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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?

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    $\begingroup$ Yeah, it's correct $\endgroup$ – Jakobian Mar 21 at 10:43
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Yes, your reasoning is correct.

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 $n$.

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Alternative argument:

$a_n \gt 0 \space \forall n \\ \Rightarrow a_{n+1} \gt a_n^2 \space \forall n \\ \Rightarrow a_n \gt a_2^{2^{n-2}} \space \forall n>2 \\ \Rightarrow a_n \gt 2^{2^{n-2}} \space \forall n>2 \\ \Rightarrow \frac{1}{a_n} < \frac{1}{2^{2^{n-2}}} \space \forall n>2 \\ \Rightarrow \lim_{n \to \infty}\frac{1}{a_n}=0$

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