Estimating the sum of fourth powers of a recursively defined sequence, $a_{n+1} = a_n - 2a_n^2$ with $a_1 = \beta$

Question

Fix $\beta\in (0,\tfrac{1}{2})$ and define recursively $a_{n+1} = a_n - 2a_n^2$ with $a_1 = \beta$. It is not hard to show that $a_n\to 0$. However, apparently $\sum_{k=1}^\infty a_k^4 \leqslant \frac{\beta}{8}$. I don't think any comparison with a geometric series would do the job as it would force $\beta$ to appear in the denominator. Any ideas for this estimate?

Answer 1

Note that $a_{n+1} - a_n = -2 a_n^2$. Squaring and summing we obtain \begin{align} 4\sum_{n=1}^\infty a_n^4 &= \sum_{n=1}^\infty (a_{n}^2 + a_{n+1}^2 - 2a_{n+1} a_n) \\&= \sum_{n=1}^\infty (a_{n}^2 - a_{n+1}^2) + 2\sum_{n=1}^\infty a_{n+1} ( a_{n+1} - a_{n})\\ &= \beta^2 -4\sum_{n=1}^\infty a_{n+1} a_n^2 \\ & \le \beta^2\end{align} where the last line is because $a_n\ge 0$. So we obtain $$ \sum_{n=1}^\infty a^4_n \le \frac{\beta^2}4$$ which easily implies $\sum_{n=1}^\infty a^4_n \le \frac{\beta}8$.

May I ask, where does the problem come from?

Edit: I have another proof. Note that $\sum_{n=1}^\infty a_n^2 = \frac12 \sum_{n=1}^\infty (a_n - a_{n+1}) = \frac{\beta}2.$ Now since $a_n$ is decreasing, $ \sum_{n=1}^\infty a_n^4 \le (\sup_{n\ge1}a_n)^2\sum_{n=1}^\infty a_n^2 = \frac{\beta^3}2 \le \frac{\beta}8$, QED.