Problem 1 - Basic Analysis: Introduction to Real Analysis

Question

Decide the convergence or divergence of the following series. a) \(\sum_{n=1}^{\infty} \frac{1}{2^{2 n+1}}\) b) \(\sum_{n=1}^{\infty} \frac{(-1)^{n}(n-1)}{n}\) c) \(\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{1 / 10}}\) d) \(\sum_{n=1}^{\infty} \frac{n^{n}}{(n+1)^{2 n}}\)

Step-by-Step Solution

Answer

a) Converges, b) Diverges, c) Converges, d) Converges.

Step 1: Determine the Convergence of Series a

The given series is \( \sum_{n=1}^{\infty} \frac{1}{2^{2n+1}} \). Notice that the general term is \( \frac{1}{2^{2n+1}} = \frac{1}{2} \cdot \frac{1}{4^n} \). This is a geometric series with the first term \( a = \frac{1}{2} \) and common ratio \( r = \frac{1}{4} \). A geometric series \( \sum a r^{n-1} \) converges if \( |r| < 1 \). Here, \( |r| = \frac{1}{4} < 1 \), so the series converges.

Step 2: Determine the Convergence of Series b

The given series is \( \sum_{n=1}^{\infty} \frac{(-1)^{n}(n-1)}{n} \). To check convergence, observe that for large \( n \), the dominant term in the numerator and denominator is \( \frac{n}{n} = 1 \), so the terms do not shrink to zero. To determine convergence or divergence, consider the test for divergence, which states that if \( \lim_{n \to \infty} a_n eq 0 \), then the series diverges. The limit here is 1, thus the series diverges.

Step 3: Determine the Convergence of Series c

The given series is \( \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{1/10}} \). This is an alternating series. Apply the Alternating Series Test, which states that the series \( \sum (-1)^n b_n \) converges if \( b_n \) is positive, decreasing, and \( \lim_{n \to \infty} b_n = 0 \). Here, \( b_n = \frac{1}{n^{1/10}} \) is positive, decreases as \( n \) increases, and \( \lim_{n \to \infty} b_n = 0 \). Thus, the series converges.

Step 4: Determine the Convergence of Series d

The given series is \( \sum_{n=1}^{\infty} \frac{n^n}{(n+1)^{2n}} \). For large \( n \), we approximate \( \left(1 + \frac{1}{n} \right)^{2n} \approx e^2 \), due to continuous compounding. So the terms look like \( \frac{n^n}{e^{2n} n^{2n}} = \left(\frac{n}{e^2 n^2} \right)^n = \left(\frac{1}{e^2 n} \right)^n \rightarrow 0 \) very fast as \( n \to \infty \). Apply the ratio test by comparing \( a_{n+1} \) and \( a_n \); you find \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 < 1 \), indicating convergence.

Key Concepts

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the "common ratio." The general form of a geometric series is:

• First term: \( a \)
• Common ratio: \( r \)
• Series: \( \sum a r^{n-1} \)

A geometric series converges, or adds up to a finite value, if the absolute value of the common ratio \( |r| \) is less than 1. If \( |r| \geq 1 \), the series diverges, meaning it does not sum to a finite number and continues on indefinitely.
To see this in action, consider the series \( \sum_{n=1}^{\infty} \frac{1}{2^{2n+1}} \), rewritten as \( \sum_{n=1}^{\infty} \frac{1}{2} \cdot \frac{1}{4^n} \). This is a geometric series with the first term \( a = \frac{1}{2} \) and common ratio \( r = \frac{1}{4} \). Since \( |r| = \frac{1}{4} < 1 \), the series converges.

Alternating Series

An alternating series is one whose terms alternate in sign. The series can be represented as \( \sum (-1)^n b_n \) or \( \sum (-1)^{n+1} b_n \), where \( b_n \) denotes the sequence of absolute values of terms. To check these series for convergence, we use the Alternating Series Test.
The Alternating Series Test tells us that an alternating series converges if:

• Each term \( b_n \) is positive.
• The sequence \( b_n \) is non-increasing: \( b_{n+1} \leq b_n \).
• The limit of \( b_n \) as \( n \to \infty \) is zero: \( \lim_{n \to \infty} b_n = 0 \).

For example, take the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{1/10}} \). In this series, \( b_n = \frac{1}{n^{1/10}} \) which is positive, decreases as \( n \) increases, and the limit as \( n \to \infty \) is zero. Thus, the alternating series converges.

Ratio Test

The Ratio Test is a convenient method for determining the convergence of a series. To apply the test, evaluate the limit:\[L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|.\]The rules for the test are:

• If \( L < 1 \), the series converges absolutely.
• If \( L > 1 \) or \( L = \infty \), the series diverges.
• If \( L = 1 \), the test is inconclusive, and another test must be used.

Consider the series \( \sum_{n=1}^{\infty} \frac{n^n}{(n+1)^{2n}} \). When applying the ratio test, you calculate the limit as \( n \to \infty \) and find that \( L = 0 \), which is less than 1. Therefore, this series converges.

Test for Divergence

Before jumping into complex tests, the Test for Divergence is a simple technique to determine if a series diverges. This test states that if the limit of the general term of a series \( a_n \) as \( n \to \infty \) is not zero, or does not exist, then the series definitely diverges.
It's a quick way to identify divergence, but it cannot prove convergence.
Let's look at the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n}(n-1)}{n} \). The general term of this series does not tend to zero as \( n \to \infty \). Instead, it approaches 1. Therefore, by the Test for Divergence, the series diverges.