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let $a\in \mathbb{R}:a>0$ and $k\in \mathbb{N}_{\geq 2}$. Furthermore $(a_n)_{n\geq 1} \subset \mathbb{R}$: $$a_1=1+a \text{ and } a_{n+1}=a_n\left(1+\frac{a-a_n^k} {ka_n^k}\right) \text{ for } n\in \mathbb{N}$$

Now I have to show that

$$\sqrt[k]{a}\leq a_{n+1}\leq a_n$$

for $n\in \mathbb{N}$. I have been looking at this problem for quite some time now, and I don't understand what's going on here. How do I know what $a_n$ even is? And where do I even start with the proof? I have never encountered that kind of sequence.

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    $\begingroup$ Did you try induction ? $\endgroup$ – Surb Jun 1 at 10:50
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    $\begingroup$ @ParabolicAlcoholic are you having trouble with the induction itself or aren't you familiar with induction? $\endgroup$ – Viktor Glombik Jun 1 at 10:54
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    $\begingroup$ I am familiar with the idea of induction, but I don't see how induction can be applied here. $\endgroup$ – ParabolicAlcoholic Jun 1 at 10:57
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    $\begingroup$ @ParabolicAlcoholic When something is defined by a recursion, using inductoin to show properties of that is usually the natural choice. Give it a try to show $a_{n+1}\le a_n$ from $a_n\ge\sqrt[k]a>0$, for a start $\endgroup$ – Hagen von Eitzen Jun 1 at 11:13
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    $\begingroup$ Sorry, but I am not making any progess. $\endgroup$ – ParabolicAlcoholic Jun 1 at 13:55
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let $a\in \mathbb{R}:a>0$ and $k\in \mathbb{N}_{\geq 2}$. Furthermore $(a_n)_{n\geq 1} \subset \mathbb{R}$: $$a_1=1+a \text{ and } a_{n+1}=a_n\left(1+\frac{a-a_n^k} {ka_n^k}\right) \text{ for } n\in \mathbb{N}$$ We wish to show that $\forall n \in \mathbb N$: $$\sqrt[k]{a}\leq a_{n+1}\leq a_n$$

We will induct on $n$, let's start with $n=1$, we need to show that: $$\sqrt[k]{a}\leq a_{2}\leq a_1$$ Observe indeed that this means: $$a_2=a_1\left(1+\frac{a-a_1^k} {ka_1^k}\right)=(1+a)\left(1+\frac{a-(1+a)^k} {k(1+a)^k}\right)=(1+a)(1 + \frac{a}{k(1+a)^k} - \frac{1}{k})$$ we know that $k\geq 2$ so $\frac{1}{k } \leq 2$ and since $a>0$ also $ \frac{1}{(1+a)^k} \leq \frac{1}{(1+a)^2}$ $$a_2 \leq (1+a)(1+ \frac{a}{2(1+a)^2} -\frac{1}{2})=(1+a) + \frac{a}{2(1+a)}-\frac{1}{2} $$ $$=(1+a) + \frac{1}{2(\frac{1}{a}+1)}-\frac{1}{2}<(1+a)+ \frac{1}{2}- \frac{1}{2}=1+a_1$$ so $a_2 \leq a_1$. Now we also need to prove that $\sqrt[k]{a} \leq a_2$ or $a \leq a_2^k$, we get: $$a_2^k=(1+a)^k(1 + \frac{a}{k(1+a)^k} - \frac{1}{k})^k > (1+a)^k(1+ \frac{a}{k}-\frac{1}{k})^k = (1+a+\frac{a}{k} +\frac{a^2}{k}-\frac{a}{k}-\frac{1}{k})^k $$ $$a_2^k > (1+a+\frac{a^2}{k}-\frac{1}{k})^k $$ We now use that $(1+x)^k \geq 1 + kx$, for natural numbers $k$ and positive $1+x$. We know the smallest value for $k$ is $2$, so $(1+a+\frac{a^2}{k}-\frac{1}{k})>(1+a-\frac{1}{2})>\frac{1}{2}+a>0$, we thus get by this Bernoulli estimate: $$a_2^k \geq 1 + k(a+\frac{a^2}{k}-\frac{1}{k})=ak + a^2> 2a +a^2 >a$$ And thus: $$\sqrt[k]{a}\leq a_{2}\leq a_1$$

Now suppose for some natural number $l$, we have the induction hypothesis: $$\sqrt[k]{a}\leq a_{l}\leq a_{l-1}$$

Your task, using this, will be to show that now we also have: $$\sqrt[k]{a}\leq a_{l+1}\leq a_{l}$$ This would complete the proof. Good luck! happy estimating.

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    $\begingroup$ You're a genius! I will be studying your beautiful answer! :) $\endgroup$ – ParabolicAlcoholic Jun 2 at 13:07
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    $\begingroup$ Notice this was only the base case, you still need to estimate to make the step from $l$ to $l+1$, as hinted by Hagen von Eitzen. $\endgroup$ – Wesley Strik Jun 2 at 13:08
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    $\begingroup$ So what we KNOW to be true is that the hypothesis holds, then our task will be to show that one step later the inequality still holds. You will have to combine the hypothesis together with the definition of your $a_{n+1}$ $\endgroup$ – Wesley Strik Jun 2 at 13:10
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    $\begingroup$ Yeah, might get back to you on that... Is it also that complex? :D $\endgroup$ – ParabolicAlcoholic Jun 2 at 13:11
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    $\begingroup$ I feel like this problem overwhelms me $\endgroup$ – ParabolicAlcoholic Jun 2 at 13:16

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