S.M. Yusuf Mathematical Methods Chapter 8 Infinite Series Solved Exercises & Convergence Tests PDF Notes - Punjab University, FBISE, QAU, UOS

Theoretical Foundations & Core Theorems

An infinite series is the formal sum of an infinite sequence of real (or complex) numbers. Mathematically, given a sequence $\{a_n\}_{n=1}^{\infty}$, the expression:

$$\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \dots + a_n + \dots$$

is defined as an infinite series. To determine whether an infinite series sums to a finite quantity, we examine its Sequence of Partial Sums (SOPS), denoted by $\{S_n\}$, where:

$$S_n = \sum_{k=1}^{n} a_k = a_1 + a_2 + \dots + a_n$$

If the limit of the sequence of partial sums exists and is finite, i.e., $\lim_{n \to \infty} S_n = S$, then the series $\sum_{n=1}^{\infty} a_n$ is said to converge to the sum $S$. If the limit does not exist or equals $\pm\infty$, the series diverges.

1. The $n$-th Term Test for Divergence

A fundamental necessary condition for convergence states that if $\sum_{n=1}^{\infty} a_n$ converges, then $\lim_{n \to \infty} a_n = 0$. Consequently, the Divergence Test states:

$$\text{If } \lim_{n \to \infty} a_n \neq 0 \quad \text{or does not exist, then } \sum_{n=1}^{\infty} a_n \text{ diverges.}$$

Caution: The converse is false. $\lim_{n \to \infty} a_n = 0$ does not guarantee convergence (e.g., the harmonic series $\sum \frac{1}{n}$).

2. Standard Reference Series

• Geometric Series Test: The series $\sum_{n=0}^{\infty} a r^n$ converges to $\frac{a}{1-r}$ if $|r| < 1$, and diverges if $|r| \ge 1$.
• Hyper-Harmonic ($p$-Series) Test: The series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if $p > 1$ and diverges if $p \le 1$.

3. Tests for Positive Term Series

When $a_n \ge 0$ for all $n$, the sequence of partial sums $\{S_n\}$ is monotonically non-decreasing. Key tests include:

• Direct Comparison Test (DCT): Let $0 \le a_n \le b_n$ for all $n \ge N$.
  • If $\sum b_n$ converges, then $\sum a_n$ converges.
  • If $\sum a_n$ diverges, then $\sum b_n$ diverges.
• Limit Comparison Test (LCT): If $a_n > 0, b_n > 0$ and $\lim_{n \to \infty} \frac{a_n}{b_n} = L$:
  • If $0 < L < \infty$, both $\sum a_n$ and $\sum b_n$ behave identically (both converge or both diverge).
  • If $L = 0$ and $\sum b_n$ converges, then $\sum a_n$ converges.
  • If $L = \infty$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.
• Cauchy's Integral Test: Let $f(x)$ be a continuous, positive, and non-increasing function on $[1, \infty)$ such that $f(n) = a_n$. The series $\sum_{n=1}^{\infty} a_n$ converges if and only if the improper integral $\int_{1}^{\infty} f(x) dx$ converges.
• D'Alembert's Ratio Test: Let $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L$.
  • If $L < 1$, the series converges absolutely.
  • If $L > 1$ (or $\infty$), the series diverges.
  • If $L = 1$, the test is inconclusive (requires Raabe's test or higher-order tests).
• Raabe's Test: Used specifically when the Ratio Test fails ($L = 1$). Compute $\lim_{n \to \infty} n \left( \frac{a_n}{a_{n+1}} - 1 \right) = \rho$.
  • If $\rho > 1$, the series converges.
  • If $\rho < 1$, the series diverges.
  • If $\rho = 1$, the test fails.
• Cauchy's Root Test: Let $\lim_{n \to \infty} (a_n)^{1/n} = L$. The series converges if $L < 1$, diverges if $L > 1$, and is inconclusive if $L = 1$.

4. Alternating Series and Absolute Convergence

An alternating series has the form $\sum_{n=1}^{\infty} (-1)^{n-1} a_n$ where $a_n > 0$.

• Leibniz's Alternating Series Test: The series converges if:
  1. $a_{n+1} \le a_n$ for all $n$ (monotonically non-increasing sequence).
  2. $\lim_{n \to \infty} a_n = 0$.
• Absolute vs. Conditional Convergence:
  • A series $\sum a_n$ is absolutely convergent if $\sum |a_n|$ converges. Absolute convergence implies ordinary convergence.
  • A series $\sum a_n$ is conditionally convergent if $\sum a_n$ converges but $\sum |a_n|$ diverges.

Formula Summary & Quick Reference

Test Name	Mathematical Condition / Limit Formula	Convergence Criteria	Divergence Criteria	Primary Use Case / Notes
$n$-th Term Test	$\lim_{n \to \infty} a_n = L$	N/A (Cannot prove convergence)	$L \neq 0$ or DNE	First line of defense for quick rejection.
Geometric Series	$\sum_{n=0}^{\infty} a r^n$	$|r| < 1 \implies S = \frac{a}{1-r}$	$|r| \ge 1$	Standard baseline reference series.
$p$-Series Test	$\sum_{n=1}^{\infty} \frac{1}{n^p}$	$p > 1$	$p \le 1$	Comparison target for algebraic functions.
Limit Comparison Test	$L = \lim_{n \to \infty} \frac{a_n}{b_n}$	$0 < L < \infty$ & $\sum b_n$ conv.	$0 < L < \infty$ & $\sum b_n$ div.	Rational expressions and radicals.
D'Alembert's Ratio Test	$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$	$L < 1$	$L > 1$	Factorials ($n!$) and exponentials ($a^n$).
Raabe's Test	$\rho = \lim_{n \to \infty} n \left( \frac{a_n}{a_{n+1}} - 1 \right)$	$\rho > 1$	$\rho < 1$	When Ratio Test yields $L = 1$.
Cauchy's Root Test	$L = \lim_{n \to \infty} (a_n)^{1/n}$	$L < 1$	$L > 1$	Expressions raised to the $n$-th power ($n^n, f(n)^n$).
Leibniz Test	Alternating series $\sum (-1)^{n-1} a_n$	$a_{n+1} \le a_n$ AND $\lim a_n = 0$	$\lim a_n \neq 0$	Alternating sign series.

Step-by-Step Solved Board Exam Questions

Problem 1 (S.M. Yusuf Ex 8.2 Variant - Limit Comparison Test)

Question: Test the convergence or divergence of the series:

$$\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2 + 3n + 1}$$

Solution:

Step 1: Identify $a_n$ and select an appropriate comparison series $b_n$.

Let the general term be:

$$a_n = \frac{\sqrt{n}}{n^2 + 3n + 1} = \frac{n^{1/2}}{n^2 + 3n + 1}$$

For large values of $n$, the dominant term in the numerator is $n^{1/2}$ and in the denominator is $n^2$. Thus, we construct $b_n$ as:

$$b_n = \frac{n^{1/2}}{n^2} = \frac{1}{n^{2 - 1/2}} = \frac{1}{n^{3/2}}$$

Step 2: Apply the Limit Comparison Test (LCT).

Evaluate $L = \lim_{n \to \infty} \frac{a_n}{b_n}$:

$$L = \lim_{n \to \infty} \left( \frac{\sqrt{n}}{n^2 + 3n + 1} \div \frac{1}{n^{3/2}} \right)$$ $$L = \lim_{n \to \infty} \left( \frac{\sqrt{n} \cdot n^{3/2}}{n^2 + 3n + 1} \right) = \lim_{n \to \infty} \left( \frac{n^2}{n^2 + 3n + 1} \right)$$

Divide numerator and denominator by $n^2$:

$$L = \lim_{n \to \infty} \frac{1}{1 + \frac{3}{n} + \frac{1}{n^2}} = \frac{1}{1 + 0 + 0} = 1$$

Step 3: Analyze the result.

Since $L = 1$, which is a finite, non-zero number ($0 < L < \infty$), both series $\sum a_n$ and $\sum b_n$ behave identically.

Now, test $\sum b_n = \sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$. This is a $p$-series with $p = \frac{3}{2}$.

Since $p = \frac{3}{2} > 1$, the auxiliary series $\sum b_n$ converges.

Conclusion: By the Limit Comparison Test, the given series $\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2 + 3n + 1}$ is convergent.

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Problem 2 (S.M. Yusuf Ex 8.4 Variant - Ratio Test Failure & Raabe's Test)

Question: Test the convergence of the series:

$$\sum_{n=1}^{\infty} \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{3 \cdot 6 \cdot 9 \cdots (3n)}$$

Solution:

Step 1: Write terms $a_n$ and $a_{n+1}$.

$$a_n = \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)}{3 \cdot 6 \cdot 9 \cdots (3n)}$$ $$a_{n+1} = \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)(3n+1)}{3 \cdot 6 \cdot 9 \cdots (3n)(3n+3)}$$

Step 2: Apply D'Alembert's Ratio Test.

$$\frac{a_{n+1}}{a_n} = \frac{1 \cdot 4 \cdot 7 \cdots (3n-2)(3n+1)}{3 \cdot 6 \cdot 9 \cdots (3n)(3n+3)} \cdot \frac{3 \cdot 6 \cdot 9 \cdots (3n)}{1 \cdot 4 \cdot 7 \cdots (3n-2)}$$ $$\frac{a_{n+1}}{a_n} = \frac{3n+1}{3n+3}$$

Taking the limit as $n \to \infty$:

$$L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{3n+1}{3n+3} = \lim_{n \to \infty} \frac{3 + \frac{1}{n}}{3 + \frac{3}{n}} = \frac{3}{3} = 1$$

Since $L = 1$, D'Alembert's Ratio Test fails (inconclusive).

Step 3: Apply Raabe's Test.

First, find $\frac{a_n}{a_{n+1}}$:

$$\frac{a_n}{a_{n+1}} = \frac{3n+3}{3n+1}$$

Now compute $\rho = \lim_{n \to \infty} n \left( \frac{a_n}{a_{n+1}} - 1 \right)$:

$$\rho = \lim_{n \to \infty} n \left( \frac{3n+3}{3n+1} - 1 \right) = \lim_{n \to \infty} n \left( \frac{3n+3 - (3n+1)}{3n+1} \right)$$ $$\rho = \lim_{n \to \infty} n \left( \frac{2}{3n+1} \right) = \lim_{n \to \infty} \frac{2n}{3n+1}$$

Divide numerator and denominator by $n$:

$$\rho = \lim_{n \to \infty} \frac{2}{3 + \frac{1}{n}} = \frac{2}{3}$$

Step 4: Conclusion.

Since $\rho = \frac{2}{3} < 1$, by Raabe's Test, the given infinite series diverges.

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Problem 3 (S.M. Yusuf Ex 8.4 Variant - Cauchy's Root Test)

Question: Test the convergence of the series:

$$\sum_{n=1}^{\infty} \left( \frac{n}{2n + 1} \right)^n$$

Solution:

Step 1: Identify the $n$-th term.

The general term contains an overall power of $n$:

$$a_n = \left( \frac{n}{2n + 1} \right)^n$$

Step 2: Apply Cauchy's Root Test.

Compute $L = \lim_{n \to \infty} (a_n)^{1/n}$:

$$L = \lim_{n \to \infty} \left[ \left( \frac{n}{2n + 1} \right)^n \right]^{1/n}$$ $$L = \lim_{n \to \infty} \frac{n}{2n + 1}$$

Step 3: Evaluate the Limit.

$$L = \lim_{n \to \infty} \frac{1}{2 + \frac{1}{n}} = \frac{1}{2 + 0} = \frac{1}{2}$$

Step 4: Conclusion.

Since $L = \frac{1}{2} < 1$, by Cauchy's Root Test, the series $\sum_{n=1}^{\infty} \left( \frac{n}{2n + 1} \right)^n$ is absolutely convergent (and therefore convergent).

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Problem 4 (S.M. Yusuf Ex 8.5 Variant - Absolute vs. Conditional Convergence)

Question: Determine whether the following alternating series is absolutely convergent, conditionally convergent, or divergent:

$$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{\sqrt{n}}$$

Solution:

Step 1: Test for Absolute Convergence.

Consider the absolute value series $\sum_{n=1}^{\infty} |a_n| = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{1}{n^{1/2}}$.

This is a $p$-series with $p = \frac{1}{2}$.

Since $p = \frac{1}{2} \le 1$, the absolute series $\sum |a_n|$ diverges.

Hence, the series is not absolutely convergent.

Step 2: Test for Ordinary Convergence using Leibniz's Test.

For the alternating series $\sum_{n=1}^{\infty} (-1)^{n-1} a_n$, where $a_n = \frac{1}{\sqrt{n}} > 0$:

1. Monotonicity Condition: Compare $a_{n+1}$ and $a_n$: $$\sqrt{n+1} > \sqrt{n} \implies \frac{1}{\sqrt{n+1}} < \frac{1}{\sqrt{n}} \implies a_{n+1} < a_n$$ Thus, $\{a_n\}$ is a monotonically decreasing sequence.
2. Limit Condition: $$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0$$

Both conditions of Leibniz's Test are satisfied. Therefore, the series $\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{\sqrt{n}}$ converges.

Step 3: Final Classification.

Because the series converges ordinarily but its absolute series diverges, the series is conditionally convergent.

Interactive Practice Quiz (Clickable MCQs)

Frequently Asked Questions (FAQs)

Q1: What is the systematic hierarchy of tests to apply when solving Chapter 8 exercises?

Answer: When analyzing a series $\sum a_n$, follow this strategic sequence:

1. Divergence Test: Compute $\lim_{n \to \infty} a_n$. If it's non-zero, stop immediately—the series diverges.
2. Structure Inspection:
   • If $a_n$ contains factorials ($n!$) or products of consecutive numbers, use the Ratio Test.
   • If $a_n = [f(n)]^n$, use Cauchy's Root Test.
   • If $a_n$ is rational or algebraic in $n$, use the Limit Comparison Test with a $p$-series.
   • If $a_n$ contains continuous functions suitable for integration, use the Integral Test.
3. Fallback for Ratio Test Failure: If the Ratio Test gives $L = 1$, switch immediately to Raabe's Test or the Limit Comparison Test.
4. Alternating Series: If signs alternate, test for absolute convergence first ($\sum |a_n|$). If that fails, apply Leibniz's Test for conditional convergence.

Q2: Why does D'Alembert's Ratio Test fail for algebraic series like $\sum \frac{1}{n^2}$ and $\sum \frac{1}{n}$?

Answer: Algebraic series constructed from polynomials or radicals have terms where $a_{n+1} / a_n$ expands to a ratio of polynomials of equal degree. For example, for $a_n = \frac{1}{n^p}$:

$$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^p = 1^p = 1$$

Because the limit is identically $1$ for all values of $p$, the Ratio Test cannot distinguish between a convergent $p$-series ($p = 2$) and a divergent $p$-series ($p = 1$). That is why polynomial and algebraic series require the Limit Comparison Test or Raabe's Test.

Q3: What is the critical distinction between Absolute Convergence and Conditional Convergence in Pakistani Board/University Exams?

Answer: The distinction is critical both theoretically and in scoring scheme:

• Absolute Convergence: A series $\sum a_n$ converges absolutely if the series of positive magnitudes $\sum |a_n|$ converges. Absolutely convergent series can be rearranged in any order without changing their sum (Riemann Rearrangement Theorem).
• Conditional Convergence: A series converges conditionally if $\sum a_n$ converges (usually via Leibniz Test), but $\sum |a_n|$ diverges. Conditionally convergent series are sensitive to rearrangement.

Exam Note: In PU and FBISE long questions, simply stating "convergent" for an alternating series when "conditionally convergent" is required will result in mark deductions. Always test $\sum |a_n|$ first.