Question

Consider the sequence \(\left\\{x_{n}\right\\}\) defined for \(n=1,2,3, \ldots\) by $$x_{n}=\sum_{k=n+1}^{2 n} \frac{1}{k}=\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2 n}.$$ a. Write out the terms \(x_{1}, x_{2}, x_{3}\). b. Show that \(\frac{1}{2} \leq x_{n}<1,\) for \(n=1,2,3, \ldots\). c. Show that \(x_{n}\) is the right Riemann sum for \(\int_{1}^{2} \frac{d x}{x}\) using \(n\) subintervals. d. Conclude that \(\lim _{n \rightarrow \infty} x_{n}=\ln 2\).

Short answer

Question: Calculate and write out the terms \(x_1\), \(x_2\), and \(x_3\) for the sequence \(x_n = \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n}\). Prove that the inequality \(\frac{1}{2} \leq x_n < 1\) holds for all \(n = 1, 2, 3, \ldots\) and show that \(x_n\) represents the right Riemann sum for \(\int_{1}^{2} \frac{1}{x} dx\) using \(n\) subintervals. Finally, conclude the limit of the sequence. Answer: The terms for the sequence are: \(x_1 = \frac{1}{2}, x_2 = \frac{1}{3} + \frac{1}{4}, x_3 = \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\). The inequality \(\frac{1}{2} \leq x_n < 1\) holds for all \(n = 1, 2, 3, \ldots\) by induction. The sequence \(x_n\) represents the right Riemann sum for \(\int_{1}^{2} \frac{1}{x} dx\) using \(n\) subintervals. As a result, the limit of the sequence is \(\lim_{n \rightarrow \infty} x_n = \ln 2\).

Step-by-step solution

(a) Calculate the terms x1, x2, x3

From the given sum, we can directly calculate the first three terms as follows: \(x_1 = \frac{1}{2}\), \(x_2 = \frac{1}{3} + \frac{1}{4}\), \(x_3 = \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\).

(b) Prove the inequality using induction

We will prove the inequality \(\frac{1}{2} \leq x_n < 1\) using induction. For the base case, we have already calculated \(x_1 = \frac{1}{2}\), which clearly meets the inequality. Now assume the inequality holds for \(x_k\): \(\frac{1}{2} \leq x_k < 1\) for some \(k \ge 1\). We need to prove the inequality holds for \(x_{k+1}\): \(x_{k+1} = \frac{1}{k+2} + \frac{1}{k+3} + \dots + \frac{1}{2(k+1)}\) Notice that for all terms, the denominators are greater than \(k+1\). Therefore, each term on the right-hand side is less than \(\frac{1}{k+1}\), and there are \(k+1\) terms in total. So, \(x_{k+1} < 1\). Now, consider the difference between \(x_{k}\) and \(x_{k+1}\): \(x_{k+1} - x_k = \big( \frac{1}{k+2} + \frac{1}{k+3} + \dots + \frac{1}{2(k+1)} \big) - \big( \frac{1}{k+1} + \frac{1}{k+2} + \dots + \frac{1}{2k} \big)\) The right-hand side simplifies to: \(\frac{1}{k+1} - \frac{1}{2k} = \frac{1}{2(k+1)}\). By induction hypothesis, we have \(\frac{1}{2} \leq x_k\), so \(x_{k+1} = x_k + \frac{1}{2(k+1)} > x_k \ge \frac{1}{2}\). Therefore, \(\frac{1}{2} \leq x_{k+1} < 1\), and the inequality holds for all \(n = 1, 2, 3, \ldots\).

(c) Show that xn is the right Riemann sum for the integral

We need to show that \(x_n\) is the right Riemann sum for \(\int_{1}^{2} \frac{1}{x} dx\) with \(n\) subintervals. We will partition the interval \([1,2]\) into \(n\) equal subintervals, each of width \(\Delta x = \frac{2-1}{n} = \frac{1}{n}\), so the endpoints of the subintervals are \(1, 1+\frac{1}{n}, 1+\frac{2}{n}, \ldots, 2\). We will choose the right endpoint of each subinterval as the representative point. The right Riemann sum will be: \(R_n = \sum_{i=1}^{n} f(x_i^*) \Delta x_i = \frac{1}{n} \sum_{i=1}^{n} \frac{1}{1+\frac{i}{n}}\). Now, we change the index by redefining \(k = n + i\), which then simplifies the sum: \(R_n = \frac{1}{n} \sum_{k=n+1}^{2n} \frac{1}{k}\). This sum is precisely the definition of \(x_n\). Therefore, \(x_n\) is the right Riemann sum for \(\int_{1}^{2} \frac{dx}{x}\) using \(n\) subintervals.

(d) Conclude the limit of the sequence

Now we know that \(x_n\) represents the right Riemann sum for the integral \(\int_{1}^{2} \frac{1}{x} dx\). This integral can be calculated as: \(\int_{1}^{2} \frac{1}{x} dx = \ln|x| \Big|_1^2 = \ln 2 - \ln 1 = \ln 2\). As \(n\) approaches infinity, the right Riemann sum \(x_n\) gets closer to the actual value of the integral. Therefore, we can conclude that \(\lim_{n \rightarrow \infty} x_n = \ln 2\).

Key concepts

Riemann sum

Understanding the concept of a Riemann sum is crucial for analyzing the behavior of sequences and the convergence of series. Think of a Riemann sum like a painter approximating a famous masterpiece with a mosaic. Each tile in the mosaic represents a portion of the integral over a small interval; similarly, each term in a Riemann sum represents the area under a curve over a slice of the domain.

The Riemann sum is a method to approximate the definite integral of a function, denoted as \(\int_a^b f(x)dx\), where \(a\) and \(b\) represent the interval over which we are integrating. The way it works is by dividing the interval \[a,b\] into smaller subintervals, each with a width we can call \(\Delta x\), and then multiplying the function value at certain points within those subintervals (usually left, right, or midpoints) by the width of the subintervals. The sum of all these products gives us a Riemann sum:
\[\sum_{i=1}^n f(c_i)\Delta x_i\]
where \(c_i\) represents the chosen point in the \(i\)-th subinterval. As the number of subintervals increases, and \(\Delta x\) gets smaller, the Riemann sum becomes a better approximation of the true definite integral, and in the case of an infinite number of infinitely small subintervals, it converges to the exact value of the integral.

Sequences and series

Sequences and series are the bread and butter of analysis, providing a foundational language for discussing convergence. A sequence is an ordered list of numbers that can continue indefinitely. We denote a sequence by \(\{a_n\}\), where \(a_n\) represents the \(n\)-th term in the sequence. It can be thought of as a function where the domain is the set of natural numbers, and the range is the set of real numbers associated with each natural number.

When we add the terms of a sequence together, we get a series. In particular, we consider the sum of a sequence:\[\sum_{n=1}^\infty a_n\]where we aim to find out whether, as we add more and more terms, the sum approaches a finite number (converges) or grows without bound (diverges). If a series converges, it means the terms of the sequence are getting smaller in such a way that the total sum is a definite, finite number. Understanding whether a series converges or diverges, and what value it converges to if applicable, is a pivotal concept in advanced mathematics, particularly in calculus and analysis.

Limit of a sequence

The limit of a sequence is the value that the terms of the sequence 'approach' as the index \(n\) increases towards infinity. For a sequence \(\{a_n\}\), we denote the limit as:\[\lim_{n \to \infty} a_n = L\]
The limit \(L\) can be thought of as the 'target' that the sequence is aiming for. If the terms of the sequence get arbitrarily close to \(L\) as \(n\) becomes very large, we say that the sequence converges to \(L\). However, if the terms do not settle down but continue to grow without bound or oscillate without approaching a particular value, we say that the sequence diverges.

In order to rigorously establish the limit of a sequence, mathematicians use the concept of an epsilon (\(\varepsilon\))-neighborhood. We say that the limit of a sequence equals \(L\) if, for every positive number \(\varepsilon\), there exists a natural number \(N\), such that for all \(n > N\), the distance between \(a_n\) and \(L\) is less than \(\varepsilon\). This precision in definition ensures that limits are not assigned based on approximations or guesses but based on unequivocal mathematical behavior as \(n\) approaches infinity.

Infinite series

An infinite series takes the concept of a sequence to the next level by considering the sum of infinitely many terms. Determining the convergence or divergence of an infinite series is a central question in mathematical analysis. For example, if we are summing the reciprocals of the natural numbers:\[\sum_{n=1}^\infty \frac{1}{n}\]we want to know if adding up these fractions, as \(n\) grows larger and larger, will lead to a number that exists within the real number system, or if the sum will grow without limit. In the case of the series of reciprocals, the sum diverges, growing beyond all bounds.

However, not all infinite series diverge. Some converge to a real number, and this can often represent a profound and surprising result, particularly when the function being summed is complex or not obviously decreasing. In the exercise provided, we're examining the convergence of a series that results in the natural logarithm of 2, which showcases how an infinite series, through the delicate interplay of its terms, can indeed sum to a precise and finite quantity. Understanding and working with infinite series is foundational for calculus, physics, and many other fields of science and engineering where continuous processes are approximated by discrete sums.