/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Are the following sequences \(z_... [FREE SOLUTION] | 91Ó°ÊÓ

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Are the following sequences \(z_{1}, z_{2}, \cdots, z_{n}, \cdots,\) bounded? Convergent? Find their limit points, (Show the details of your work.) $$z_{n}=(3+4 i)^{n} / n !$$

Short Answer

Expert verified
The sequence is bounded, convergent, and its limit point is 0.

Step by step solution

01

Simplify the Sequence

The sequence is given by \(z_n = \frac{(3+4i)^n}{n!}\). To analyze this sequence, we start by considering the behavior of the numerator and denominator separately. The numerator \((3+4i)^n\) grows quickly, but we need to compare it to the growth of the denominator \(n!\).
02

Determine the Limit of the Sequence

To find the limit of \(z_n\), consider the terms \(\left|\frac{(3+4i)^n}{n!}\right| = \frac{\left|3+4i\right|^n}{n!}\). First, calculate \(\left|3+4i\right| = \sqrt{3^2 + 4^2} = 5\). Then the sequence simplifies to \(\lim_{n \to \infty} \frac{5^n}{n!}\).
03

Apply the Limit Test

The ratio \(\frac{5^n}{n!}\) as \(n\) approaches infinity goes to zero because factorial \(n!\) grows faster than the exponential \(5^n\). So, \(\lim_{n \to \infty} z_n = \lim_{n \to \infty} \frac{5^n}{n!} = 0\).
04

Convergence and Boundedness

Since \(z_n\) approaches zero, the sequence is convergent. A sequence that converges must be bounded. Therefore, the sequence is both bounded and convergent. The only limit point of this sequence is 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bounded Sequences
A sequence is considered to be bounded if there exists a real number that serves as a limit beyond which the values of the sequence do not go. More formally, a sequence \(a_n\) is bounded if there exist two real numbers \(M\) and \(N\) such that \(N \leq a_n \leq M\) for all \(n\). For complex sequences, a sequence \(((z_n))\) is bounded if the absolute values of the terms \( |z_n| \) are bounded.

In the case of the sequence \( z_n = \frac{(3+4i)^n}{n!} \), the term \( \left|\frac{(3+4i)^n}{n!}\right| = \frac{5^n}{n!} \) as shown in the solution, becomes smaller as \(n\) grows larger. Since this value gets indefinitely close to zero and never exceeds some bound, we assert that \(z_n\) is bounded by any number greater than zero.
Convergent Sequences
A sequence \(((a_n))\) is said to be convergent if it approaches a specific value, called the limit, as \(n\) goes to infinity. Mathematically, a sequence \(a_n\) is convergent if there exists a number \(L\) such that for every \(\epsilon > 0\), there exists an integer \(N\) such that \(|a_n - L| < \epsilon\) for all \(n > N\).

In the exercise sequence \( z_n = \frac{(3+4i)^n}{n!} \), the limit was found by comparing the growth rates of \((3+4i)^n\) and \(n!\). While \( (3+4i)^n \) grows exponentially, \( n! \) grows faster. This led to the conclusion that \( \lim_{n \to \infty} \frac{5^n}{n!} = 0 \), implying \(z_n\) converges to 0.
Limit Points
Limit points, also known as accumulation points, of a sequence are values that the terms of the sequence get arbitrarily close to as the sequence progresses. A sequence can have multiple limit points, or just one, or even none, depending on its properties.

For the sequence given by \( z_n = \frac{(3+4i)^n}{n!} \), we established that the entire sequence converges to 0. This indicates that 0 is the only limit point of this sequence. To say a sequence has a limit point of 0 means that for any \( \epsilon > 0 \), there are infinitely many terms in the sequence within \( \epsilon \) distance of 0.
  • For convergent sequences like ours, the limit (0 in this case) is often the sole limit point.
  • Non-convergent sequences might have a set of limit points instead.

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