Problem 22 - Calculus: Early Transcendental Functions

Question

Compute the sum and the limit of the sum as \(n \rightarrow \infty.\) $$\sum_{i=1}^{n} \frac{1}{n}\left[\left(\frac{2 i}{n}\right)^{2}+4\left(\frac{i}{n}\right)\right]$$

Step-by-Step Solution

Answer

The sum of the series is \( \frac{4}{3} + \frac{2}{n} + \frac{2}{n^{2}} \) and the limit as \(n \rightarrow \infty\) is \( \frac{4}{3}\).

Step 1: Evaluate the sum

To begin, we can rewrite the sum in terms of \(n\) and \(i\). We calculate the sum using the formula: \(\frac{1}{n}\sum_{i=1}^{n} \left[\left(\frac{2i}{n}\right)^{2}+4\left(\frac{i}{n}\right)\right]\). Now breakdown this sum into two parts: \(\frac{1}{n}\sum_{i=1}^{n}\left(\frac{2i}{n}\right)^{2} + \frac{1}{n}\sum_{i=1}^{n}4\left(\frac{i}{n}\right)\). On simplification this can be further written as \(\frac{4}{n^{3}}\sum_{i=1}^{n}i^{2} + \frac{4}{n^{2}}\sum_{i=1}^{n}i\). We know that \(\sum_{i=1}^{n}i^{2} = \frac{n(n+1)(2n+1)}{6}\) and \(\sum_{i=1}^{n}i = \frac{n(n+1)}{2}\). Using these formulas, we get the total sum as: \( \frac{4}{n^{2}}\left(\frac{2n+1}{6}\right) + \frac{2(n+1)}{n} \) which simplifies to \( \frac{4}{3} + \frac{2}{n} + \frac{2}{n^{2}} \).

Step 2: Calculate the limit as \(n \rightarrow \infty\)

Now that we have the sum, we calculate the limit when \(n\) approaches infinity: \( \lim_{n \rightarrow \infty} \left[\frac{4}{3} + \frac{2}{n} + \frac{2}{n^{2}}\right]\). As \(n\) goes to infinity, terms with \(n\) in the denominator will go to zero, resulting in the limit as \( \frac{4}{3}\).

Key Concepts

Limit of a Sum

The concept of the "Limit of a Sum" is crucial in understanding how we can approximate the area under curves, or solve problems involving infinite series. In the given exercise, we compute the limit of a sum as \(n\) approaches infinity. Essentially, we're looking at how the sum behaves as the number of terms increases indefinitely. This approach is common in finding definite integrals.

The sum provided in the exercise is:

• \( \sum_{i=1}^{n} \frac{1}{n}\left[\left(\frac{2i}{n}\right)^{2}+4\left(\frac{i}{n}\right)\right] \)

To find the limit, the sum is broken into simpler parts, and known summation formulas are applied. The expression is simplified to \( \frac{4}{3} + \frac{2}{n} + \frac{2}{n^{2}} \).

As \(n\) becomes very large, the terms \(\frac{2}{n}\) and \(\frac{2}{n^{2}}\) approach zero. Thus, the limit evaluates to \(\frac{4}{3}\). The process of taking the limit helps transition from a discrete sum to a continuous function evaluation, closely linked with definite integrals.

Definite Integral

A definite integral represents the accumulated area under a curve within a specific interval \([a, b]\). It is the continuous counterpart to the sum of areas of rectangles used in Riemann sums. The transition from a Riemann sum to a definite integral is done using limits.

In our exercise, the sum and its limit correspond to an integral of a function over a certain interval. The rewritten sum \( \frac{1}{n}\sum \left(\left(\frac{2i}{n}\right)^2+4\left(\frac{i}{n}\right)\right) \), after simplification and limiting process, gives an integral form:

• \( \int_0^1 (4x^2 + 4x) \, \text{d}x \)

Here, \(x = \frac{i}{n}\) and the interval of integration is \([0, 1]\), reflecting the units of division used per \(n\) rectangles.

Thus, the limit of the sum approach in the problem connects directly to computing a definite integral, giving us a powerful tool for calculating area, volume, and other quantities in continuous settings.

Infinite Series

An infinite series is a sum of infinitely many terms. Unlike a finite sum, an infinite series is concerned more with the behavior as the number of terms tends towards infinity. It covers concepts like convergence, divergence and partial sums.

While our exercise focuses on transitioning a finite sum to a continuous function limit, it also hints at how infinite series might operate, if approached similarly. Consider an infinite sequence \(\sum_{i=1}^{\infty} a_i\), where \(a_i\) could be terms like \(\frac{1}{n^2}\) or \(\frac{1}{n}\).

Convergence analysis helps understand if the entire sum reaches a finite limit. If, as \(n\) approaches infinity, the partial sums settle towards a single value, the series converges. If not, it diverges.

In contexts like the problem given, these infinite series concepts assure us that as terms contribute less towards the total sum (as seen when \(n\) grows), the approach towards limits becomes vital in mathematical analysis.