2, For $\frac{1}{1}$ , + $\frac{1}{2}$ , + $\frac{1}{3}$ , + $\frac{1}{4}$ , + ... eventually accumulates to two, thus two food samples need be prepared.

The required number is $\lim_{n\rightarrow\infty}H_{n}$ , where $H_{n}=\sum_{r=1}^n\frac{1}{r}$

This limit doesn't exist as a real number (as $H_n$ is monotonically increasing and unbounded above). In fact, $\lim_{n\rightarrow\infty}\left(H_n-log_en\right)=\gamma\approx0.577$ Where $\gamma$ is the Euler-Macheroni constant. Also, $H_5\approx2.2833>2.25$

Hence, the answer is undefined (as long as we limit our discussion to the real number system).

EDIT: I see that the options have been changed, the answer is now $\infty$

1+1/2+[1/3+1/4]+[1/5+1/6+1/7+1/8]+[1/9...

>

1+1/2+[1/4+1/4]+[1/8+1/8+1/8+1/8]+[1/16...

1+1/2+1/2+1/2+1/2...

We can see that this diverges, increasing at a rate of about 1 unit every time the number of terms quadruples.

Since this diverges, the former must also diverge, therefore, the answer is infinity.

(Proof invented by this guy )