$a_0 \in (0,1)$; $a_{k+1} = a_k - \frac{1}{2}a_k^2$; Want to find $\lim_{k \to \infty} ka_k$.
It is not too difficult to show $a_k \to 0$ as $k \to \infty$ using the monotone convergence theorem. But I cannot seem to make any further progress. Attempting to find $a_k$ in terms of $a_0$ by induction seems difficult.
First, one can easily show that $a_k \neq 0$ for every $k$. Hence, one has $$\frac{1}{a_{k+1}} - \frac{1}{a_k} = \frac{1}{a_k - \frac{1}{2}a_k^2} - \frac{1}{a_k}= \frac{1}{a_k} \left( \frac{1}{1-\frac{1}{2}a_k}-1\right)$$
and because $a_k \rightarrow 0$, you get from here the following expansion $$\frac{1}{a_{k+1}} - \frac{1}{a_k} = \frac{1}{a_k} \left( 1 + \frac{1}{2}a_k + o(a_k)-1\right) = \frac{1}{2} + o(1)$$
Hence, the sequence $\displaystyle{\left(\frac{1}{a_{k+1}} - \frac{1}{a_k} \right)}$ tends to $\displaystyle{\frac{1}{2}}$ : you can now use Cesaro's theorem with this sequence, to get that $$\lim_{k \rightarrow +\infty} \frac{1}{ka_k} = \frac{1}{2}$$
and finally, $$\boxed{\lim_{k \rightarrow +\infty} ka_k = 2}$$
