# Justify an approximation of $-\sum_{n=2}^\infty H_n\left(\frac{1}{\zeta(n)}-1\right)$, where $H_n$ denotes the $n$th harmonic number

For integers $n\geq 1$, let $\mu(n)$ the Möbius function, see this MathWorld, and let $H_n$ the $n$th harmonic number as $H_n=1+1/2+\ldots+1/n$. Also we denote the Riemann zeta function with $\zeta(s)$.

Question. Do you know how to justify a good (five or six rights decimals) of $$-\sum_{n=2}^\infty H_n\left(\frac{1}{\zeta(n)}-1\right)\,?$$ Thanks you in advance.

I know that Abel's identity implies for real numbers $x>1$ that $$-\sum_{1\leq n\leq x} H_n\left(\frac{1}{\zeta(n)}-1\right)=\left(\sum_{1\leq n\leq x}H_n\right)\cdot\left(1-\frac{1}{\zeta(x)}\right)-\int_1^x\left(\sum_{1\leq n\leq t}H_n\right)\frac{\zeta'(t)dt}{(\zeta(t))^2},$$ but after some non-rigorous calculations I think that previous calculation doesn't provide me a way to find such approximation due that (I believe that is due it, that the main term of Abel's identity vanishes) $$\lim_{x\to\infty}(x\log x-x+\gamma x)\cdot\left(1-\frac{1}{\zeta(x)}\right)=0.$$ I think that we need asymptotic for harmonic numbers (I know Euler's theorem) and also asymptotics for the Riemann zeta function.

• $\frac{1}{\zeta(n)}-1$ decays to zero pretty fast, hence it is enough to compute a partial sum, without even invoking summation by parts or Abel summation. Is such constant ($\approx -1.221959874493$) relevant for some reason? – Jack D'Aurizio Feb 20 '18 at 15:22
• Many thanks @JackD'Aurizio I don't know a good reason to compute it, just the exercise to how calculate a good approximation (five or six decimals, justifying the method). – user243301 Feb 20 '18 at 15:28