Problem 13 - Precalculus with Limits

Question

In Exercises 13-20, (a) rewrite the sum as a rational function \(S(n)\), (b) use \(S(n)\) to complete the table, and (c) find \(\lim_{n \to \infty} S(n)\). $$\displaystyle\sum_{i=1}^{n} \dfrac{i^3}{n^4}$$

Step-by-Step Solution

Answer

The sum in terms of a rational function \(S(n)\) is \(\frac{(n+1)^2}{4n^2}\) and as \(n\) goes to infinity, the limit of \(S(n)\) is \(\frac{1}{4}\).

Step 1: Rewrite the sum as a rational function \(S(n)\)

Let's express the series sum \(\displaystyle\sum_{i=1}^{n} \frac{i^3}{n^4}\) as a rational function of \(n\). To do this, note that \(\displaystyle\sum_{i=1}^{n} i^3\) can be expressed using the formula \(\frac{n^2*(n+1)^2}{4}\). Hence we have, \[S(n) = \frac{1}{n^4} \displaystyle \sum_{i=1}^{n} i^3 = \frac{1}{n^4} \cdot \frac{n^2*(n+1)^2}{4} = \frac{(n+1)^2}{4n^2}\]

Step 2: Use \(S(n)\) to complete the table

The specific values for the table have not been provided in the exercise, but the value of \(S(n)\) can be calculated for any desired value of \(n\) by substituting the value of \(n\) into the rational function \(\frac{(n+1)^2}{4n^2}\). For example, for \(n=2\), \(S(2)=\frac{9}{16}\), for \(n=3\), \(S(3)=\frac{16}{9}\), and so on.

Step 3: Find \(\lim_{n \to \infty} S(n)\)

Next, the limit of \(S(n)\) as \(n\) approaches infinity is found by substituting \(n=\infty\) into the function: \[\lim_{n \to \infty} S(n) = \lim_{n \to \infty} \frac{(n+1)^2}{4n^2} = \frac{1}{4}\]. This is because as \(n\) approaches infinity, the \((n+1)^2\) and \(4n^2\) terms dominate, and their ratio approaches 1, hence, the function simplifies to \(\frac{1}{4}\).

Key Concepts

Infinite Series

In an infinite series, we deal with the sum of infinitely many terms. This series converges if the sum approaches a finite number as the number of terms increases.
In our exercise, the original sum is expressed as \(\sum_{i=1}^{n} \frac{i^3}{n^4}\). Here, we are not looking at an infinite number of terms right away, but we can use the concept of infinite series as \(n\) becomes very large.
Understanding infinite series helps us see the behavior of the sum when \(n\) approaches infinity. If it results in a finite number, the series is convergent. If not, it's divergent.
In our example, we see that as \(n\) grows really large, the infinite series approaches a specific value suggesting convergence.

Limit Calculation

Limit calculation is a fundamental concept in calculus used to determine the value that a function approaches as the input approaches a particular point, often infinity.
It helps us understand the behavior of functions over very large or tiny scales. We calculate limits to predict exact results when direct computation is not feasible.
In the textbook example, the goal is to calculate \(\lim_{n \to \infty} S(n)\), examining how the rational function \(S(n) = \frac{(n+1)^2}{4n^2}\) behaves as \(n\) becomes infinitely large.
Here, we observe that both the numerator \((n+1)^2\) and denominator \(4n^2\) grow large, but the leading terms dictate the limit. Thus, we simplify the function to deduce that it approaches \(\frac{1}{4}\).

Power Series

A power series is an infinite series of the form \(\sum_{n=0}^{\infty} a_n x^n\), where the terms include powers of a variable.
Power series are incredibly useful because they can represent complicated functions in a form that is easy to analyze and manipulate.
In our exercise, although the expression \(\sum_{i=1}^{n} \frac{i^3}{n^4}\) is not a power series on the face of it, the manipulation to express it as \(S(n)\) employs a similar idea.
Breaking down the term into parts that can scale with \(n\) helps us assess their behavior better, akin to focusing on leading terms in power series.

Sum of Cubes

The "sum of cubes" formula, \(\sum_{i=1}^{n} i^3 = \frac{n^2(n+1)^2}{4}\), allows for a compact representation of the sum of cubed integers.
Cubing numbers amplifies their magnitude, making calculations more complex, but formulas like these simplify computation significantly.
In the given exercise, this formula becomes crucial when transforming the sum \(\sum_{i=1}^{n} \frac{i^3}{n^4}\) into a rational function \(S(n)\).
Utilizing this formula helps break down and simplify the sum, allowing the application of further steps to analyze the limit as \(n\) approaches infinity.
Thus, the sum of cubes is not just a neat mathematical trick, but a powerful tool in calculus for simplifying and evaluating sequences and series.