Let a_1 = 1 and a_(n+1) = (1 - 1/(4n^2))a_n, for n greater than or equal to 1. (a) Show that...

Question:

Let {eq}a_1 = 1 {/eq} and {eq}a_{n+1} = (1-\frac{1}{4n^2})a_n {/eq}, for {eq}n \geq 1 {/eq}.

(a) Show that {eq}\displaystyle \lim_{n \rightarrow \infty} a_n {/eq} exists.

(b) What do you think that {eq}\displaystyle \lim_{n \rightarrow \infty} a_n {/eq} is?

Determining Limits of Sequences:

This problem involves finding the limit of a given sequence. The idea is to use the ratio test or simply analyze the sequence as it keeps decreasing. If the ratio of the term and the preceding term is always less than unity, then the sequence must converge. Using this idea we also try to find the limiting value of the sequence.

Answer and Explanation:

Given, a sequence

{eq}\displaystyle a_1 = 1 {/eq} and {eq}\displaystyle a_{n+1} = (1 - \frac{1}{4n^2} ) a_n {/eq} for n greater than 1

(a) We need to show that {eq}\displaystyle \lim_{n \to \infty} a_n {/eq} exists.

If we analyze the sequence properly, we can observe that each subsequent term eventually is multiplied by a number which is less than 1. Thus, since we multiply a number less than 1, infinite times while finding the limit of {eq}\displaystyle a_n {/eq} we know that it always is less than 1 and thus converges.

(b) Determine value of {eq}\displaystyle a_n {/eq}

Since a number smaller than 1 is multiplied infinite times, it makes sense that the number {eq}\displaystyle a_n {/eq} must go to {eq}\displaystyle 0 {/eq} as {eq}\displaystyle \lim_{n \to \infty} x^n = 0 {/eq} when {eq}\displaystyle x < 1 {/eq}

Even if {eq}\displaystyle x = 0.999999999999 {/eq} , this holds.

Thus, we can say that {eq}\displaystyle \lim_{n \to \infty} a_n \to 0 {/eq}


Learn more about this topic:

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How to Determine the Limits of Functions

from Math 104: Calculus

Chapter 6 / Lesson 4
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