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Let $a_1 = 0$ and $$a_{n+1} = \sqrt{5+2a_n}, n>0$$ Show that $\lim_{n\to\infty}a_n$ exists and find the limit.

Manual computation shows that the limit will be approximately 3.5 but how would i go about proving it?

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I got rid of the confusion between $A$ and $a$ when I edited, but I left the confusion between $a(n)$ and $a_n$; you might want to decide which you really want. –  Brian M. Scott Jan 12 '13 at 22:00
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3 Answers

up vote 6 down vote accepted

If you can show that the sequence converges, it’s easy to discover what the limit must be. Suppose that $\langle a_n:n\in\Bbb Z^+\rangle$ does converge, and let $L$ be the limit. Then

$$L=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\sqrt{5+2a_n}=\sqrt{5+2\lim_{n\to\infty}a_n}=\sqrt{5+2L}\;,$$

since the function $f(x)=\sqrt{5+2x}$ is continuous. Thus, $L$ satisfies the quadratic $L^2=5+2L$, and we must have

$$L=\frac{2\pm\sqrt{24}}2=1\pm\sqrt6\;.$$

Evidently $a_n\ge 0$ for all $n\in\Bbb Z^+$, so in fact $L=1+\sqrt6$.

One way to show that the sequence converges is to show that it is monotone increasing and bounded. This can be proved by induction on $n$. Suppose that $0\le a_n<1+\sqrt6$; can you carry out the induction step to show that $$0\le a_n\le a_{n+1}<1+\sqrt6\;?$$

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First show that $0 \leq a(n) \leq 1 + \sqrt6$ by induction.

This is because, we have$$a(n+1) = \sqrt{5+2a(n)} \leq \sqrt{5+2 + 2 \sqrt6} = \sqrt{(1+\sqrt6)^2} = 1 + \sqrt6$$

Now show using induction that $a(n)$ is an increasing sequence. The same procedure as above will work.

Hence, now we have a monotone increasing sequence that is bounded above. Hence, $a(n)$ converges to a limit. Hence, if the limit is $L$, then we have $$\lim_{n \to \infty} a(n+1) = \lim_{n \to \infty} \sqrt{5+2a(n)}$$ i.e. $$L = \sqrt{5+2L} \implies L = 1 + \sqrt6 \,\,\,\,\,\,\,\,\,\, (\because \sqrt{x}\text{ is continuous for }x>0)$$

The choice of $1+\sqrt6$ in the very first step was of-course by reverse engineering.

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I am actually interested in this problem since I have never made a induction proof earlier. So here is my try.

Since we don't have a closed form of $a_{n}$ we instead prove that $a_{n+1}<1+\sqrt{6}$ since it is obvious that $a_{n}\leq a_{n+1}$.

First step: Since $n>0$ we want to test that the statement holds for $n=1$:

$a_{2}=\sqrt{5}<1+\sqrt{6}$ since we are given that $a_{1}=0$.

Inductive step: Assume now that $n=k$ holds (for arbitrary $k>1,k\in \mathbb{N}$)

If we now can prove that for $n=k+1$ also is true, then we are done with the proof.

$a_{k+2}=\sqrt{5+2a_{k+1}}$ = $\sqrt{5+2\sqrt{5+2a_{k}}}$. From here I not sure how to proceed because I don't know where I am supposed to end.

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Can someone fill in and give a hint of my try above? –  EricAm Jan 13 '13 at 18:28
    
You can bump the question by editing your initial answer itself. No need to add a new answer. –  user17762 Jan 13 '13 at 23:12
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Haven't I already answered your question in my answer above? –  user17762 Jan 13 '13 at 23:13
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