The expression $\frac{k^{2}}{1.001^{k}}$ reaches its maximum value with which natural number $k$?
We are looking for the value of $k$ for which the term of the sequence

$$
\frac{1^{2}}{1,001}, \frac{2^{2}}{1,001^{2}}, \frac{3^{2}}{1,001^{3}}, \ldots, \frac{k^{2}}{1,001^{k}}, \frac{(k+1)^{2}}{1,001^{k+1}}, \ldots
$$

is maximum. We want to compare the sizes of any two terms of this sequence. The simplest way is to compare any term with the next one. It turns out that it is easier to compare two fractions with positive equal denominators. For example, the third term is greater than the second because

$$
\frac{2^{2}}{1,001^{2}}=\frac{2^{2} \times 1,001}{1,001^{3}}0$ for any $n$, to check if

$$
\frac{k^{2}}{1,001^{k}}=\frac{k^{2} \times 1,001}{1,001^{k+1}}2001>1000$. In particular,

$$
\frac{2000^{2}}{1,001^{2000}}\frac{2002^{2}}{1,001^{2002}}
$$

Thus, the given sequence strictly increases for $1 \leq k \leq 2001$ and then strictly decreases for $k \geq 2001$. Therefore, the largest term of this sequence is achieved with $k=2001$.