Problem 37 - A Course in Calculus and Real Analysis

Question

Consider the sequence whose \(n\) th term is given by the following. In each case, determine the limit of the sequence by expressing the \(n\) th term as a Riemann sum for a suitable function. (i) \(\frac{1}{n^{17}} \sum_{i=1}^{n} i^{16}\), (ii) \(\frac{1}{n^{5 / 2}} \sum_{i=1}^{n} i^{3 / 2}\), (iii) \(\sum_{i=1}^{n} \frac{1}{\sqrt{i n+n^{2}}}\), (iv) \(\frac{1}{n}\left\\{\sum_{i=1}^{n}\left(\frac{i}{n}\right)+\sum_{i=n+1}^{2 n}\left(\frac{i}{n}\right)^{3 / 2}+\sum_{i=2 n+1}^{3 n}\left(\frac{i}{n}\right)^{2}\right\\}\).

Step-by-Step Solution

Answer

The limits of the sequences are: (i) \(\frac{1}{17}\) (ii) \(\frac{2}{5}\) (iii) 2 (iv) \(\frac{31}{2}\)

Step 1: Identify the function

Here the Riemann sum is \(\sum_{i=1}^{n} i^{16}\), so consider the Riemann sum for the function \(f(x) = x^{16}\). We have: $$ \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}f\left(\frac{i}{n^{16}}\right) = \int_0^1 x^{16} dx $$

Step 2: Calculate the limit as n approaches infinity

Now, integrate \(f(x) = x^{16}\): $$ \lim_{n\to\infty}\frac{1}{n^{17}}\sum_{i=1}^{n}i^{16} = \int_0^1 x^{16} dx = \left[\frac{x^{17}}{17}\right]_0^1 = \frac{1}{17} $$ (ii) \(\frac{1}{n^{5/2}} \sum_{i=1}^{n} i^{3/2}\)

Step 1: Identify the function

Here the Riemann sum is \(\sum_{i=1}^{n} i^{3/2}\), so consider the Riemann sum for the function \(f(x) = x^{3/2}\). We have: $$ \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}f\left(\frac{i}{n^{2/5}}\right) = \int_0^1 x^{3/2} dx $$

Step 2: Calculate the limit as n approaches infinity

Now, integrate \(f(x) = x^{3/2}\): $$ \lim_{n\to\infty}\frac{1}{n^{5/2}}\sum_{i=1}^{n}i^{3/2} = \int_0^1 x^{3/2} dx = \left[\frac{2}{5}x^{5/2}\right]_0^1 = \frac{2}{5} $$ (iii) \(\sum_{i=1}^{n} \frac{1}{\sqrt{i n+n^{2}}}\)

Step 1: Identify the function

Here the Riemann sum is \(\sum_{i=1}^{n} \frac{1}{\sqrt{i n+n^{2}}}\), so consider the Riemann sum for the function \(f(x) = \frac{1}{\sqrt{x}}\). We have: $$ \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}f\left(\frac{i}{n}\right) = \int_0^1 \frac{1}{\sqrt{x}} dx $$

Step 2: Calculate the limit as n approaches infinity

Now, integrate \(f(x) = \frac{1}{\sqrt{x}}\): $$ \lim_{n\to\infty}\sum_{i=1}^{n}\frac{1}{\sqrt{i n+n^{2}}} = \int_0^1 \frac{1}{\sqrt{x}} dx = \left[2\sqrt{x}\right]_0^1 = 2 $$ (iv) $\frac{1}{n}\left\\{\sum_{i=1}^{n}\left(\frac{i}{n}\right)+\sum_{i=n+1}^{2 n}\left(\frac{i}{n}\right)^{3 / 2}+\sum_{i=2 n+1}^{3 n}\left(\frac{i}{n}\right)^{2}\right\\}$

Step 1: Identify the function

The sum has three sub-sums within it, so we split the sum into three parts. Consider the Riemann sum for the functions: 1. \(f_1(x) = x\) 2. \(f_2(x) = x^{3/2}\) 3. \(f_3(x) = x^2\)

Step 2: Calculate the limit as n approaches infinity

Now, integrate \(f_1(x) = x\) in the range \([0, 1]\), \(f_2(x) = x^{3/2}\) in the range \([1, 2]\), and \(f_3(x) = x^2\) in the range \([2, 3]\): $$ \lim_{n\to\infty}\frac{1}{n}\left\\{\sum_{i=1}^{n}\left(\frac{i}{n}\right)+\sum_{i=n+1}^{2n}\left(\frac{i}{n}\right)^{3/2}+\sum_{i=2n+1}^{3n}\left(\frac{i}{n}\right)^{2}\right\\} = \int_0^1 x dx + \int_1^2 x^{3/2} dx + \int_2^3 x^2 dx $$ $$ = \left[\frac{x^2}{2}\right]_0^1 + \left[\frac{2}{5}x^{5/2}\right]_1^2 + \left[\frac{x^3}{3}\right]_2^3 = \frac{1}{2} + \frac{30}{5} - \frac{2}{5} + \frac{19}{3} - \frac{8}{3} = \frac{7}{2} + 6 + \frac{11}{3} = \frac{31}{2} $$ So, the limits of the sequences are: (i) \(\frac{1}{17}\) (ii) \(\frac{2}{5}\) (iii) 2 (iv) \(\frac{31}{2}\)

Key Concepts

Sequence Convergence

When exploring sequences, convergence is a central concept. A sequence converges if it approaches a specific value, known as the limit, as the number of terms increases indefinitely. Picture this as a race to a finishing line: the values of the sequence keep getting closer and closer to a single number.
The role of Riemann sums in this context is crucial. They help break down the sequence into smaller parts that can be easily understood and analyzed. By transforming the sequence into a sum that resembles an integral, we can better determine its behavior as it approaches infinity.
Divergent sequences, on the other hand, don't settle at a particular value. Recognizing whether a sequence converges or diverges is key in calculus, especially in problems involving integration and limits.

Limit of a Sequence

The limit of a sequence is the value that the sequence approaches as the index grows larger. It's like setting a target and watching the values move closer to this target as you count further along the sequence.

• If a sequence converges to a limit, it suggests a predictable pattern in its progression.
• The limit can sometimes be calculated exactly, while other times, it may require approximation techniques.

Using Riemann sums helps transition from a sequence context to an integral context, allowing us to apply integration methods to find limits. This bridging is particularly evident when dealing with sequences expressed in terms of discrete sums, which resemble integrals.
This method provides a clearer picture of how the sequence behaves at infinity, simplifying complex sequences into more manageable forms to find their ultimate limits.

Calculus Integration

Calculus ties together various mathematical concepts, with integration being a fundamental component. Integration is essentially the reverse process of differentiation, and it is used to compute areas under curves, among other things.

• In the context of Riemann sums, integration helps convert sums of sequence terms into continuous functions over a specific interval.
• This process simplifies calculations and provides a holistic view of how a sequence behaves over an interval.

When sequences are expressed as Riemann sums, integration becomes the tool for determining their limits easily. Imagine transforming a series of discrete steps into a smooth curve - this is what integration achieves.
By integrating functions derived from Riemann sums, we can unlock valuable insights into both continuous and discrete behaviors, paving the way to solve more complex analytical problems.