Proving divergence of series using a recursive relation

Question

I have been thinking for an hour about this problem but could not find any way to solve it. Let's $0\lt a_n \lt a_{n+1}+a_{n}^2$, prove that $\sum_{n=1}^{\infty}a_n$ is divergent.

Any hints and suggestions would be appreciated.

Thanks!

Answer 1

Suppose $\displaystyle\sum_{n=1}^{\infty}a_n$ converges. Then there exists an $N$ such that $0 < a_n < \dfrac{1}{2}$ for all $n \ge N$.

Define a new sequence $\{b_n\}_{n=N}^{\infty}$ by $b_N = a_N$ and $b_{n+1} = b_n-b_n^2$ for all $n \ge N$.

Clearly, $0 < b_N \le a_N$. Now suppose $0 < b_n \le a_n$ for some integer $n \ge N$.

Since $x-x^2$ is positive and increasing on $[0,\frac{1}{2}]$, $0 < b_{n+1} = b_n-b_n^2 \le a_n-a_n^2 < a_{n+1}$.

So by induction, we have $0 < b_n \le a_n$ for all $n \ge N$.

Therefore, by direct comparison, $\displaystyle\sum_{n=N}^{\infty}b_n$ also converges.

Since $b_n$ is an decreasing positive sequence and $\dfrac{1}{x^2}$ is decreasing for $x > 0$, we have:

$m-N = \displaystyle\sum_{n=N}^{m-1}1$ $= \displaystyle\sum_{n=N}^{m-1}\dfrac{b_n-b_{n+1}}{b_n^2}$ $= \displaystyle\sum_{n=N}^{m-1}\int_{b_{n+1}}^{b_n}\dfrac{1}{b_n^2}\,dx $ $\ge \displaystyle\sum_{n=N}^{m-1}\int_{b_{n+1}}^{b_n}\dfrac{1}{x^2}\,dx$ $= \displaystyle\int_{b_m}^{b_N}\dfrac{1}{x^2}\,dx$ $= \dfrac{1}{b_m} - \dfrac{1}{b_N}$.

Rearranging gives $\dfrac{1}{b_m} \le m-N+\dfrac{1}{b_N}$, i.e. $b_m \ge \dfrac{1}{m-N+\frac{1}{b_N}}$ for all $m \ge N$.

Hence, $\displaystyle\sum_{m=N}^{\infty}b_m$ diverges by limit comparison to $\displaystyle\sum_{m=N}^{\infty}\dfrac{1}{m}$. This is a contradiction.

Therefore, $\displaystyle\sum_{n=1}^{\infty}a_n$ diverges.

Answer 2

Suppose that the series $a_n$ is convergent. Then $a_n\to 0$, and there exists $N$ such that $0<a_n<1$ for $n\geq N$. We have for $n\geq N$ the inequality $\displaystyle \frac{a_{n+1}}{a_n}>1-a_n>0$. We get $\displaystyle \frac{a_{n+1}}{a_N}>(1-a_N)\cdots(1-a_n)$, and $$\sum_{k=N}^{n}(-\log(1-a_k))\geq \log a_N-\log(a_{n+1})$$ Now $-\log(1-a_k)\sim a_k$, the series are positive, hence the series $-\log(1-a_k)$ is convergent. Hence there exists $M$ such that $\log a_{n+1}\geq M$ for all $n\geq N$. This imply that $a_{n+1}$ do not $\to 0$ if $n\to +\infty$, a contradiction. Hence the series $a_n$ is divergent.