TITLE: Find $k\in \mathbb N$ that maximizes $f(k)=\frac{k^2}{1.001^k}$ (no calculus or calculators)
QUESTION [8 upvotes]: Given: $k\in \mathbb N$ and $\displaystyle f(k)=\frac{k^2}{1.001^k}.$
Find: the value of $k$ that maximizes $f(k)$

This is a question asked in math contest precalculus level. No calculators allowed.
Using calculus and a calculator for logs it is easy to see that the answer is $k=2001$. But how could we address the problem using an approach feasible for a math competition at the precalculus level without calculators?
Certainly it is true that for the maximizer $k$ it holds that:
$$\frac{(k-1)^2}{1.001^{k-1}}<\frac{k^2}{1.001^k}\ \text{and}\ \ \frac{k^2}{1.001^k}>\frac{(k+1)^2}{1.001^{k+1}}$$
But my developments of this condition are not leading to a productive direction for a solution obtained without a calculator.
Hints and solutions are appreciated. Sorry if this is a duplicate.

REPLY [5 votes]: 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 $k=2001$

REPLY [5 votes]: 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 $k>2000$.