S.M. Yusuf Mathematical Methods Chapter 8 Infinite Series Solved Exercises & Convergence Tests PDF Notes - Punjab University, FBISE, QAU, UOS
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.
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 $\sum \frac{1}{n}$).
When $a_n \ge 0$ for all $n$, the sequence of partial sums $\{S_n\}$ is monotonically non-decreasing. Key tests include:
An alternating series has the form $\sum_{n=1}^{\infty} (-1)^{n-1} a_n$ where $a_n > 0$.
| 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. |
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.
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.
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).
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 $a_n = \frac{1}{\sqrt{n}} > 0$:
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.
Q1: The hyper-harmonic series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if and only if:
Q2: If D'Alembert's Ratio Test yields $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 1$, what can be concluded?
Q3: Raabe's Test states that a positive term series converges if $\rho = \lim_{n \to \infty} n \left( \frac{a_n}{a_{n+1}} - 1 \right)$ satisfies:
Q4: What is the sum of the telescoping infinite series $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$?
Q5: Which test is most suitable to check the absolute convergence of $\sum_{n=1}^{\infty} \left(\frac{2n+1}{3n+4}\right)^n$?
Q6: If $\lim_{n \to \infty} a_n = 0$ for a series $\sum a_n$, what can be stated about its convergence?
Q7: The series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$ is known to be:
Q8: For the series $\sum_{n=1}^{\infty} \frac{n^2}{3n^2 + 1}$, the $n$-th term test shows that the series:
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:
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:
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.