Answered: assume that each sequence converges and…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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assume that each sequence converges and find
its limit. a1 = -4, an+1 = √(8 + 2an)

Expert Solution

Step 1

The given sequence $a_{1} = - 4$ and $a_{n + 1} = \sqrt{8 + 2 a_{n}}$ .

We have to find the limit of the sequence.

Step 2

The given sequence $a_{1} = - 4$ and $a_{n + 1} = \sqrt{8 + 2 a_{n}}$ .

$a_{2} = \sqrt{8 + 2 a_{1}} = \sqrt{8 + 2 (- 4)} = 0 a_{3} = \sqrt{8 + 2 a_{2}} = \sqrt{8 + 2 (0)} = 2.8284 a_{4} = \sqrt{8 + 2 a_{3}} = \sqrt{8 + 2 (2.8284)} = 3.6955 a_{5} = \sqrt{8 + 2 a_{4}} = \sqrt{8 + 2 (3.6955)} = 3.9231 . . . . . .$

The sequence $a_{n}$ is convergent, hence assume that $\lim_{n \to \infty} a_{n} = k$ .

Since $\lim_{n \to \infty} a_{n} = k$ , the limit of $\lim_{n \to \infty} a_{n + 1} = k$ .

Now,

$\begin{array}{rcl} a_{n + 1} & = & \sqrt{8 + 2 a_{n}} \\ \lim_{n \to \infty} (a_{n + 1}) & = & \lim_{n \to \infty} \sqrt{8 + 2 a_{n}} \\ k & = & \sqrt{8 + 2 k} \\ k^{2} & = & 8 + 2 k \\ k^{2} - 2 k - 8 & = & 0 \\ k^{2} + 2 k - 4 k - 8 & = & 0 \\ k (k + 2) - 4 (k + 2) & = & 0 \\ (k - 4) (k + 2) & = & 0 \\ k - 4 & = & 0, k + 2 = 0 \\ k & = & 4, k = - 2 \end{array}$

The limit is either $k = 4$ or $k = - 2$ .

From the sequence terms, the sequence is increasing and positive.

Thus, the sequence does not converge to $k = - 2$ .

Therefore the sequence converges to $k = 4$ .

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