Proof of a zeta function limit

Question

At following page from Mathworld I found an interesting limit but I can not get its proof as it figures as a personal communication from B. Cloître (B. Cloître, pers. comm., Oct. 4, 2005):

$$\gamma=\lim_{x\rightarrow\infty}\bigl(\zeta(\zeta(x))-2^x+(4/3)^x+1\bigr)$$

_{[eq. (36), typo $z\mapsto x$ corrected]}

Could you advise me the best way to get its proof, or even better providing it directly here?

Thanks in advance

Answer 1

Use the expansion $$\zeta(s)=\frac{1}{s-1}+\gamma+{\cal O}(s-1),$$ hence, since $\zeta(x)=1+2^{-x}+3^{-x}+4^{-x}+5^{-x}+\cdots$, you have $$\zeta(\zeta(x))=\frac{2^x}{1+(2/3)^x+(2/4)^x+(2/5)^x+\cdots}+\gamma+{\cal O}(2^{-x})$$ $$\qquad=2^x[1-(2/3)^x-(2/4)^x]+\gamma+R(x),$$ where $\lim_{x\rightarrow\infty} R(x)=0$. The desired limit follows, $$\lim_{x\rightarrow\infty}[\zeta(\zeta(x))-2^x+(4/3)^x+1-\gamma]=0.$$