displaystyle \sum_{p=2}^\infty \dfrac{1}{p(p-1)} = \sum_{p=2}^\infty \left( \dfrac{1}{p-1} - \dfrac{1}{p} \right) = 1$ ,

We obtain: $\displaystyle \sum_{p=2}^\infty \dfrac{p^n - 1}{p^{n+1}(p-1)} = 1 - \sum_{p=1}^\infty \dfrac{1}{p(p+1)^{n+1}} \quad ...(2)$ .

From $(1)$ and $(2)$ , we obtain:

$\ln(f(n)) = n^\alpha \ln \left( 1+\dfrac1n - \dfrac1n \sum_{p=1}^\infty \dfrac{1}{p(p+1)^{n+1}} \right)$

Since $g(x) = \dfrac{1}{x^{n+1}}$ is concave up on $[1, \infty)$ , we have:

$0 \leq \sum_{p=1}^\infty \dfrac{1}{p(p+1)^{n+1}} \leq \sum_{p=1}^\infty \dfrac{1}{(p+1)^{n+1}} \leq \int_1^\infty \dfrac{dx}{x^{n+1}} = \dfrac{1}{n}$

Hence, $\ln(f(n)) = n^\alpha \ln \left(1 + \dfrac1n + O \left( \dfrac1n \right) \right) \approx n^{\alpha - 1}$ as $n \to \infty$ .

So, when $\alpha = \dfrac12; \ln(f(n)) = 0 \Rightarrow f(n) =B = 1$ .

So, when $\alpha = 1; \ln(f(n)) = 1 \Rightarrow f(n) = A = e$ .

So $A+B = e+1 \approx \boxed{3.718}$ .

First, let’s consider the sum as $n \rightarrow \infty$

$\zeta \left( 2 \right) +\zeta \left( 3 \right) +\zeta \left( 4 \right) +...+\zeta \left( n+1 \right)$

which can be re-arranged and expressed as follows

$n+\displaystyle \sum _{ k=2 }^{ \infty }{ \displaystyle \sum _{ n=2 }^{ \infty }{ \frac { 1 }{ { k }^{ n } } } } =n+1$

Hence, we have

$f\left( n \right) ={ \left( \dfrac { n+1 }{ n } \right) }^{ { n }^{ a } }$

which gets us the answer we seek

$f\left( n,1 \right) +f\left( n,\frac { 1 }{ 2 } \right) =e+1=3.718...$