Find $k\in \mathbb N$ that maximizes $f(k)=\frac{k^2}{1.001^k}$ (no calculus or calculators)

We know that it has to satisfy $$\frac{k^2}{1.001^k} > \frac{(k+1)^2}{1.001^{(k+1)}}\implies 1.001k^2 > k^2+2k+1 \implies 0.001k^2-2k-1 > 0 \implies k^2-2000k -1000 > 0$$ so $k>2000$.

Now, on the second equation, $$\frac{k^2}{1.001^k} < \frac{(k+1)^2}{1.001^{(k+1)}}\implies 1.001k^2 < k^2+2k+1 \implies 0.001k^2-2k-1 < 0 \implies k^2-2000k -1000 < 0$$

so $k<2002$.

And then, the maximum value is when $k=2001$

The first inequality gives (cancel a factor of $1.001^{k-1}$) \begin{eqnarray*} 1001(k-1)^2 &<& 1000k^2 \\ k^2-2002k+1001 &<&0 \\ (k-1001)^2 &<&1001^2-1001 <1001^2 \\ k &<& 2002 \end{eqnarray*} Similarly the other inequality gives $k>2000$.