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Would anybody have a proof (or at least a sketch) of the following approximation?

Considering $1/\sum_{i=2}^n\zeta(i)$ for $n=25,50,100$ we get:

$1/\sum_{i=2}^{25}\zeta(i)=0.04000000004768466\dots$

$1/\sum_{i=2}^{50}\zeta(i)=0.020000000000000000355271\dots$

$1/\sum_{i=2}^{100}\zeta(i)=0.0100000000000000000000000000000000788860905221\dots$

so that, if I got this right, we could guess (at least for $n\in 25\mathbb{N^*}$) the approximation

$1/\sum_{i=2}^n\zeta(i)=\frac{1}{n}+10^{-2-8\left \lfloor{\frac{n}{25}}\right \rfloor}u_n$ where $u_n\in[0,1)$.

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    Well, $\sum_{i=2}^{n}\zeta(i)$ is very close to $n$ since $$\sum_{i\geq 2}\left(\zeta(i)-1\right)=1.$$ – 2017-02-06
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    Yes, good point! I'll try to see if I can get the answer with this, but if somebody has it then please feel free to post it. – 2017-02-06

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From the integral representation of the Riemann $\zeta$ function we have: $$\sum_{k\geq 2}\left(\zeta(k)-1\right) = \sum_{k\geq 2}\frac{1}{(k-1)!}\int_{0}^{+\infty}\frac{x^{k-1}}{e^x-1}e^{-x}\,dx=\int_{0}^{+\infty}e^{-x}\,dx=1 \tag{1}$$ hence: $$ \sum_{i=2}^{N}\zeta(i) = (N-1)+\sum_{i\geq 2}\left(\zeta(i)-1\right)-\sum_{i>N}\left(\zeta(i)-1\right)=N-\sum_{i>N}\left(\zeta(i)-1\right)\tag{2} $$ equals: $$ N-\left(2^{-N}+\frac{1}{2}3^{-N}+\frac{1}{3}4^{-N}+\frac{1}{4}5^{-N}+\ldots\right)\tag{3}$$ from which it is simple to derive accurate bounds for $\frac{1}{\sum_{i=2}^{N}\zeta(i)}\approx\frac{1}{N}$.