S.M. Yusuf Mathematical Methods Chapter 8 Infinite Series Solved Exercises & Convergence Tests | PU UOS KU FBISE Notes

Welcome to the ultimate preparation guide for Chapter 8: Infinite Series from the standard textbook Mathematical Methods by S.M. Yusuf and Abdul Majeed. This unit is a cornerstone for BS Mathematics, BSc, and Higher Secondary/Intermediate examination boards across Pakistan, including Punjab University (PU), University of Sargodha (UOS), University of Karachi (KU), FBISE, and all Punjab Boards.

Key Definitions & Theoretical Framework

1. Infinite Series and Sequence of Partial Sums

An expression of the form:

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

is called an infinite series. To determine whether an infinite series converges or diverges, we construct its Sequence of Partial Sums (SOPS) $\{S_n\}$, defined as:

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

• Convergence: If $\lim_{n \to \infty} S_n = S$ exists as a finite real number, the series $\sum a_n$ is said to converge, and $S$ is called the sum of the series.
• Divergence: If $\lim_{n \to \infty} S_n$ does not exist or tends to $\pm \infty$, the series diverges.

2. Absolute vs. Conditional Convergence

• Absolute Convergence: A series $\sum a_n$ is absolutely convergent if the series of absolute values $\sum |a_n|$ converges. Absolute convergence implies convergence, i.e., if $\sum |a_n|$ converges, then $\sum a_n$ also converges.
• Conditional Convergence: A series $\sum a_n$ is conditionally convergent if $\sum a_n$ converges, but $\sum |a_n|$ diverges.

3. Overview of Fundamental Convergence Tests

Test Name	Mathematical Statement / Condition	Conclusion
$n$-th Term Test for Divergence	Calculate $L = \lim_{n \to \infty} a_n$	If $L \neq 0$, the series diverges. (If $L = 0$, test is inconclusive).
Basic Comparison Test (BCT)	Let $0 \le a_n \le b_n$ for all $n \ge k$.	1. If $\sum b_n$ converges, $\sum a_n$ converges. 2. If $\sum a_n$ diverges, $\sum b_n$ diverges.
Limit Comparison Test (LCT)	Calculate $L = \lim_{n \to \infty} \frac{a_n}{b_n}$ ($a_n, b_n > 0$)	If $0 < L < \infty$, both $\sum a_n$ and $\sum b_n$ behave identically (both converge or both diverge).
D'Alembert's Ratio Test	Calculate $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$	- $L < 1$: Absolutely convergent. - $L > 1$ or $L = \infty$: Divergent. - $L = 1$: Test inconclusive.
Cauchy's Root Test	Calculate $L = \lim_{n \to \infty} |a_n|^{1/n}$	- $L < 1$: Absolutely convergent. - $L > 1$ or $L = \infty$: Divergent. - $L = 1$: Test inconclusive.
Cauchy's Integral Test	Let $a_n = f(n)$, where $f(x)$ is positive, continuous, decreasing on $[1, \infty)$.	$\sum_{n=1}^{\infty} a_n$ and $\int_{1}^{\infty} f(x) dx$ both converge or both diverge.
Leibniz's Alternating Series Test (AST)	For $\sum (-1)^{n-1} a_n$ ($a_n > 0$): 1. $a_{n+1} \le a_n$ for all $n$ 2. $\lim_{n \to \infty} a_n = 0$	If both conditions are met, the alternating series converges.

Important Standard Formulas

When working through Chapter 8 exercises, memory of these standard benchmark series and limit evaluation formulas is essential:

• Geometric Series Benchmark: $$\sum_{n=0}^{\infty} a r^n = a + ar + ar^2 + \dots$$
  • Converges if $|r| < 1$, with sum $S = \frac{a}{1 - r}$.
  • Diverges if $|r| \ge 1$.
• Hyper-Harmonic ($p$-series) Test Benchmark: $$\sum_{n=1}^{\infty} \frac{1}{n^p} = 1 + \frac{1}{2^p} + \frac{1}{3^p} + \dots$$
  • Converges if $p > 1$.
  • Diverges if $p \le 1$ (When $p = 1$, it is the standard Harmonic Series, which diverges).
• Standard Limits for Series Analysis:
  • $\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n = e^x$
  • $\lim_{n \to \infty} n^{1/n} = 1$
  • $\lim_{n \to \infty} \frac{n!}{n^n} = 0$
  • $\lim_{n \to \infty} \frac{x^n}{n!} = 0 \quad \text{for any real } x$

Step-by-Step Solved Board/Exam Questions

Question 1: (Limit Comparison Test - S.M. Yusuf Ex 8.2)

Problem: Determine the convergence or divergence of the series $\sum_{n=1}^{\infty} \frac{n^2 + 5}{n^4 + 3n + 1}$.

Solution:

Let the given series be $\sum a_n$, where:

$$a_n = \frac{n^2 + 5}{n^4 + 3n + 1} > 0 \quad \forall n \ge 1$$

To choose an appropriate auxiliary series $\sum b_n$, inspect the highest powers of $n$ in the numerator and denominator:

$$b_n = \frac{n^2}{n^4} = \frac{1}{n^2}$$

Now apply the Limit Comparison Test (LCT):

$$L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \left( \frac{n^2 + 5}{n^4 + 3n + 1} \cdot \frac{n^2}{1} \right)$$ $$L = \lim_{n \to \infty} \frac{n^4 + 5n^2}{n^4 + 3n + 1}$$

Divide numerator and denominator by $n^4$:

$$L = \lim_{n \to \infty} \frac{1 + \frac{5}{n^2}}{1 + \frac{3}{n^3} + \frac{1}{n^4}} = \frac{1 + 0}{1 + 0 + 0} = 1$$

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

The auxiliary series $\sum b_n = \sum_{n=1}^{\infty} \frac{1}{n^2}$ is a $p$-series with $p = 2 > 1$, which is convergent.

Conclusion: By the Limit Comparison Test, the given series $\sum_{n=1}^{\infty} \frac{n^2 + 5}{n^4 + 3n + 1}$ is also convergent.

---

Question 2: (D'Alembert's Ratio Test - S.M. Yusuf Ex 8.4)

Problem: Test the convergence of the series $\sum_{n=1}^{\infty} \frac{n!}{n^n}$.

Solution:

Let the given series be $\sum a_n$, where:

$$a_n = \frac{n!}{n^n}$$

Find $a_{n+1}$ by replacing $n$ with $n+1$:

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

Form the ratio $\frac{a_{n+1}}{a_n}$:

$$\frac{a_{n+1}}{a_n} = \frac{n!}{(n+1)^n} \cdot \frac{n^n}{n!} = \left( \frac{n}{n+1} \right)^n = \left( \frac{1}{\frac{n+1}{n}} \right)^n = \frac{1}{\left(1 + \frac{1}{n}\right)^n}$$

Apply limit as $n \to \infty$:

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

Using the standard limit $\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$:

$$L = \frac{1}{e}$$

Since $e \approx 2.718$, we have $L = \frac{1}{2.718} < 1$.

Conclusion: By D'Alembert's Ratio Test, the given series $\sum_{n=1}^{\infty} \frac{n!}{n^n}$ is convergent.

---

Question 3: (Cauchy's Root Test - S.M. Yusuf Ex 8.4)

Problem: Test the convergence of the series $\sum_{n=1}^{\infty} \left( \frac{n}{2n + 1} \right)^n$.

Solution:

Let $a_n = \left( \frac{n}{2n + 1} \right)^n$. Because the whole expression is raised to the $n$-th power, we apply Cauchy's Root Test:

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

Divide numerator and denominator by $n$:

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

Since $L = \frac{1}{2} < 1$, the test guarantees convergence.

Conclusion: By Cauchy's Root Test, the series $\sum_{n=1}^{\infty} \left( \frac{n}{2n + 1} \right)^n$ is absolutely convergent (and hence convergent).

---

Question 4: (Alternating Series & Conditional Convergence - S.M. Yusuf Ex 8.5)

Problem: Determine whether the series $\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{\sqrt{n}}$ is absolutely convergent, conditionally convergent, or divergent.

Solution:

Step 1: Test for Absolute Convergence
Consider the series of absolute values $\sum_{n=1}^{\infty} |a_n|$:

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

This is a $p$-series with $p = 1/2$. Since $p = 1/2 \le 1$, the absolute series $\sum |a_n|$ diverges.
Therefore, the series is not absolutely convergent.

Step 2: Test original series using Leibniz's Alternating Series Test (AST)
The alternating series is $\sum_{n=1}^{\infty} (-1)^{n-1} a_n$, where $a_n = \frac{1}{\sqrt{n}} > 0$. Check the two conditions:

1. Monotonic Decrease: $\sqrt{n+1} > \sqrt{n} \implies \frac{1}{\sqrt{n+1}} < \frac{1}{\sqrt{n}} \implies a_{n+1} < a_n$. (Condition satisfied).
2. Zero Limit: $\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0$. (Condition satisfied).

Since both conditions of AST are satisfied, the series $\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{\sqrt{n}}$ converges.

Conclusion: Because the series converges but fails absolute convergence, it is conditionally convergent.

Top 8 Solved MCQs with Rationale

1. The $p$-series $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ is:
   (A) Divergent
   (B) Absolutely Convergent
   (C) Conditionally Convergent
   (D) Oscillatory
   Correct Answer: (B) Absolutely Convergent
   Rationale: For a $p$-series $\sum \frac{1}{n^p}$, it converges if $p > 1$. Here $p = 3/2 = 1.5 > 1$. Since all terms are positive, convergence implies absolute convergence.


2. If $\lim_{n \to \infty} a_n = 5$, then the infinite series $\sum_{n=1}^{\infty} a_n$:
   (A) Converges to 5
   (B) Converges to 0
   (C) Diverges
   (D) Requires Ratio Test
   Correct Answer: (C) Diverges
   Rationale: By the $n$-th Term Test for Divergence, if $\lim_{n \to \infty} a_n \neq 0$, the series must diverge. Here the limit is $5 \neq 0$.


3. The sum of the infinite geometric series $1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \dots$ is:
   (A) $\frac{1}{2}$
   (B) $\frac{2}{3}$
   (C) $\frac{4}{3}$
   (D) $\frac{3}{2}$
   Correct Answer: (B) $\frac{2}{3}$
   Rationale: First term $a = 1$, common ratio $r = -\frac{1}{2}$. Since $|r| < 1$, $S = \frac{a}{1 - r} = \frac{1}{1 - (-1/2)} = \frac{1}{3/2} = \frac{2}{3}$.


4. If D'Alembert's Ratio Test yields $L = 1$, the test is:
   (A) Convergent
   (B) Divergent
   (C) Absolutely Convergent
   (D) Inconclusive
   Correct Answer: (D) Inconclusive
   Rationale: When $L = 1$, the Ratio Test provides no information. Another test (such as LCT, Integral Test, or Raabe's Test) must be applied.


5. The alternating harmonic series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$ is:
   (A) Absolutely Convergent
   (B) Conditionally Convergent
   (C) Divergent
   (D) Unbounded
   Correct Answer: (B) Conditionally Convergent
   Rationale: The series converges by Leibniz's AST, but its absolute series $\sum \frac{1}{n}$ is the harmonic series which diverges ($p=1$). Thus it is conditionally convergent.


6. Using Cauchy's Root Test for $\sum_{n=1}^{\infty} \left(\frac{3n+1}{2n+5}\right)^n$, the limit $L$ equals:
   (A) $\frac{2}{3}$
   (B) $\frac{3}{2}$
   (C) $1$
   (D) $0$
   Correct Answer: (B) $\frac{3}{2}$
   Rationale: $L = \lim_{n \to \infty} (a_n)^{1/n} = \lim_{n \to \infty} \frac{3n+1}{2n+5} = \frac{3}{2}$. Since $L = 1.5 > 1$, the series diverges.


7. Which of the following tests is most suitable for series involving factorials like $(2n)!$?
   (A) Cauchy's Root Test
   (B) D'Alembert's Ratio Test
   (C) Integral Test
   (D) Geometric Series Test
   Correct Answer: (B) D'Alembert's Ratio Test
   Rationale: Factorials simplify efficiently when forming the ratio $\frac{a_{n+1}}{a_n}$, as $(n+1)! = (n+1)n!$.


8. The series $\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$ can be shown to converge using:
   (A) $n$-th Term Test
   (B) Limit Comparison Test with $\sum \frac{1}{n^2}$
   (C) Cauchy's Root Test
   (D) Both (B) and Integral Test
   Correct Answer: (D) Both (B) and Integral Test
   Rationale: It easily compares to $\sum \frac{1}{n^2}$ ($p=2 > 1$) via LCT, and $f(x) = \frac{1}{x^2+1}$ integrates easily to $\arctan(x)$, making both tests valid.