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 $\infty \cdot 0 = 0$.
 A: For $x \ne 0, 1$, we have
$$\ln{a_k} = 2 \ln{k} + \ln{x} + k \ln{(1-x^2)}$$
Now $\ln{x}$ is fixed and $\ln{(1 - x^2)}$ is a fixed negative number. Noting that $k \to \infty$ much faster than $\ln{k} \to \infty$, we see that $\ln{a_k} \to -\infty$, which is equivalent to stating that $a_k \to 0$.
The endpoint cases are easier to handle.

To make my statement more precise, let $a > 0$ and $b > 0$ and consider the function $f(k) = a k - b \ln{k}$; we wish to show that $\lim_{k \to \infty} f(k) = \infty$.
We can say that $$e^{f(k)} = \frac{e^{ak}}{k e^b}$$
Now apply L'Hospital's rule to find that 
$$\lim_{k\to\infty} e^{f(k)} = \lim_{k\to \infty} \frac{ae^{ak}}{e^b} = \infty$$ 
In particular, this implies that $f(k) \to \infty$.
A: 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 $(1 - \epsilon^2) < 1$. 
A: Not really. You (should) know that for any $\alpha >1$ and fixed $n$, $$\frac{k^n}{\alpha ^k}\to 0$$ as $k\to\infty$. The proof of this may be carried out by applying L'Hôpital's rule to $$\lim_{x\to +\infty}\dfrac{x^n}{e^x}$$ $n$ times, then noting $\alpha ^k=e^{k\log \alpha }$.
Now, since $x\in [0,1]$ we may write $x=1-\varepsilon$, where $1>\varepsilon >0$. Then $x^2=1-2(\varepsilon-\varepsilon^2)$ so that $1-x^2=2(\varepsilon-\varepsilon^2)<1$, since $\varepsilon-\varepsilon^2\leqslant \dfrac 1 4$ on $[0,1]$.
A: If $x=0$ or $x=1$ then $a_k = 0$. So, suppose $x \in (0,1)$. Then $\theta = 1-x^2 \in (0,1)$ also, and $a_k = x k \theta^k$.
We notice that $a_{k+1} = (1+\frac{1}{k}) \theta a_k$. If we choose $N$ large enough so that $(1+\frac{1}{k}) \theta \le \lambda <1$ for $k \ge N$, then we have $a_{k+1} \le \lambda a_k$, and hence $a_k \to 0$.
