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Given a sequence $\{a_n\}$, with $n\geq1$, where $$a_{1}=4,$$ and $$a_{n+1}=\sqrt{a_{n} +20}.$$ Prove via induction that, for all $n \geq 1$, $$a_{n+1}>a_{n}.$$

  • Apparent Convergence

The sequence appears to be increasing, and possibly bounded at 5. How may I show convergence, and find its limit.

  • Regarding Notations

Additionally are there any beginners guide on using appropriate notations?

Thank you for your assistance.

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marked as duplicate by Martin Sleziak, user127.0.0.1, Davide Giraudo, M Turgeon, TMM Mar 4 at 20:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
so far im thinking: for large values of 'n': a_{n}= a_{n+1}= L. Then L = sqrt(L+20). solving we have L^2 - L - 20= 0. hence L=5 the upper limit? –  student101 Sep 24 '12 at 11:29

4 Answers 4

up vote 5 down vote accepted

First let's guess what the limit is (if exists): say $a_n\to\alpha$, then, by continuity of the operations $+$ and $\sqrt{\ \ }$, we have $$\alpha = \sqrt{\alpha+20},\ \text{ i.e., }\ \alpha^2=\alpha+20$$ Its roots are $-4$ and $5$. Since $a_n>0$ always, only $\alpha=5$ can be valid.

Now use induction, with additional hypothesis that $a_n\le 5$ to prove $a_{n+1}>a_n$ and still $a_{n+1}\le 5$.

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The induction proofs write themselves. We show first that $a_{n+1}\gt a_n$ for all $n$. The result is true at $n=1$. Suppose that we know that $a_{k+1}\gt a_k$ for some particular $k$. We need to show that $a_{k+2}\gt a_{k+1}$. This is straightforward: $$a_{k+2}=\sqrt{a_{k+1}+20}\gt \sqrt{a_k+20}=a_{k+1}.$$

We prove, again by induction, that $a_n\lt 5$ for all $n$. This is true at $n=1$. Suppose that we know that $a_k\lt 5$. We show that $a_{k+1}\lt 5$. This is easy: $$a_{k+1}=\sqrt{a_k+20}\lt \sqrt{25}=5.$$

The sequence $(a_n)$ is increasing and bounded above by $5$, so it has a limit. That the limit is $5$ is proved as in the answer by Berci Pecsi.

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EDIT: This is still false. For example, the sequence $4, 10, 10, 10, 10, \ldots$ satisfies $a_{n+1}>\sqrt{a_n+20}$ for all $n$, and $a_1=4$, but not $a_{n+1}>a_n$.

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sorry, missed the conditions of the sequence. now added. –  student101 Sep 24 '12 at 11:06
    
once again made an error >< should be an equal to sign for sequence –  student101 Sep 24 '12 at 11:24

Consider another sequence $b_{n}$ with the same definition

$$b_{n+1}=\sqrt{b_{n}+20}$$

Since the function $f(x)=\sqrt{x+20} $ is an increasing function,

If $b_{1}>a_{1}$ , then obvioulsy $b_{2}>a_{2}$ and so on an $b_{n}>a_{n}$ for all $n$ ,

Let $b_{1}=5$ , then $b_{n}=5$ for all n. Since $b_{1}>a_{1}$ , we have $b_{n}>a_{n}\implies5>a_{n}$ .

Since $a_{n}$ is monotonically increasing and has an upper bound, it must converge.

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