1. ## 2 more questions about sequences..

i have a sequence which is defined by the following:

a(1) ∈ (0, 0.5)

a(n+1) = a(n) - 0.5a(n)^(n+1)

1: Is it possible to form the ''equation'' of the sequence a(n) with the above conditions? if yes, how?
2: with n -> infinite, is it possible to find a limit? Aren't there more than one possibilities?

sorry for the fuzz..

2. ## Re: 2 more questions about sequences..

Originally Posted by nappysnake
i have a sequence which is defined by the following:

$a_1 \in (0, 0.5)$

$a_{n+1} = a_n - 0.5a_n^{n+1}$

1: Is it possible to form the ''equation'' of the sequence $(a_n)$ with the above conditions? if yes, how?
2: with $n \to\infty$, is it possible to find a limit? Aren't there more than one possibilities?
This is an interesting sequence!

The answer to 1) has to be No: it is not possible (as far as I can see) to give an explicit formula for $a_n.$ You just have to work with the recursive formula that gives you $a_{n+1}$ in terms of $a_n.$

For 2) also, it does not seem possible to find the limit explicitly. The best that can be done is show that the limit must exist, and that it can be narrowed down to lie in some interval. The value of the limit will depend on the first term of the sequence.

From the equation $a_{n+1} = a_n - \tfrac12a_n^{n+1}$, you can show by induction that $a_{n+1} and also that $a_{n+1}>0.$ Thus $(a_n)$ is a decreasing sequence of positive terms. There is a theorem which says that any decreasing sequence that is bounded below must converge to a limit. So we know that this sequence must converge to some limit $\ell\geqslant0.$ The big question is whether this limit is actually equal to 0, or whether the sequence converges to a strictly positive limit. In fact, it is true that $\ell>0.$

To see that, write $a=a_1$ for the first term of the sequence, so that $0 Since the sequence decreases, it follows that $a_n for all n. Therefore $a_n-a_{n+1} = \tfrac12a_n^{n+1} < \tfrac12a^{n+1}$ for each n=1,2,3,... . It follows that

\begin{aligned}a-a_n &= (a_1-a_2)+(a_2-a_3)+\ldots+(a_{n-1}-a_n) \\&< \tfrac12(a^2+a^3+\ldots+a^n) = \frac{a^2- a^{n+1}}{2(1-a)}\end{aligned}

(sum of geometric series). If you now let $n\to\infty$ on both sides, you find that $a-\ell\leqslant \frac{a^2}{2(1-a)}.$ Therefore

$\ell\geqslant a-\frac{a^2}{2(1-a)} = \frac{a(2-3a)}{2(1-a)} > 0.$

Thus the limit of the sequence is strictly positive, and lies between $\tfrac{a(2-3a)}{2(1-a)}$ and $a.$ I don't see any way of narrowing it down more closely than that.

3. ## Re: 2 more questions about sequences..

could you please show me how you perform induction in the above sequence?? everything else is pretty much clearer now, thanks!

4. ## Re: 2 more questions about sequences..

Originally Posted by nappysnake
could you please show me how you perform induction in the above sequence?? everything else is pretty much clearer now, thanks!
It is a very simple induction, to prove the fact that $0 For n=1 that is true by the initial assumption on $a_1.$ Since $a_{n+1} = a_n - \tfrac12a_n^{n+1} = a_n(1-\tfrac12a_n^n)$, it follows from the inductive hypothesis $0 that $0 (because the factor $1-\tfrac12a_n^n$ certainly lies between 0 and 1). That completes the inductive step.

That is the only inductive proof that is needed. Once you know that $a_n>0$ it immediately follows from the equation $a_{n+1} = a_n - \tfrac12a_n^{n+1}$ that $a_{n+1}, and therefore the sequence $(a_n)$ is decreasing.