Problem 20 - Calculus: Early Transcendental Functions

Question

Compute the sum and the limit of the sum as \(n \rightarrow \infty.\) $$\sum_{i=1}^{n} \frac{1}{n}\left[\left(\frac{i}{n}\right)^{2}-5\left(\frac{i}{n}\right)\right]$$

Step-by-Step Solution

Answer

The sum \(S_n\) comes out to be \(\frac{1}{3}-\frac{5}{2}\), and the limit as \(n \rightarrow \infty\) is \(-\frac{13}{6}\).

Step 1: Summation

To compute the sum, first observe that the summation can be split up as two different sums. So, the original sum \(S_n\) can be written as: \[ S_n = \sum_{i=1}^{n} \frac{1}{n}\left(\frac{i}{n}\right)^{2}- \sum_{i=1}^{n} 5\frac{1}{n}\left(\frac{i}{n}\right)\]. Now, use the formula for the sum of squares of the first \(n\) natural numbers: \(\frac{n(n + 1)(2n + 1)}{6}\), and the formula for the sum of first \(n\) natural numbers: \(\frac{n(n + 1)}{2}\) to compute the two sums separately.

Step 2: Limit

Having computed the sum, next task is to compute the limit as \(n \rightarrow \infty.\) The limit of a sequence is the value that the terms of a sequence \('approach'\) as \(n \rightarrow \infty.\) For the computed sum, divide each term by \(n^3\), where \(n^3\) comes from considering the highest power of \(n\) in the denominator. Then take the limit as \(n \rightarrow \infty\). Here apply the rule that limit of a difference of two functions is the difference of their limits.

Key Concepts

Summation

Summation is a fundamental concept in calculus and involves adding terms together. In this exercise, we broke down a complex summation into simpler parts to manage it easier. It allows us to calculate the total result for a set of terms. For example, splitting a sum into two parts can help us use known formulas to find values more easily. This approach simplifies calculations by tackling each part separately. Summation is not just about adding numbers; it involves recognizing patterns and applying formulas, such as the sum of squares of natural numbers.

Limit of a sequence

In calculus, the limit of a sequence describes the behavior of a sequence as it progresses towards infinity. It helps us understand how a sequence evolves over time. When working with limits, we investigate the value that the terms of a sequence approach when the index increases. Calculating the limit of a sequence involves algebraic manipulation to simplify the expression. In this exercise, dividing each term by the highest power of n in the denominator helps us evaluate the limit as n approaches infinity. This method allows us to see which terms eventually dominate and what their sum converges to.

Natural numbers

Natural numbers are the simplest numbers that start from 1 and continue upwards: 1, 2, 3, and so on. They are fundamental in mathematics, especially in summation problems, like the one in this exercise. Natural numbers are essential because they provide the basic building blocks for more complex calculations. In mathematical formulas, such as those for summing sequences, these numbers are crucial. For instance, the sum of the first n natural numbers and their squares relies on natural number formulas, allowing for efficient computation.

Sum of squares

The sum of squares is a specific type of summation involving squaring each natural number before adding them up. It is often represented in formulas that simplify finding the aggregate result of a sequence of natural numbers squared. The formula, \(\frac{n(n + 1)(2n + 1)}{6}\), provides an efficient way to calculate this sum without directly computing each square. This concept not only simplifies calculations but also reveals patterns and relationships among numbers, enhancing our understanding of mathematical sequences and their limits. By using this concept, we manage large sums more effectively.