<s>[INST] How can I show that this sequence converges to $0$? For $x \in [0,1]$, let $a_k = k^2x(1-x^2)^k$. How can I show that $\lim_{k \to \infty}a_k = 0$? I don't know what to do because $k^2 \to \infty$ and $(1-x^2)^k \to 0$, so I would have to prove that $\infty \cdot 0 = 0$.
 [/INST] We see that $\frac{a_{k+1}}{a_k} = \frac{(k+1)^2x(1-x^2)^{k+1}}{k^2x(1-x^2)^{k}} = (1 + \frac{2}{k} + \frac{1}{k^2})(1-x^2)$. 
Since $x < 1$, $1 - x^2 < 1$, say equal to $1 - \epsilon$, with $\epsilon > 0$.
Since $\lim_{k \to \infty} (1 + \frac{2}{k} + \frac{1}{k^2}) = 1$, we can take $k$ large enough so that $(1 + \frac{2}{k} + \frac{1}{k^2}) < 1 + \epsilon$. 
Hence $\frac{a_{k+1}}{a_k} \leq (1+ \epsilon)(1 - \epsilon) = 1 - \epsilon^2 < 1$ for large enough $k$. Say for $k > N$.
Then we have, for $k > N$, $a_k = a_N\frac{a_{N+1}}{a_N} \cdots \frac{a_k}{a_{k-1}} \leq a_N(1- \epsilon^2)^{k - N}$
This tends to zero clearly as $k \to \infty$, since $(1 - \epsilon^2) < 1$. 
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