How can I show the convergence and find limit.

Let $a_1=0$ and $$a_{n+1}=\frac{a_n^2+3}{2(a_n+1)}~,~n \geq 1$$ . Show that $\{\ a_n \}$ converges and find its limit.

Trial: Here \begin{align} a_1 &=0\\ \text{and}~ a_2 &=\frac{3}{2} \\ \text{and}~ a_3 &=\frac{21}{20} \end{align} . Here I find no relationship between $a_n ~\text{and}~ a_{n+1}$. Please help.

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I first saw the following slick method in Bartle and Sherbert's Introduction to Real Analysis (the slickness due to them, not me).

Prove that $(a_n)$ is bounded below and eventually decreasing as follows:

1) Using the relation $a_n^2-(2a_{n+1})a_{n}+(3-2a_{n+1})=0$ and the Quadratic formula, show that $a_{n+1}\ge1$ for $n>1$.

2) Use the result of 1) to show that the quantity $a_n-a_{n+1}$ is non-negative for $n>1$.

Of course, then, it follows that $(a_n)$ converges, and you can find the value of the limit using your recurrence relation.

The above is meant to be a (big) hint). The details are given below.

From the definition of $a_{n+1}$, you have $$a_n^2-(2a_{n+1})a_{n}+(3-2a_{n+1})=0.$$ So, $a_{n}$ is a solution of the quadratic equation $x^2-(2a_{n+1})x +(3-2a_{n+1})=0$. We must then have $$(2a_{n+1})^2-4\cdot 1\cdot(3-2a_{n+1})\ge0.$$ This implies that $$a_{n+1}^2 + 2a_{n+1}\ge3.$$ And then, as $a_{n+1}>0$, it follows that $a_{n+1}\ge1$ for $n>1$.

Now, consider $$a_n-a_{n+1} = a_n-{a_n^2+3\over 2(a_n+1)} ={(a_n+3)(a_n-1)\over 2(a_n+1)}.$$

Since $a_n\ge1$ for $n>1$, we have $a_n-a_{n+1}\ge 0$ for $n>1$.

We have shown that $(a_n)$ is an eventually decreasing sequence that is bounded below. It thus converges to some $L$. Using your recurrence relation, you can find the value of $L$.

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If (a_n) converges and L is the limit then when n is very large, $a_n$ and $a_{n+1}$ are both very close to L. Now think about your recurrence relation but with $a_n$ and $a_{n+1}$ both replaced by L

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this was not meant to be an answer –  Adam Rubinson Jan 22 '13 at 13:54
In this way I can find the limit but how I show that $\{\ a_n \}$ converges. –  Argha Jan 22 '13 at 14:02
Apply the epsilon definition for convergence to both a_n and a_n+1 in your original equation and fiddle around with it –  Adam Rubinson Jan 22 '13 at 14:14
Given epsilon there is an N such that |a_n - L| < epsilon for all n>N. Therefore we also have |a_n+1 - L| < epsilon for all n>N... –  Adam Rubinson Jan 22 '13 at 14:16

Okay what do we need for conergence? If we can show that the sequence is bounded and monotone we know that it converges.

• bounded: Obviously we have $0\leq a_n$. I will show that $a_n<10$ (which is rather arbitary) by induction. We have $a_0<10$. Now we make the step $n\rightarrow n+1$. \begin{align} a_{n+1} &= \frac{a_n^2+3}{2(a_n+1)} <\frac{a_n^2+2a_n+3}{2(a_n+1)} \\ &=\frac{(a_n+1)^2+2}{2(a_n+1)} = \frac{(a_n+1)^2}{2(a_n+1)}+\frac{2}{2(a_n+1)} \\ &= \frac{a_n+1}{2} + \frac{2}{2(a_n+1)} \leq^* \frac{10+1}{2}+\frac{2}{2(0+1)} \\ &= 5.5 +1 <10\end{align} Where we use $0\leq a_n<10$ at $^ *$.
• monotone: here we show, that $\forall n > 0$ we have $\displaystyle \frac{a_{n+1}}{a_n}<1$. I think this is something you should proof.

For the last step: We proofed that our sequence converges. So there exists an $a=\lim_{n\rightarrow \infty} a_n$ Knowing this, you can solve $a_{\infty} = \frac{a_{\infty}^2+3}{2(a_{\infty}+1)}$ for $a_{\infty}$.

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