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.
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:
$$S_n = \sum_{k=1}^{n} a_k = a_1 + a_2 + \dots + a_n$$| 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. |
When working through Chapter 8 exercises, memory of these standard benchmark series and limit evaluation formulas is essential:
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.
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.
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).
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 $\sum_{n=1}^{\infty} |a_n|$:
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:
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.