Problem 38 - A Course in Calculus and Real Analysis

Question

Do \(\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \frac{1}{\sqrt{i+n}}\) and \(\lim _{n \rightarrow \infty} \frac{1}{n^{18}} \sum_{i=1}^{n} i^{16}\) exist? If yes, find them.

Step-by-Step Solution

Answer

The first limit does not exist as the sum diverges. The second limit exists and is equal to \(\frac{1}{17}\).

Step 1: Using comparison test

Since the summand is positive, we can use the comparison test to determine if the sum converges. Recall that the comparison test states that if \(0 \le a_n \le b_n\) for all \(n\) and the sum \(\sum_{n=1}^\infty b_n\) converges, then \(\sum_{n=1}^\infty a_n\) converges as well. Conversely, if \(\sum_{n=1}^\infty a_n\) diverges, then \(\sum_{n=1}^\infty b_n\) diverges. Here, we have \(a_i = \frac{1}{\sqrt{i+n}}\) which is the summand for the first limit. Notice that for every \(i \ge 1\) and \(n \ge 1\), we can write the inequality: \(\frac{1}{\sqrt{i+n}} \ge \frac{1}{\sqrt{2n}}\) Thus: \(\sum_{i=1}^n \frac{1}{\sqrt{i+n}} \ge \sum_{i=1}^n \frac{1}{\sqrt{2n}} = n \cdot \frac{1}{\sqrt{2n}} = \frac{\sqrt{n}}{\sqrt{2}}\) As \(n \to \infty\), \(\frac{\sqrt{n}}{\sqrt{2}} \to \infty\). By the comparison test, this means the sum \(\sum_{i=1}^n \frac{1}{\sqrt{i+n}}\) diverges. Therefore, the first limit does not exist. #Step 2: Analyzing the second expression#

Step 2: Using Riemann sum

The second expression can be interpreted as a Riemann sum. Recall that given a function \(f(x)\), the Riemann sum approximation of its definite integral over the interval \([a, b]\) using \(n\) equally spaced partitions can be written as: \(\sum_{i=1}^n f(x_i) \Delta x = \sum_{i=1}^n f(a + i \cdot \Delta x) \Delta x\) where \(\Delta x = \frac{b - a}{n}\) and \(x_i = a + i \cdot \Delta x\). The second expression is given by: \(\frac{1}{n^{18}} \sum_{i=1}^{n} i^{16}\) Let's rewrite this expression in the form of a Riemann sum: \(\sum_{i=1}^n \left( \frac{i}{n} \right)^{16} \cdot \frac{1}{n} = \sum_{i=1}^n \left( \frac{i \cdot \Delta x}{\Delta x} \right)^{16} \cdot \Delta x\) We have a Riemann sum with \(f(x) = x^{16}\), \(\Delta x = \frac{1}{n}\), and \(x_i = \frac{i}{n} = \frac{i \cdot \Delta x}{\Delta x}\). So, our expression approximates the definite integral of the function \(f(x) = x^{16}\) over the interval \([0, 1]\): \(\int_0^1 x^{16} \,\mathrm{d}x\) #Step 3: Finding the limit of the second expression#

Step 3: Integrating the function and taking the limit

To find the limit of the second expression, we can find the definite integral mentioned in step 2: \(\int_0^1 x^{16} \,\mathrm{d}x = \left[\frac{x^{17}}{17}\right]_0^1 = \frac{1^{17}}{17}-\frac{0^{17}}{17} = \frac{1}{17}\) So, as \(n \to \infty\), the Riemann sum converges to the definite integral: \(\lim_{n\to\infty} \frac{1}{n^{18}} \sum_{i=1}^{n} i^{16} = \frac{1}{17}\) #Final answer# The first limit does not exist, while the second limit exists and is \(\frac{1}{17}\).