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I want to show that the sequence $t_n$ is decreasing.

$t_1$=6

$t_{n+1}$=$\sqrt{2+3t_n}$

The sequence $(t_n)$ from n=1 to n= infinity

Now what i did was simply $t_{n+1}$=<$t_n$ So if i proceed and plug in equation $\sqrt{2+3t_n}$-$t_n$=<0 it seems that this inequality doesnt hold for $t_n$>=3 where 3 is a lower bound because the sequence is clearly decreasing. Can someone help me out..

Jack D'Aurizio
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Mark
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3 Answers3

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$f(x)=\sqrt{2+3x}$ is concave and increasing over $I=[2,6]$. Additionally, it is a contraction over $I$ (we have $\frac{1}{3}\leq f'(x)\leq\frac{5}{9}$). By the Banach fixed point theorem and concavity it follows that $t_n$ is decreasing towards the only solution of $f(x)=x$ in $I$.

enter image description here

Once you draw it, it is pretty clear: the blue curve is the graph of $f(x)$ and the purple line is the graph of $y=x$.

Jack D'Aurizio
  • 353,855
  • Who will ever understand what you wanted to show – Mark Dec 24 '17 at 19:26
  • I am tempted, but I will refrain from giving a harsh reply. What I want to show is the last line: t_n is decreasing towards the expected value. – Jack D'Aurizio Dec 24 '17 at 19:53
  • Haha dont do it, what i mean is that so much people in here give Phd answers which we as beginners never grasp or understand. Thats why it is important to stay in the frame of the question and try to use those definitions if ofcourse otherwise stated – Mark Dec 24 '17 at 20:12
  • @Mark Answers on MSE are meant to help (a) you, the OP, but also (b) others who may happen to research the same question later on and come upon this topic, quite possibly in different contexts with different backgrounds. Because of (b) it is advisable that you not dismiss offhand what is otherwise a good answer. – dxiv Dec 25 '17 at 07:43
  • @dxiv ok i will thank you sir can you comment on my last comment i put under your question... still waiting – Mark Dec 25 '17 at 15:56
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Hint:   using that $\sqrt{a}-\sqrt{b}=\cfrac{a-b}{\sqrt{a}+\sqrt{b}}\,$ for non-negative $a,b\,$:

$$ \require{cancel}\;t_{n+1}-t_n = \sqrt{2+3t_n} - \sqrt{2+3t_{n-1}}=\cfrac{\bcancel{2}+3t_n-(\bcancel{2}+3t_{n-1})}{\sqrt{2+3t_n} + \sqrt{2+3t_{n-1}}}=\cfrac{3(t_{n}-t_{n-1})}{\sqrt{2+3t_n} + \sqrt{2+3t_{n-1}}} $$

Since the denominator is positive, it follows that the differences between consecutive terms $t_{n+1}-t_n$ and $t_{n}-t_{n-1}$ have the same sign, and since $t_2 - t_1 \lt 0$ it follows that the entire sequence is decreasing.

dxiv
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  • Good, but did you see where the fault of my reasoning was??? – Mark Dec 24 '17 at 17:44
  • @Mark Your approach works if you choose the lower bound to be $t$ instead of $3$ where $t$ is the root of $t=\sqrt{2+3t},$, as explained in Michael's answer and comments. – dxiv Dec 24 '17 at 17:53
  • thank you!!, but also should i always use the root as a lower bound!! – Mark Dec 24 '17 at 18:43
  • @Mark It doesn't hurt to try (though it may not always simplify calculatons). Reasoning is that if the sequence converges then its limit must satisfy the equation $t=\sqrt{2+3t}$, which is obtained by passing the recurrence relation to the limit. And if the sequence is decreasing, then its limit is also a lower bound. Of course, you don't know that the sequence is in fact convergent and decreasing, so you can't use that in the proof directly. But you can still use it as a heuristic to "guess" what a sensible lower bound would be, and then prove that it is indeed a lower bound. – dxiv Dec 24 '17 at 18:50
  • yes so that is what i mean, i guessed that the sequence will decrease and i saw in my calculator that the limit will approach 3,56 and so according to the definition i used 3 as a lower bound for the induction proof to show that $t_{n+1}$-3>=0 and i could do that , but then i had to show the sequence decreasing and there i got into troubles because i was still using the statement $t_n$>=3 ... you see where it went wrong – Mark Dec 24 '17 at 19:05
  • @Mark where it went wrong It didn't "go wrong", it just didn't work out algebraically. That's because $,3,$ was not the best choice for a lower bound. The same difficulty would have been with trying to prove that $,\pi,$ is a lower bound for example, which it is in fact, but it doesn't follow directly from the quadratic inequality. – dxiv Dec 25 '17 at 18:30
  • yes i see, we see that the sequence decreases but to work that out with algebra i got stuck because i thought any lower bound could work to prove that the sequence is decreasing, but that would only work if i proved convergence then i can choose any lower bound below the limit of the sequence. Am i right? – Mark Dec 25 '17 at 18:35
  • @Mark The question asks to prove that the sequence is decreasing. Technically, you don't need to prove anything about a lower bound, unless your proof of monotonicity otherwise requires it. For example, my answer doesn't establish, or make use of, any lower bound. And if you wanted to prove convergence, instead, it would still suffice to prove that the sequence is decreasing since it is trivially bounded below by $0$. – dxiv Dec 25 '17 at 18:45
  • yes i see what you mean, but i want to stick to the teachers method, which was the fact that he said show in the in the inequality that tn+1-tn<=0 , but my lower bound of 3 got me in trouble so it was better for me to just choose the limit as a lower bound because in that case any number above the limit will statisfy the inequality and the limit statisfy it also but not 3 in order to show that the sequence decreases – Mark Dec 25 '17 at 18:53
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Since $$t_{n+1}-t_n=\sqrt{2+3t_n}-t_n=\frac{2+3t_n-t_n^2}{\sqrt{2+3t_n}+t_n}=\frac{\left(\frac{3+\sqrt{17}}{2}-t_n\right)\left(\frac{-3+\sqrt{17}}{2}+t_n\right)}{\sqrt{2+3t_n}+t_n},$$ it's enough to prove that $\frac{3+\sqrt{17}}{2}<t_n,$ which is true by induction because $$t_{n+1}-\frac{3+\sqrt{17}}{2}=\sqrt{2+3t_n}-\frac{3+\sqrt{17}}{2}=\frac{2+3t_n-\frac{9+17+6\sqrt{17}}{4}}{\sqrt{2+3t_n}+\frac{3+\sqrt{17}}{2}}=\frac{3\left(t_n-\frac{3+\sqrt{17}}{2}\right)}{\sqrt{2+3t_n}+\frac{3+\sqrt{17}}{2}}$$ and $t_1=6>\frac{3+\sqrt{17}}{2}.$

Michael Rozenberg
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  • As you see this inequality $$\frac{2+3t_n-t_n^2}{\sqrt{2+3t_n}+t_n}$$=<0 doesnt hold when we choose a lower bound of lets say $t_n$>=3 , because if we plug in 3 for $t_n$ we will get a positive number and that cannot be less than or equal to zero... – Mark Dec 24 '17 at 16:31
  • But $t_1=6$ and $t_n>\frac{3+\sqrt{17}}{2}$ hods by induction. See please the second part of my proof. – Michael Rozenberg Dec 24 '17 at 16:36
  • But the sequence is decreasing to approximatly 3,56 when n goes to infinity. Now according to definition we can choose a lower bound of less than 3,56 And i choose $t_n$=>3 so now why does the inequality of $t_{n+1}-t_n=<0$ doesnt hold for 3??? – Mark Dec 24 '17 at 16:40
  • The sequence converges to $\frac{3+\sqrt{17}}{2}$ exactly and i proved that $t_n>\frac{3+\sqrt{17}}{2}>3$. – Michael Rozenberg Dec 24 '17 at 16:42
  • yes i know, but why can i not choose 3 as lower bound?? According to definition i can but then then the inequality doest hold.. – Mark Dec 24 '17 at 16:43
  • Because if so then you can not solve your problem. Why do you look for wrong solutions if you have a right solution already? – Michael Rozenberg Dec 24 '17 at 16:44
  • what do you mean? You mean i cannot choose 3 as a lower bound to show that the sequence is decreasing? – Mark Dec 24 '17 at 16:45
  • Yes, of course! It does not help for your problem solving. – Michael Rozenberg Dec 24 '17 at 16:46
  • i dont get it? But then the inequality should hold but it doesnt if i plug in 3 so how can that be? – Mark Dec 24 '17 at 16:48
  • $3$ it's a lower bound, but it does not help. $\frac{3+\sqrt{17}}{2}$ helps. – Michael Rozenberg Dec 24 '17 at 16:49
  • so you mean i cannot plug in a 3??? To show the inequality holds?? – Mark Dec 24 '17 at 16:53
  • I think, you can not. At least, I don't see it. See please my solution. I think it's much better. – Michael Rozenberg Dec 24 '17 at 16:55
  • yes i see but now what you mean is that i should use the limit in this case (3+$\sqrt17$)/2 for inequalities in order to show if a sequence decreasing or increasing? – Mark Dec 24 '17 at 16:59
  • We proved that for all natural $n$ we have:$t_{n+1}<t_n$ and $t_n>\frac{3+\sqrt{17}}{2},$ which says that $t$ converges – Michael Rozenberg Dec 24 '17 at 17:01
  • yes but is that the method i should use from now on to prove a sequence is decreasing or increasing, so should i use the limit as benchmark to show that?? Because i thought i just have to use the lower bound or upper bound you see what i mean ??? – Mark Dec 24 '17 at 17:03
  • Now, you can find the limit. You can not use the limit before you proved that this limit exists. Now, you can wright $\lim\limits_{n\rightarrow+\infty}t_{n+1}=\lim\limits_{n\rightarrow+\infty}\sqrt{2t_n+3}$ and you can find the limit. – Michael Rozenberg Dec 24 '17 at 17:05
  • yes so first you prove the limit by induction where you can use the 3 as a lower bound .. and show that for all n $t_n$>=3 aand then determine with the exact limit number decreasing or increasing?? – Mark Dec 24 '17 at 17:08
  • My first line gives that $t_{n+1}<t_n$ because in the second line is proved by induction that $t_n>\frac{3+\sqrt{17}}{2}.$ – Michael Rozenberg Dec 24 '17 at 17:10
  • i see but you dont get me, i thought that if i choose a lower bound 3 because at first i made a guess that when n goes to infinity then it means that the limit will be (3+sqrt17)/2 , so then i thought i can prove by induction that $t_n$>=3 , which i could and that is true the inequality holds, but then i had problems wirh proving if the sequence decreases because i thought that i could still use $t_n$>=3 which i could not use , you see what i mean now – Mark Dec 24 '17 at 17:22
  • I understood. But this way was false. For the proof of $t_{n+1}>t_n$ the inequality $t_n>3$ does not help. – Michael Rozenberg Dec 24 '17 at 17:27
  • yes so i should change strategy and just choose the limit as a lower bound in the future so i dont come into trouble like this again!!! – Mark Dec 24 '17 at 17:35
  • Yes, it's better of course. But even this thing sometimes does not help. – Michael Rozenberg Dec 24 '17 at 17:38
  • how can that be??? – Mark Dec 24 '17 at 17:39
  • what strategy should i deploy that always works? – Mark Dec 24 '17 at 17:39
  • We have no the general strategy. It depend on the problem. See the dxiv 's solution. It's another idea, how to solve your problem. – Michael Rozenberg Dec 24 '17 at 17:42
  • i see i see !!! Thank you!!!!!! – Mark Dec 24 '17 at 17:53
  • You are welcome! Good luck! – Michael Rozenberg Dec 24 '17 at 17:53