Sequence & Series

Sequence and Series — Practice Questions & Solutions

> Perfect for: Semester exams, GATE, ESE, and competitive exam preparation. Step-by-step solutions with exam-oriented explanations.

Module 1: Sequence and Series — Definitions & Examples

Q1. Define sequence and series. Give examples of convergent, divergent, and oscillating sequences.

Solution:

Sequence: A sequence is a function whose domain is the set of natural numbers $\mathbb{N}$. It is denoted by $\{a_n\}_{n=1}^{\infty}$ or simply $\{a_n\}$.

Series: A series is the sum of the terms of a sequence. If $\{a_n\}$ is a sequence, then $S = a_1 + a_2 + a_3 + \dots = \sum_{n=1}^{\infty} a_n$ is an infinite series.

Examples:

Q2. Show that the sequence $a_n = \frac{n^2+1}{2n^2+3}$ converges to .

Solution:

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n^2+1}{2n^2+3}$

Divide numerator and denominator by $n^2$:

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

$\boxed{\lim_{n \to \infty} \frac{n^2+1}{2n^2+3} = \frac{1}{2}}$

The sequence converges to $\frac{1}{2}$.

Q3. Test the convergence of the series $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$.

Solution:

Use partial fraction decomposition:

$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$

The $N$-th partial sum is:

$S_N = \sum_{n=1}^{N} \left(\frac{1}{n} - \frac{1}{n+1}\right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{N} - \frac{1}{N+1}\right)$

This is a telescoping series:

$S_N = 1 - \frac{1}{N+1}$

$\lim_{N \to \infty} S_N = \lim_{N \to \infty} \left(1 - \frac{1}{N+1}\right) = 1$

$\boxed{\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = 1 \text{ (Convergent)}}$

Module 2: Comparison Test

Q4. Test the convergence of $\sum_{n=1}^{\infty} \frac{1}{n^2+1}$.

Solution:

Let $a_n = \frac{1}{n^2+1}$ and compare with .

$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{n^2}{n^2+1} = 1 \quad \text{(finite and non-zero)}$

Since $\sum b_n = \sum \frac{1}{n^2}$ is a , it converges.

By the Limit Comparison Test, $\sum a_n$ also converges.

$\boxed{\sum_{n=1}^{\infty} \frac{1}{n^2+1} \text{ is convergent}}$

Q5. Test the convergence of $\sum_{n=1}^{\infty} \frac{n+1}{n^2+3}$.

Solution:

Let $a_n = \frac{n+1}{n^2+3}$. Compare with .

$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{n(n+1)}{n^2+3} = \lim_{n \to \infty} \frac{n^2+n}{n^2+3} = 1$

Since $\sum b_n = \sum \frac{1}{n}$ is the harmonic series ($p=1$), it diverges.

By the Limit Comparison Test, $\sum a_n$ also diverges.

$\boxed{\sum_{n=1}^{\infty} \frac{n+1}{n^2+3} \text{ is divergent}}$

Q6. Test the convergence of $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^3+2n}}$.

Solution:

Compare with $b_n = \frac{1}{n^{3/2}}$.

$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{n^{3/2}}{\sqrt{n^3+2n}} = \lim_{n \to \infty} \sqrt{\frac{n^3}{n^3+2n}} = 1$

Since $\sum \frac{1}{n^{3/2}}$ is a p-series with p=\frac{3}{2} > 1, it converges.

$\boxed{\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^3+2n}} \text{ is convergent}}$

Module 3: Ratio Test (D'Alembert's Test)

Q7. State and apply D'Alembert's ratio test to $\sum_{n=1}^{\infty} \frac{n^2}{2^n}$.

Solution:

D'Alembert's Ratio Test: For a series $\sum a_n$ with positive terms, compute:

$L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n}$

• If L < 1: series converges
• If L > 1: series diverges
• If $L = 1$: test fails

Application:

$a_n = \frac{n^2}{2^n}, \quad a_{n+1} = \frac{(n+1)^2}{2^{n+1}}$

$\frac{a_{n+1}}{a_n} = \frac{(n+1)^2}{2^{n+1}} \cdot \frac{2^n}{n^2} = \frac{(n+1)^2}{2n^2}$

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

Since L = \frac{1}{2} < 1:

$\boxed{\sum_{n=1}^{\infty} \frac{n^2}{2^n} \text{ is convergent}}$

Q8. Test the convergence of $\sum_{n=1}^{\infty} \frac{n!}{n^n}$.

Solution:

$a_n = \frac{n!}{n^n}, \quad a_{n+1} = \frac{(n+1)!}{(n+1)^{n+1}}$

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

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

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

Since L = \frac{1}{e} < 1:

$\boxed{\sum_{n=1}^{\infty} \frac{n!}{n^n} \text{ is convergent}}$

Q9. Test the convergence of $\sum_{n=1}^{\infty} \frac{3^n \cdot n!}{n^n}$.

Solution:

$a_n = \frac{3^n \cdot n!}{n^n}$

$\frac{a_{n+1}}{a_n} = \frac{3^{n+1}(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{3^n \cdot n!} = \frac{3(n+1) \cdot n^n}{(n+1)^{n+1}}$

$= \frac{3}{\left(1+\frac{1}{n}\right)^n}$

$L = \lim_{n \to \infty} \frac{3}{\left(1+\frac{1}{n}\right)^n} = \frac{3}{e} \approx \frac{3}{2.718} \approx 1.104$

Since L = \frac{3}{e} > 1:

$\boxed{\sum_{n=1}^{\infty} \frac{3^n \cdot n!}{n^n} \text{ is divergent}}$

Q10. Test the convergence of $\sum_{n=1}^{\infty} \frac{1}{n}$ using ratio test. What does this show?

Solution:

$a_n = \frac{1}{n}, \quad a_{n+1} = \frac{1}{n+1}$

$\frac{a_{n+1}}{a_n} = \frac{n}{n+1}$

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

$\boxed{L = 1 \Rightarrow \text{Ratio test fails}}$

This shows that the ratio test is inconclusive for the harmonic series. We must use the p-series test or integral test to prove divergence.

Module 4: Raabe's Test

Q11. State Raabe's test and apply it to $\sum_{n=1}^{\infty} \frac{1}{n^2}$.

Solution:

Raabe's Test: For a series $\sum a_n$ with positive terms, if $\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = 1$ (ratio test fails), compute:

$R = \lim_{n \to \infty} n\left(\frac{a_n}{a_{n+1}} - 1\right)$

• If R > 1: converges
• If R < 1: diverges
• If $R = 1$: test fails

Application to $\sum \frac{1}{n^2}$:

First check ratio test:

$\frac{a_n}{a_{n+1}} = \frac{(n+1)^2}{n^2} = \left(1+\frac{1}{n}\right)^2 \to 1$

Ratio test fails. Apply Raabe's test:

$n\left(\frac{a_n}{a_{n+1}} - 1\right) = n\left[\left(1+\frac{1}{n}\right)^2 - 1\right] = n\left[1 + \frac{2}{n} + \frac{1}{n^2} - 1\right]$

$= n\left[\frac{2}{n} + \frac{1}{n^2}\right] = 2 + \frac{1}{n}$

$R = \lim_{n \to \infty} \left(2 + \frac{1}{n}\right) = 2$

Since R = 2 > 1:

$\boxed{\sum_{n=1}^{\infty} \frac{1}{n^2} \text{ is convergent}}$

Q12. Test the convergence of $\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}$.

Solution:

$a_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}$

$a_{n+1} = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)(2n+1)}{2 \cdot 4 \cdot 6 \cdots (2n)(2n+2)}$

$\frac{a_{n+1}}{a_n} = \frac{2n+1}{2n+2}$

$L = \lim_{n \to \infty} \frac{2n+1}{2n+2} = 1 \quad \text{(Ratio test fails)}$

Apply Raabe's test:

$\frac{a_n}{a_{n+1}} = \frac{2n+2}{2n+1}$

$n\left(\frac{a_n}{a_{n+1}} - 1\right) = n\left(\frac{2n+2}{2n+1} - 1\right) = n\left(\frac{1}{2n+1}\right) = \frac{n}{2n+1}$

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

Since R = \frac{1}{2} < 1:

$\boxed{\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \text{ is divergent}}$

Q13. Test the convergence of $\sum_{n=1}^{\infty} \frac{n}{(n+1)(n+2)(n+3)}$.

Solution:

Use comparison test directly:

$a_n = \frac{n}{(n+1)(n+2)(n+3)} \approx \frac{n}{n^3} = \frac{1}{n^2} \text{ for large } n$

Compare with $b_n = \frac{1}{n^2}$:

$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{n^3}{(n+1)(n+2)(n+3)} = 1$

Since $\sum \frac{1}{n^2}$ converges (p=2 > 1):

$\boxed{\sum_{n=1}^{\infty} \frac{n}{(n+1)(n+2)(n+3)} \text{ is convergent}}$

Module 5: Fourier Series

Q14. Find the Fourier series of $f(x) = x$ in the interval $(-\pi, \pi)$.

Solution:

The Fourier series is:

$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx)$

where:

$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx$

$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos nx \, dx$

$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin nx \, dx$

Compute $a_0$:

$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x \, dx = 0 \quad \text{(odd function over symmetric interval)}$

Compute $a_n$:

$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \cos nx \, dx = 0 \quad \text{(odd $\times$ even = odd)}$

Compute $b_n$:

$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \sin nx \, dx = \frac{2}{\pi} \int_{0}^{\pi} x \sin nx \, dx$

Using integration by parts: $u = x$, $dv = \sin nx \, dx$

$= \frac{2}{\pi}\left[\left(-\frac{x \cos nx}{n}\right)_0^{\pi} + \frac{1}{n}\int_0^{\pi} \cos nx \, dx\right]$

$= \frac{2}{\pi}\left[-\frac{\pi \cos n\pi}{n} + 0 + \frac{1}{n}\left(\frac{\sin nx}{n}\right)_0^{\pi}\right]$

$= \frac{2}{\pi}\left[-\frac{\pi (-1)^n}{n} + 0\right] = \frac{2(-1)^{n+1}}{n}$

$\boxed{f(x) = x = 2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin nx = 2\left(\sin x - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} - \dots\right)}$

Q15. Find the Fourier series of $f(x) = x^2$ in $(-\pi, \pi)$. Hence deduce that $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$.

Solution:

$f(x) = x^2$ is an even function, so $b_n = 0$.

Compute $a_0$:

$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \, dx = \frac{2}{\pi} \int_{0}^{\pi} x^2 \, dx = \frac{2}{\pi} \cdot \frac{\pi^3}{3} = \frac{2\pi^2}{3}$

Compute $a_n$:

$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \cos nx \, dx = \frac{2}{\pi} \int_{0}^{\pi} x^2 \cos nx \, dx$

Using integration by parts twice:

$\int x^2 \cos nx \, dx = \frac{x^2 \sin nx}{n} + \frac{2x \cos nx}{n^2} - \frac{2 \sin nx}{n^3}$

Evaluating from $0$ to $\pi$:

$= \left(0 + \frac{2\pi \cos n\pi}{n^2} - 0\right) - (0) = \frac{2\pi (-1)^n}{n^2}$

$a_n = \frac{2}{\pi} \cdot \frac{2\pi (-1)^n}{n^2} = \frac{4(-1)^n}{n^2}$

Fourier series:

$\boxed{x^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx}$

Deduction:

At $x = \pi$:

$\pi^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos n\pi = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{(-1)^n \cdot (-1)^n}{n^2}$

$\pi^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{1}{n^2}$

$\frac{2\pi^2}{3} = 4\sum_{n=1}^{\infty} \frac{1}{n^2}$

$\boxed{\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}}$

Q16. Find the Fourier series of $f(x) = \begin{cases} 0 & -\pi < x < 0 \\ 1 & 0 < x < \pi \end{cases}$

Solution:

$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx = \frac{1}{\pi} \int_{0}^{\pi} 1 \, dx = 1$

$a_n = \frac{1}{\pi} \int_{0}^{\pi} \cos nx \, dx = \frac{1}{\pi}\left[\frac{\sin nx}{n}\right]_0^{\pi} = 0$

$b_n = \frac{1}{\pi} \int_{0}^{\pi} \sin nx \, dx = \frac{1}{\pi}\left[-\frac{\cos nx}{n}\right]_0^{\pi}$

$= -\frac{1}{n\pi}[\cos n\pi - \cos 0] = -\frac{1}{n\pi}[(-1)^n - 1]$

$= \begin{cases} \frac{2}{n\pi} & n \text{ odd} \\ 0 & n \text{ even} \end{cases}$

$\boxed{f(x) = \frac{1}{2} + \frac{2}{\pi}\left(\sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \dots\right)}$

Module 6: Half Range Fourier Series

Q17. Find the half-range sine series of $f(x) = x$ in $(0, \pi)$.

Solution:

For half-range sine series, we extend $f(x)$ as an odd function on $(-\pi, \pi)$.

$f(x) = \sum_{n=1}^{\infty} b_n \sin nx$

where:

$b_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) \sin nx \, dx = \frac{2}{\pi} \int_{0}^{\pi} x \sin nx \, dx$

Using integration by parts:

$= \frac{2}{\pi}\left[\left(-\frac{x \cos nx}{n}\right)_0^{\pi} + \frac{1}{n}\int_0^{\pi} \cos nx \, dx\right]$

$= \frac{2}{\pi}\left[-\frac{\pi \cos n\pi}{n} + \frac{\sin nx}{n^2}\Big|_0^{\pi}\right]$

$= \frac{2}{\pi}\left[-\frac{\pi (-1)^n}{n} + 0\right] = \frac{2(-1)^{n+1}}{n}$

$\boxed{f(x) = x = 2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin nx}$

Q18. Find the half-range cosine series of $f(x) = x$ in $(0, \pi)$.

Solution:

For half-range cosine series, extend $f(x)$ as an even function on $(-\pi, \pi)$.

$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos nx$

where:

$a_0 = \frac{2}{\pi} \int_{0}^{\pi} x \, dx = \frac{2}{\pi} \cdot \frac{\pi^2}{2} = \pi$

$a_n = \frac{2}{\pi} \int_{0}^{\pi} x \cos nx \, dx$

Integration by parts:

$= \frac{2}{\pi}\left[\left(\frac{x \sin nx}{n}\right)_0^{\pi} - \frac{1}{n}\int_0^{\pi} \sin nx \, dx\right]$

$= \frac{2}{\pi}\left[0 + \frac{\cos nx}{n^2}\Big|_0^{\pi}\right] = \frac{2}{\pi n^2}[\cos n\pi - 1]$

$= \frac{2}{\pi n^2}[(-1)^n - 1] = \begin{cases} -\frac{4}{\pi n^2} & n \text{ odd} \\ 0 & n \text{ even} \end{cases}$

$\boxed{f(x) = x = \frac{\pi}{2} - \frac{4}{\pi}\left(\cos x + \frac{\cos 3x}{9} + \frac{\cos 5x}{25} + \dots\right)}$

Q19. Find the half-range sine series of $f(x) = \pi - x$ in $(0, \pi)$.

Solution:

$b_n = \frac{2}{\pi} \int_{0}^{\pi} (\pi - x) \sin nx \, dx$

Split:

$= \frac{2}{\pi}\left[\pi \int_0^{\pi} \sin nx \, dx - \int_0^{\pi} x \sin nx \, dx\right]$

First integral:

$\int_0^{\pi} \sin nx \, dx = \left[-\frac{\cos nx}{n}\right]_0^{\pi} = \frac{1 - (-1)^n}{n}$

Second integral (from Q17):

$\int_0^{\pi} x \sin nx \, dx = \frac{\pi (-1)^{n+1}}{n}$

So:

$b_n = \frac{2}{\pi}\left[\frac{\pi(1-(-1)^n)}{n} - \frac{\pi(-1)^{n+1}}{n}\right]$

$= \frac{2}{n}\left[1 - (-1)^n + (-1)^n\right] = \frac{2}{n}$

$\boxed{f(x) = \pi - x = 2\sum_{n=1}^{\infty} \frac{\sin nx}{n}}$

Q20. Find the half-range cosine series of $f(x) = (\pi - x)^2$ in $(0, \pi)$.

Solution:

$a_0 = \frac{2}{\pi} \int_{0}^{\pi} (\pi - x)^2 \, dx$

Let $u = \pi - x$, $du = -dx$:

$= \frac{2}{\pi} \int_{0}^{\pi} u^2 \, du = \frac{2}{\pi} \cdot \frac{\pi^3}{3} = \frac{2\pi^2}{3}$

$a_n = \frac{2}{\pi} \int_{0}^{\pi} (\pi - x)^2 \cos nx \, dx$

Let $u = \pi - x$:

$= \frac{2}{\pi} \int_{0}^{\pi} u^2 \cos n(\pi - u) \, du = \frac{2}{\pi} \int_{0}^{\pi} u^2 (-1)^n \cos nu \, du$

Wait, $\cos n(\pi - u) = \cos n\pi \cos nu + \sin n\pi \sin nu = (-1)^n \cos nu$

$= \frac{2(-1)^n}{\pi} \int_{0}^{\pi} u^2 \cos nu \, du$

From Q15, we know:

$\int_{0}^{\pi} u^2 \cos nu \, du = \frac{2\pi (-1)^n}{n^2}$

So:

$a_n = \frac{2(-1)^n}{\pi} \cdot \frac{2\pi (-1)^n}{n^2} = \frac{4}{n^2}$

$\boxed{f(x) = (\pi-x)^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{\cos nx}{n^2}}$

Module 7: Mixed Exam-Level Problems

Q21. Test the convergence of $\sum_{n=1}^{\infty} \frac{(n+1)(n+2)}{n^2 \cdot 2^n}$.

Solution:

Use Ratio Test:

$a_n = \frac{(n+1)(n+2)}{n^2 \cdot 2^n}$

$a_{n+1} = \frac{(n+2)(n+3)}{(n+1)^2 \cdot 2^{n+1}}$

$\frac{a_{n+1}}{a_n} = \frac{(n+2)(n+3)}{(n+1)^2 \cdot 2^{n+1}} \cdot \frac{n^2 \cdot 2^n}{(n+1)(n+2)}$

$= \frac{(n+3) \cdot n^2}{2(n+1)^3} = \frac{n^3(1+\frac{3}{n})}{2n^3(1+\frac{1}{n})^3}$

L = \lim_{n \to \infty} \frac{1+\frac{3}{n}}{2(1+\frac{1}{n})^3} = \frac{1}{2} < 1

$\boxed{\text{The series is convergent}}$

Q22. Test the convergence of $\sum_{n=1}^{\infty} \frac{n^n}{n! \cdot 3^n}$.

Solution:

Ratio Test:

$a_n = \frac{n^n}{n! \cdot 3^n}, \quad a_{n+1} = \frac{(n+1)^{n+1}}{(n+1)! \cdot 3^{n+1}}$

$\frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{(n+1)! \cdot 3^{n+1}} \cdot \frac{n! \cdot 3^n}{n^n}$

$= \frac{(n+1)^{n+1}}{(n+1) \cdot n^n \cdot 3} = \frac{(n+1)^n}{3n^n} = \frac{1}{3}\left(1+\frac{1}{n}\right)^n$

L = \lim_{n \to \infty} \frac{1}{3}\left(1+\frac{1}{n}\right)^n = \frac{e}{3} \approx \frac{2.718}{3} \approx 0.906 < 1

$\boxed{\text{The series is convergent}}$

Q23. Test the convergence of $\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 5 \cdot 8 \cdots (3n-1)}$.

Solution:

$a_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 5 \cdot 8 \cdots (3n-1)}$

$a_{n+1} = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)(2n+1)}{2 \cdot 5 \cdot 8 \cdots (3n-1)(3n+2)}$

$\frac{a_{n+1}}{a_n} = \frac{2n+1}{3n+2}$

L = \lim_{n \to \infty} \frac{2n+1}{3n+2} = \frac{2}{3} < 1

$\boxed{\text{The series is convergent}}$

Q24. Find the Fourier series of $f(x) = e^x$ in $(-\pi, \pi)$.

Solution:

$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} e^x \, dx = \frac{1}{\pi}(e^{\pi} - e^{-\pi}) = \frac{2\sinh \pi}{\pi}$

$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} e^x \cos nx \, dx$

Using $\int e^{ax} \cos bx \, dx = \frac{e^{ax}(a\cos bx + b\sin bx)}{a^2+b^2}$:

$= \frac{1}{\pi}\left[\frac{e^x(\cos nx + n\sin nx)}{1+n^2}\right]_{-\pi}^{\pi}$

At $x = \pi$: $\cos n\pi = (-1)^n$, $\sin n\pi = 0$

At $x = -\pi$: $\cos(-n\pi) = (-1)^n$, $\sin(-n\pi) = 0$

$= \frac{1}{\pi(1+n^2)}[e^{\pi}(-1)^n - e^{-\pi}(-1)^n] = \frac{2(-1)^n \sinh \pi}{\pi(1+n^2)}$

Similarly:

$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} e^x \sin nx \, dx = \frac{1}{\pi}\left[\frac{e^x(\sin nx - n\cos nx)}{1+n^2}\right]_{-\pi}^{\pi}$

$= \frac{1}{\pi(1+n^2)}[e^{\pi}(0 - n(-1)^n) - e^{-\pi}(0 - n(-1)^n)]$

$= \frac{-n(-1)^n}{\pi(1+n^2)}(e^{\pi} - e^{-\pi}) = \frac{-2n(-1)^n \sinh \pi}{\pi(1+n^2)}$

$\boxed{e^x = \frac{\sinh \pi}{\pi}\left[1 + 2\sum_{n=1}^{\infty} \frac{(-1)^n}{1+n^2}(\cos nx - n\sin nx)\right]}$

Q25. Expand $f(x) = x\sin x$ as a Fourier series in $(-\pi, \pi)$.

Solution:

Note: $f(x) = x\sin x$ is an even function since $f(-x) = (-x)\sin(-x) = (-x)(-\sin x) = x\sin x = f(x)$.

Therefore $b_n = 0$.

$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x\sin x \, dx = \frac{2}{\pi} \int_{0}^{\pi} x\sin x \, dx$

$= \frac{2}{\pi}[-x\cos x + \sin x]_0^{\pi} = \frac{2}{\pi}[\pi] = 2$

$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x\sin x \cos nx \, dx = \frac{2}{\pi} \int_{0}^{\pi} x\sin x \cos nx \, dx$

Using $2\sin A \cos B = \sin(A+B) + \sin(A-B)$:

$= \frac{1}{\pi} \int_{0}^{\pi} x[\sin(n+1)x - \sin(n-1)x] \, dx$

For $n = 1$:

$a_1 = \frac{1}{\pi} \int_{0}^{\pi} x\sin 2x \, dx = \frac{1}{\pi}\left[-\frac{x\cos 2x}{2} + \frac{\sin 2x}{4}\right]_0^{\pi} = \frac{1}{\pi}\left[-\frac{\pi}{2}\right] = -\frac{1}{2}$

For $n \neq 1$:

$\int_0^{\pi} x\sin mx \, dx = \left[-\frac{x\cos mx}{m}\right]_0^{\pi} + \frac{1}{m}\int_0^{\pi} \cos mx \, dx$

$= -\frac{\pi \cos m\pi}{m} = -\frac{\pi(-1)^m}{m}$

So:

$a_n = \frac{1}{\pi}\left[-\frac{\pi(-1)^{n+1}}{n+1} + \frac{\pi(-1)^{n-1}}{n-1}\right] = (-1)^{n+1}\left[\frac{-1}{n+1} + \frac{1}{n-1}\right]$

$= (-1)^{n+1}\left[\frac{-(n-1) + (n+1)}{(n+1)(n-1)}\right] = (-1)^{n+1} \cdot \frac{2}{n^2-1}$

$\boxed{x\sin x = 1 - \frac{\cos x}{2} + 2\sum_{n=2}^{\infty} \frac{(-1)^{n+1}}{n^2-1}\cos nx}$

Quick Reference: Convergence Tests

Fourier Series Formulas

Exam Tips & Common Mistakes

1. Always check symmetry first — if $f(x)$ is even, $b_n = 0$; if odd, $a_n = 0$. This saves half the work.

Summary

This unit covers the essential toolkit for sequences, series, and Fourier analysis:

• Sequences: Definitions, convergence, and limit computation
• Series: Partial sums, convergence, and divergence criteria
• Comparison Test: Direct and limit forms for positive-term series
• Ratio Test (D'Alembert): The go-to test for factorials and exponentials
• Raabe's Test: The rescue when ratio test fails ($L=1$)
• Fourier Series: Full-range expansions on $(-\pi, \pi)$
• Half-Range Series: Sine and cosine expansions on $(0, \pi)$

Master these 25 solved problems and you'll be fully prepared for any Sequence and Series examination!