5

Given $a_1 = 0.5 $ and $ a_{n+1} = a_n - a_n^2,$ determine the convergence of $ \sum_{n=1}^{\infty} a_n $ .

I have found out that $\lim_{n \to \infty} a_n = 0$, so the necessary condition holds, and d'Alembert cannot help me.

If I try Raabe criterion, I need something about $a_n$ asymptotic, but I could not find it.

Mittens
  • 46,352
Anton Shcherbina
  • 111
  • Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – cpiegore Mar 28 '23 at 16:08
  • What have you tried? Can you think of a way to bound the terms starting at $a_n$, either from above or below, such that you can determine their sum? You may not need the exact asymptotic behavior if you can find a useful bound. – mjqxxxx Mar 28 '23 at 16:28
  • It`s trivial if we have a proper bound , but .. how to find it out ? I got no idea – Anton Shcherbina Mar 28 '23 at 17:14
  • Does this answer your question? Find asymptotic of recurrence sequence. Essentially $a_n \approx 1/n.$ – Somos Mar 29 '23 at 00:01
  • Did you read the MSE question 506723 that I linked to? It is about the exact same sequence denoted by $x_n$. It is shown that $ x_n \ge \frac{1}{n + 1 + \log n + \gamma}$ and so $\sum a_n$ diverges by comparison to the Harmonic series. – Somos Apr 02 '23 at 21:16

3 Answers3

4

If you know that $a_n\to0$, then there exists $n_0$ such that $a_n<\frac12$ for all $n\ge n_0$. For such $n$ we have $1/(1-a_n)<2$, and $$ \frac1{a_{n+1}}=\frac1{a_n(1-a_n)}=\frac1{a_n}+\frac1{1-a_n}<\frac1{a_n}+2.\tag1 $$ Subtract $\frac1{a_n}$ from both sides, and create a telescoping sum from $n=n_0$ to $n=N-1$: $$ \frac1{a_{N}}-\frac1{a_{n_0}}=\sum_{n=n_0}^{N-1}\left(\frac1{a_{n+1}}-\frac1{a_n}\right)<2(N-n_0)\tag2 $$ Rearrange to find for all $N> n_0$: $$a_N>\frac1{\frac1{a_{n_0}}+2(N-n_0)},\tag3$$ hence $\sum a_n$ diverges by comparison to the harmonic series.

Note the final rearrangement (3) requires $a_n>0$ for all large $n$. Indeed, you should be able to prove $a_n\in(0,1)$ for all $n$.

grand_chat
  • 40,909
1

Here is another solution based on the following criteria:

Let $(a_n)_{n \geq 1}$ be a decreasing sequence of positive reals. Let \begin{align} b_n = \frac{1}{a_{n+1}} - \frac{1}{a_n}, n \geq 1 \end{align} If $\sum_na_n<\infty$, the $(b_n)$ is unbounded.

A proof of this can be found in this posting which incidentally uses Raabe's theorem.

Now, notice that $a_{n+1}=a_n-a^2_n$, $a_0=\frac12\in[0,1]$ defines monotone decreasing sequence of nonnegative numbers and that $a_n\xrightarrow{n\rightarrow\infty}0$. $$\frac{1}{a_{n+1}}-\frac{1}{a_n}=\frac{1}{a_n(1-a_n)}-\frac{1}{a_n}=\frac{1}{1-a_n}\leq 2$$ Hence, $\sum_na_n$ diverges.

Mittens
  • 46,352
0

Obviously $a_n$ decreases, and set $\epsilon=a_k$ for some $k$, then $a_{k+1}=a_k(1-a_k)=(1-\epsilon)a_k$ and $a_{k+2}>(1-\epsilon)a_{k+1}=(1-\epsilon)^2a_k$ since $a_{k+1}<a_k=\epsilon$, therefore the finite sum of $\sum_{j\geq k}a_j$ can approximate $a_k(\frac{1}{1-(1-\epsilon)})=a_k/\epsilon=1$, since $\epsilon$ is fixed. Then notice that this holds for any k (thus can take k sufficiently large), you have $\sum_{i\geq 1}a_i$ do not converge by the Cauchy criterion.

Q Lee
  • 108
  • @Gonçalo I mean by "finite sum" that the terms of this sum are finite, because the Cauchy criterion do not allow adding infinite terms together. – Q Lee Mar 29 '23 at 01:23