Suppose that $\{a_n\}$ be a sequence of real number and $a_1\in(0,1)$ and $a_{n+1}=a_n-a_n^2$. Show that $\sum a_n$ is a divergent series.

Suppose that $\{a_n\}$ be a sequence of real number and $a_1\in(0,1)$ and $a_{n+1}=a_n-a_n^2$. Show that $\sum a_n$ is a divergent series.

I know that $a_n$ are decreasing sequences. Now how we can proceed from here?

Solution 1:

It is straightforward to check that $(a_n)$ is positive and decreasing. From this, we get

$$ \frac{1}{a_{n+1}} = \frac{1}{a_n} + \frac{1}{1-a_n} \leq \frac{1}{a_n} + \frac{1}{1-a_1} $$

Writing $ c = \frac{1}{1-a_1}$ for simplicity and applying the above inequality repeatedly leads to

$$ \frac{1}{a_n} \leq \frac{1}{a_1} + c(n-1) \qquad\text{or equivalently}\qquad \frac{1}{c(n-1)+a_1^{-1}} \leq a_n. $$

So by the comparison test, $\sum a_n$ diverges.

Addendum. We can actually say more about the asymptotic behavior of $(a_n)$.

1. Since $(a_n)$ is positive and decreasing, it converges to some value $\ell \geq 0$. Taking limit to the recurrence relation then gives $\ell = \ell - \ell^2$, and solving this gives $\ell = 0$.

2. By the Stolz–Cesàro theorem, $$ \lim_{n\to\infty} \frac{a_n^{-1}}{n} = \lim_{n\to\infty} \frac{a_{n+1}^{-1} - a_n^{-1}}{(n+1) - n} = \lim_{n\to\infty} \frac{1}{1-a_n} = 1, $$ and so, $a_n^{-1} \sim n$.

3. By the Stolz–Cesàro theorem again, $$ \lim_{n\to\infty} \frac{a_n^{-1} - n}{\log n} = \lim_{n\to\infty} \frac{(a_{n+1}^{-1} - n - 1) - (a_n^{-1} - n)}{\log(n+1) - \log n} = \lim_{n\to\infty} \frac{a_n}{(1-a_n)\log(1+\frac{1}{n})} = 1, $$ and so, $a_n^{-1} = n + (1 + o(1))\log n$.

4. Define $\varepsilon_n = a_n^{-1} - n - \log n$. Then \begin{align*} \varepsilon_{n+1} - \varepsilon_n &= \frac{a_n}{1-a_n} - \log\left(1+\frac{1}{n}\right) \\ &= a_n - \frac{1}{n} + \frac{a_n^2}{1-a_n} + \left( \frac{1}{n} - \log\left(1+\frac{1}{n}\right) \right) \\ &= \frac{1}{n + \mathcal{O}(\log n)} - \frac{1}{n} + \mathcal{O}\left(\frac{1}{n^2}\right) \\ &= \mathcal{O}\left(\frac{\log n}{n^2}\right). \end{align*} So it follows that $(\varepsilon_n)$ converges, and hence, $a_n^{-1} = n + \log n + \mathcal{O}(1)$.

5. Let $f(x) = x - x^2$ and define $C(x)$ by $$ C(x) = \lim_{n\to\infty} \left( \frac{1}{f^{\circ(n-1)}(x)} - n - \log n \right) $$ for $x \in (0, 1)$, where $f^{\circ n}$ denotes the $n$-fold composition of $f$. In light of the previous step, we know that this limit converges to a finite value. Moreover, \begin{align*} C(f(x)) &= \lim_{n\to\infty} \left( \frac{1}{f^{\circ(n-1)}(f(x))} - n - \log n \right) \\ &= \lim_{n\to\infty} \left( \frac{1}{f^{\circ n}(x)} - n - \log n \right) \\ &= \lim_{n\to\infty} \left( \frac{1}{f^{\circ (n-1)}(x)} - (n-1) - \log (n-1) \right) \\ &= C(x) + 1. \end{align*}