Convergent Recursion

Calculus Level 1

The sequence $\{a_n\}$ follows the recursion $a^2_{n+1}=2a_n+3$ with $a_1=7.$

Determine $\displaystyle \lim_{n \to \infty} a_n$ .

The answer is 3.

1 solution

Chew-Seong Cheong
Apr 20, 2015

$\begin{aligned} L & = \lim_{n \to \infty} {a_n} \\ \lim_{n \to \infty} a_{n+1} & = \lim_{n \to \infty} a_n \\ \lim_{n \to \infty} {\sqrt{2a_n+3}} & = L \\ \sqrt{2L+3} & = L \\ L^2 & = 2L+3 \\ L^2-2L-3 & = 0 \\ (L+1)(L-3) & = 0 \\ \implies L & = \boxed{3} \end{aligned}$

Note that from $a_{n+1}^2 = 2a_n + 3$ and $a_1=7$ , $a_n > 0$ for all $n$ .

Moderator note:

Nicely done. For completeness, you should show why $L+1 \ne 0$ .

Since each term is square root of some number, that why L is not equal to -1. Right?

Puneet Pinku - 4 years ago

$a_{n+1}^2 = 2a_n + 3$ and $a_1 = 7$ , implies that $a_n > 0$ for all $n$ .

Chew-Seong Cheong - 4 years ago

Just one thing, a[n-1] =L??

Puneet Pinku - 4 years ago

Only when $n \to \infty$ .

Chew-Seong Cheong - 4 years ago

So you are saying that a(n) =a(n-1) when n is infinity.

Puneet Pinku - 4 years ago

@Puneet Pinku – Yes, when the series converges.

Chew-Seong Cheong - 4 years ago

@Puneet Pinku – Actually, it is because, when $n \to \infty$ , $n \pm a \to n$ for $a << \infty$ .

Chew-Seong Cheong - 4 years ago

Convergent Recursion

The answer is 3.

1 solution

Moderator note:

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