S.M. Yusuf Mathematical Methods Chapter 8: Infinite Series Solved Exercises & Convergence Tests

1. Introduction & Key Concepts of Infinite Series

In Mathematical Methods by S.M. Yusuf (Chapter 8), an infinite series is defined as the formal sum of an infinite sequence of real numbers $a_1, a_2, a_3, \dots$, denoted as:

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

The convergence or divergence of the series depends on the sequence of its partial sums $S_n = \sum_{k=1}^{n} a_k$. If $\lim_{n \to \infty} S_n = S$ exists as a finite real number, the series converges to $S$; otherwise, it diverges.

Primary Tests for Convergence

• $n$-th Term Test for Divergence: If $\lim_{n \to \infty} a_n \neq 0$, then the series $\sum a_n$ diverges. Note: If $\lim_{n \to \infty} a_n = 0$, the test is inconclusive.
• $p$-Series Test: The hyper-harmonic series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if $p > 1$ and diverges if $p \le 1$.
• Limit Comparison Test (LCT): For positive term series $\sum a_n$ and $\sum b_n$, if $\lim_{n \to \infty} \frac{a_n}{b_n} = c$ where $0 < c < \infty$, then both series either converge or diverge together.
• D'Alembert's Ratio Test: Let $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$.
  • If $L < 1$, the series converges absolutely.
  • If $L > 1$, the series diverges.
  • If $L = 1$, the test fails.
• Cauchy's $n$-th Root Test: Let $L = \lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty} |a_n|^{1/n}$.
  • If $L < 1$, the series converges.
  • If $L > 1$, the series diverges.
  • If $L = 1$, the test fails.
• Integral Test: If $f(x)$ is continuous, positive, and decreasing on $[1, \infty)$ such that $f(n) = a_n$, then $\sum_{n=1}^{\infty} a_n$ and $\int_{1}^{\infty} f(x) dx$ behave identically.
• Alternating Series Test (Leibniz Test): An alternating series $\sum (-1)^{n-1} u_n$ ($u_n > 0$) converges if:
  1. $u_{n+1} \le u_n$ for all $n$.
  2. $\lim_{n \to \infty} u_n = 0$.

2. Step-by-Step Solved Exercises

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

Solution:

Let $a_n = \frac{n!}{n^n}$. Then the next term $a_{n+1}$ is:

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

Applying D'Alembert's Ratio Test:

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

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

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

Since $e \approx 2.718$, $L = \frac{1}{e} < 1$. By D'Alembert's Ratio Test, the series converges absolutely.

Example 2: Test the convergence of $\sum_{n=1}^{\infty} \frac{1}{n^2 + 5n + 6}$

Solution:

Let $a_n = \frac{1}{n^2 + 5n + 6}$. For large values of $n$, the dominant term in the denominator is $n^2$. Choose $b_n = \frac{1}{n^2}$.

Using the Limit Comparison Test (LCT):

$$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \left( \frac{1}{n^2 + 5n + 6} \cdot \frac{n^2}{1} \right) = \lim_{n \to \infty} \frac{n^2}{n^2(1 + \frac{5}{n} + \frac{6}{n^2})} = 1$$

Since $c = 1$ is a finite non-zero number, both series behave identically.

By $p$-series test, $\sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} \frac{1}{n^2}$ converges because $p = 2 > 1$. Hence, the given series converges by the Limit Comparison Test.

Example 3: Test the convergence of $\sum_{n=2}^{\infty} \frac{1}{n \ln(n)}$

Solution:

Let $f(x) = \frac{1}{x \ln(x)}$ for $x \ge 2$. Note that $f(x)$ is positive, continuous, and monotonically decreasing on $[2, \infty)$. We apply the Integral Test:

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

Since $\frac{d}{dx}[\ln(x)] = \frac{1}{x}$, using substitution $u = \ln(x)$:

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

Since the improper integral diverges to $\infty$, the series $\sum_{n=2}^{\infty} \frac{1}{n \ln(n)}$ diverges by the Integral Test.

Example 4: Determine whether $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n}}$ is absolutely convergent, conditionally convergent, or divergent.

Solution:

Step 1: Absolute Convergence Check
Consider the series of absolute values $\sum_{n=1}^{\infty} \left| \frac{(-1)^{n-1}}{\sqrt{n}} \right| = \sum_{n=1}^{\infty} \frac{1}{n^{1/2}}$.
This is a $p$-series with $p = 1/2 \le 1$, so it diverges. Therefore, the series is not absolutely convergent.

Step 2: Alternating Series Test (Leibniz Test)
Here $u_n = \frac{1}{\sqrt{n}} > 0$.
1. Monotonicity: Since $\sqrt{n+1} > \sqrt{n}$, we have $u_{n+1} = \frac{1}{\sqrt{n+1}} < \frac{1}{\sqrt{n}} = u_n$. The terms are decreasing.
2. Limit check: $\lim_{n \to \infty} u_n = \lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0$.

Both conditions of Leibniz Test are satisfied. Thus, the series converges.
Conclusion: The series is conditionally convergent.

3. Important Multiple Choice Questions (MCQs)

1. The series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if and only if:
   • A) $p < 1$
   • B) $p > 1$
   • C) $p = 1$
   • D) $p \le 1$

   Rationale: By the $p$-series test, $\sum \frac{1}{n^p}$ converges strictly when the exponent $p > 1$.

2. If $\lim_{n \to \infty} a_n \neq 0$, then the series $\sum a_n$:
   • A) Converges
   • B) Diverges
   • C) Converges conditionally
   • D) Is inconclusive

   Rationale: The $n$-th term test states that a non-zero limit of the general term guarantees divergence.

3. In D'Alembert's Ratio Test, if $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 1$, the test is:
   • A) Convergent
   • B) Divergent
   • C) Inconclusive (Fails)
   • D) Absolutely Convergent

   Rationale: When $L = 1$, the ratio test yields no conclusion and another test must be applied.

4. Cauchy's Root Test is most suitable for series where $a_n$ involves:
   • A) Factorials ($n!$)
   • B) $n$-th powers ($f(n)^n$)
   • C) Logarithmic terms only
   • D) Partial fractions

   Rationale: The $n$-th root test simplifies expressions involving $n$-th powers through $\sqrt[n]{(g(n))^n} = g(n)$.

5. The alternating harmonic series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$ is:
   • A) Divergent
   • B) Absolutely Convergent
   • C) Conditionally Convergent
   • D) Oscillatory

   Rationale: It converges by the Leibniz test, but its absolute counterpart $\sum \frac{1}{n}$ (harmonic series) diverges.

6. For an infinite geometric series $\sum_{n=0}^{\infty} a r^n$ to converge, the common ratio $r$ must satisfy:
   • A) $|r| > 1$
   • B) $|r| < 1$
   • C) $r = 1$
   • D) $r \ge 1$

   Rationale: A geometric series converges to $\frac{a}{1-r}$ if and only if $|r| < 1$.

7. If $\sum a_n$ is a series of positive terms and $\lim_{n \to \infty} \frac{a_n}{b_n} = 5$, then:
   • A) $\sum a_n$ converges only
   • B) $\sum b_n$ diverges only
   • C) Both series behave identically
   • D) The test is inconclusive

   Rationale: By the Limit Comparison Test, if the limit is a finite positive constant ($0 < c < \infty$), both series have the same convergence behavior.

8. The sum of the infinite geometric series $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$ is:
   • A) $1$
   • B) $2$
   • C) $3/2$
   • D) $\infty$

   Rationale: Here $a = 1$ and $r = 1/2$. The sum is $S = \frac{a}{1-r} = \frac{1}{1 - 1/2} = 2$.

Rule