S.M. Yusuf Mathematical Methods Chapter 8 Infinite Series Solved Exercises PDF & Convergence Tests | BS Math, Punjab University, FBISE Guide

Theoretical Foundations & Core Theorems

An infinite series is the formal sum of an infinite sequence of real (or complex) numbers, denoted by:

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

To evaluate whether an infinite sum yields a finite value, we construct the sequence of partial sums $\{S_k\}$, where $S_k = \sum_{n=1}^{k} a_n = a_1 + a_2 + \dots + a_k$. If the limit $\lim_{k \to \infty} S_k = S$ exists as a finite number, the series is said to converge to $S$. Otherwise, the series diverges.

1. Necessary Condition for Convergence ($n$-th Term Test)

If $\sum_{n=1}^{\infty} a_n$ converges, then $\lim_{n \to \infty} a_n = 0$. Consequently, if $\lim_{n \to \infty} a_n \neq 0$ or does not exist, the series diverges. Warning: $\lim_{n \to \infty} a_n = 0$ is a necessary condition, but not sufficient (e.g., the harmonic series $\sum \frac{1}{n}$ diverges even though $\frac{1}{n} \to 0$).

2. Fundamental Benchmark Series

• Geometric Series: $\sum_{n=0}^{\infty} a r^n$ converges to $\frac{a}{1-r}$ if $|r| < 1$, and diverges if $|r| \ge 1$.
• $p$-Series (Hyperharmonic 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

• Basic Comparison Test (BCT): 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 $\lim_{n \to \infty} \frac{a_n}{b_n} = L$, where $0 < L < \infty$, then both series perform identically (either both converge or both diverge).
• D'Alembert's Ratio Test: Let $a_n > 0$ and $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L$.
  • If $L < 1$, the series converges absolutely.
  • If $L > 1$ (or $\infty$), the series diverges.
  • If $L = 1$, the test is inconclusive.
• Cauchy's Root Test: Let $a_n \ge 0$ and $\lim_{n \to \infty} (a_n)^{1/n} = L$.
  • If $L < 1$, the series converges.
  • If $L > 1$, the series diverges.
  • If $L = 1$, the test fails.
• Cauchy's Integral Test: Let $f(x)$ be a continuous, positive, and decreasing 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.

4. Alternating Series & Absolute Convergence

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

• Leibniz Rule (Alternating Series Test): The series converges if:
  1. $b_{n+1} \le b_n$ for all $n$ (monotonically decreasing sequence).
  2. $\lim_{n \to \infty} b_n = 0$.
• Absolute vs. Conditional Convergence:
  • $\sum a_n$ is absolutely convergent if $\sum |a_n|$ converges.
  • $\sum a_n$ is conditionally convergent if $\sum a_n$ converges but $\sum |a_n|$ diverges.

Formula Summary & Quick Reference

Test / Theorem	Mathematical Formula / Condition	Primary Application / Usage
$n$-th Term Test	$\lim_{n \to \infty} a_n = L \neq 0 \implies \text{Diverges}$	First quick sanity check for divergence.
$p$-Series Test	$\sum \frac{1}{n^p} \quad \begin{cases} \text{Conv.} & p > 1 \\ \text{Div.} & p \le 1 \end{cases}$	Benchmark test for comparison test problems.
Limit Comparison Test	$L = \lim_{n \to \infty} \frac{a_n}{b_n}, \quad 0 < L < \infty$	Rational expressions and algebraic roots in $n$.
D'Alembert's Ratio Test	$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$	Expressions containing factorials ($n!$) or powers ($a^n$).
Cauchy's Root Test	$L = \lim_{n \to \infty} |a_n|^{1/n}$	Expressions where the entire term is raised to power $n$.
Cauchy's Integral Test	$\int_{1}^{\infty} f(x) dx \text{ where } f(n) = a_n$	Expressions involving logarithmic terms like $\ln(n)$.
Leibniz Test	$b_{n+1} \le b_n \quad \text{and} \quad \lim_{n \to \infty} b_n = 0$	Alternating series containing $(-1)^n$ or $(-1)^{n-1}$.

Step-by-Step Solved Board Exam Questions

Question 1 (Ex 8.4 - Ratio Test Application)

Problem: Determine the convergence or divergence of the infinite series:

$$\sum_{n=1}^{\infty} \frac{n!}{(2n)!}$$

Solution:

Let the general term be $a_n = \frac{n!}{(2n)!}$. Since $a_n > 0$ for all $n \ge 1$, we apply D'Alembert's Ratio Test.

Step 1: Formulate $a_{n+1}$

$$a_{n+1} = \frac{(n+1)!}{(2(n+1))!} = \frac{(n+1)!}{(2n+2)!}$$

Step 2: Evaluate the limit ratio

$$L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \left( \frac{(n+1)!}{(2n+2)!} \times \frac{(2n)!}{n!} \right)$$

Step 3: Simplify factorials

$$\frac{(n+1)!}{n!} = n+1, \quad \frac{(2n)!}{(2n+2)!} = \frac{(2n)!}{(2n+2)(2n+1)(2n)!} = \frac{1}{(2n+2)(2n+1)}$$ $$L = \lim_{n \to \infty} \frac{n+1}{(2n+2)(2n+1)} = \lim_{n \to \infty} \frac{n+1}{2(n+1)(2n+1)} = \lim_{n \to \infty} \frac{1}{2(2n+1)}$$

Step 4: Compute limit

$$L = \frac{1}{\infty} = 0$$

Since $L = 0 < 1$, by D'Alembert's Ratio Test, the series $\sum_{n=1}^{\infty} \frac{n!}{(2n)!}$ converges absolutely.

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Question 2 (Ex 8.3 - Integral Test Application)

Problem: Test the convergence of the series using Cauchy's Integral Test:

$$\sum_{n=2}^{\infty} \frac{1}{n \ln(n)}$$

Solution:

Step 1: Check conditions for the Integral Test

Let $f(x) = \frac{1}{x \ln(x)}$ for $x \in [2, \infty)$.

• $f(x) > 0$ for $x \ge 2$.
• $f(x)$ is continuous for $x \ge 2$.
• $f'(x) = -\frac{\ln(x) + 1}{(x \ln(x))^2} < 0$ for $x \ge 2$, so $f(x)$ is monotonically decreasing.

Step 2: Evaluate the Improper Integral

$$I = \int_{2}^{\infty} \frac{1}{x \ln(x)} \, dx = \lim_{t \to \infty} \int_{2}^{t} \frac{1}{x \ln(x)} \, dx$$

Using substitution $u = \ln(x) \implies du = \frac{1}{x} \, dx$:

$$\int \frac{1}{x \ln(x)} \, dx = \ln|\ln(x)|$$

Applying boundaries:

$$I = \lim_{t \to \infty} \left[ \ln(\ln(t)) - \ln(\ln(2)) \right]$$

Step 3: Conclusion

As $t \to \infty$, $\ln(\ln(t)) \to \infty$. Thus, $I = \infty$ (diverges).

By Cauchy's Integral Test, the series $\sum_{n=2}^{\infty} \frac{1}{n \ln(n)}$ diverges.

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Question 3 (Ex 8.5 - Absolute vs Conditional Convergence)

Problem: Test the series for absolute and conditional convergence:

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

Solution:

Step 1: Test for Absolute Convergence

Consider the series of absolute values:

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

This is a $p$-series with $p = \frac{1}{2} \le 1$. Hence, the absolute series $\sum \frac{1}{\sqrt{n}}$ diverges. Therefore, the given series is not absolutely convergent.

Step 2: Apply Leibniz Rule (Alternating Series Test)

Here, $b_n = \frac{1}{\sqrt{n}} > 0$.

1. Monotonicity: $\sqrt{n+1} > \sqrt{n} \implies \frac{1}{\sqrt{n+1}} < \frac{1}{\sqrt{n}} \implies b_{n+1} < b_n$. Thus, $b_n$ is decreasing.
2. Limit Check: $\lim_{n \to \infty} b_n = \lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0$.

Both conditions of Leibniz Rule are satisfied. Thus, the series converges.

Conclusion: The series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{\sqrt{n}}$ is conditionally convergent.

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Question 4 (Ex 8.4 - Cauchy's Root Test Application)

Problem: Test the convergence of the series:

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

Solution:

Let $a_n = \left( \frac{n}{2n + 1} \right)^n$. Since $a_n > 0$, we apply Cauchy's Root Test.

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

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

Step 2: Evaluate the limit

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

Step 3: Conclusion

Since $L = \frac{1}{2} < 1$, Cauchy's Root Test confirms that the series converges absolutely.

Interactive Practice Quiz (Clickable MCQs)

Frequently Asked Questions (FAQs)

Q1: Why does $\lim_{n \to \infty} a_n = 0$ NOT guarantee that a series converges?

Answer: The condition $\lim_{n \to \infty} a_n = 0$ means the individual terms are getting infinitely small, but it does not tell us how fast they are shrinking. If the terms do not shrink fast enough, their partial sums can grow without bound. The classic counterexample is the Harmonic Series $\sum_{n=1}^{\infty} \frac{1}{n}$. Here, $\lim_{n \to \infty} \frac{1}{n} = 0$, yet the sum diverges to infinity because the accumulation of terms grows logarithmically ($\ln k \to \infty$).

Q2: How do I decide whether to use the Ratio Test or the Root Test for S.M. Yusuf Chapter 8 questions?

Answer: Use these guidelines based on the algebraic form of $a_n$:

• Use the Ratio Test ($\lim \left|\frac{a_{n+1}}{a_n}\right|$): Whenever the general term contains factorials (e.g., $n!$, $(2n)!$) or products of consecutive terms. Factorials cancel out cleanly when dividing $a_{n+1}$ by $a_n$.
• Use the Root Test ($\lim |a_n|^{1/n}$): Whenever the entire term $a_n$ is raised to an $n$-th power (e.g., $(f(n))^n$ or $a_n = \left(\frac{n}{3n+1}\right)^n$). Taking the $1/n$ power eliminates the outer exponent directly.

Q3: What is the exact difference between Absolute Convergence and Conditional Convergence?

Answer:

• Absolute Convergence: A series $\sum a_n$ converges absolutely if the series of absolute values $\sum |a_n|$ converges. Absolute convergence is stronger; an absolutely convergent series remains convergent regardless of how its terms are rearranged (Riemann Series Theorem).
• Conditional Convergence: A series $\sum a_n$ converges conditionally if $\sum a_n$ converges, but its absolute series $\sum |a_n|$ diverges. A prime example is $\sum \frac{(-1)^{n-1}}{n}$, which converges by the Leibniz rule, but whose absolute series $\sum \frac{1}{n}$ diverges.