But since $ \lim_{n \to \infty} b_{n+1} = \lim_{n \to \infty} b_n = L $, we must have

February 22, 2026

$But since $ \lim_{n \to \infty} b_{n+1} = \lim_{n \to \infty} b_n = L $, we must have$

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First, observe that $ f(u) < u $ for all $ u > 0 $, since $ \frac{u^4}{4} > 0 $. Thus, the sequence $ b_n $ is strictly decreasing and bounded below by 0. Therefore, it converges to some limit $ L \geq 0 $.
Taking limits on both sides of the recurrence:
\lim_{n \to \infty} b_{n+1} = \lim_{n \to \infty} f(b_n) = f(L).
L = f(L) = L - \frac{L^4}{4} \Rightarrow \frac{L^4}{4} = 0 \Rightarrow L = 0.
Thus, the sequence converges to $ \boxed{0} $.
To confirm convergence, note that $ f(u) $ is continuous and decreasing on $ (0, \infty) $, and $ b_n \in (0,1] $ for all $ n $, so the sequence is monotonic and bounded, hence convergent.

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