# $a_1=1, a_{n+1}=a_n+a_n^2$ for all $n \geq 1$. Prove that $\lim_{n \to \infty}\frac{1}{a_n}=0$.

Let $$\{a_n\}$$ be the sequence of real numbers such that $$a_1=1, a_{n+1}=a_n+a_n^2$$ for all $$n \geq 1$$. Prove that $$\lim_{n \to \infty}\frac{1}{a_n}=0$$.

Sol: $$\{a_n\}$$ is monotonically increasing so it either converges to a number greater than $$1$$ or it diverges to $$\infty$$ (monotonicity prevents it from oscillating too). If it converges to $$l$$ then $$l=l+l^2$$ implies $$l=0$$ which is a contradiction. Hence it diverges and $$\lim_{n \to \infty} a_n=\infty$$ implies $$\lim_{n \to \infty}\frac{1}{a_n}=0$$. Is this correct?

• Yeah, it's correct – Jakobian Mar 21 at 10:43

A simpler argument: it is clear that $$a_n \ge 1$$ for all $$n$$. An easy inductive argument gives: $$a_n \ge n$$ for all $$n$$. Hence
$$0 < \frac{1}{a_n} \le \frac{1}{n}$$
for all $$n$$.
$$a_n \gt 0 \space \forall n \\ \Rightarrow a_{n+1} \gt a_n^2 \space \forall n \\ \Rightarrow a_n \gt a_2^{2^{n-2}} \space \forall n>2 \\ \Rightarrow a_n \gt 2^{2^{n-2}} \space \forall n>2 \\ \Rightarrow \frac{1}{a_n} < \frac{1}{2^{2^{n-2}}} \space \forall n>2 \\ \Rightarrow \lim_{n \to \infty}\frac{1}{a_n}=0$$