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1. Algebra — Equations, Progressions and Series

Logarithmic series

Logarithmic series expansions provide fundamental polynomial representations for transcendental functions like $\ln(1+x)$ and $\ln\left(\frac{1+x}{1-x}\right)$ within specific intervals of convergence.

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Definition and statement

Standard Logarithmic Series Expansion (Mercator Series / Taylor Expansion)
For $-1 < x \le 1$, the function $f(x) = \ln(1+x)$ is expanded into the convergent power series: $$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n}$$

📐 Geometric Interpretation

Geometrically, the logarithmic series represents the approximation of the area under the hyperbola $y = \frac{1}{1+t}$ from $t=0$ to $t=x$ by alternating polynomial curves whose accuracy increases as higher-order terms are included within the interval of convergence $(-1, 1]$.

⚙️ Algebraic & Structural Meaning

It establishes a bridge between transcendental logarithmic functions and rational algebraic polynomials, enabling limit evaluations, infinite series summation, and numerical approximation of natural logarithms.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Interval of Convergence (Ratio Test / D'Alembert's Test) -1 < x \le 1 (Half-open interval) Strictly Mandatory / Necessary For $|x| > 1$, the general term does not tend to zero, making the series divergent and yielding meaningless infinite values.
Continuity and Differentiability at x = 1 x = 1 (Endpoint inclusion via Abel's Theorem) Conditionally Convergent At $x = -1$, the series reduces to the harmonic series $-(1 + \frac{1}{2} + \frac{1}{3} + \dots)$, which diverges to $-\infty$. Thus $x = -1$ must be excluded.
Uniform Convergence on compact subsets [-r, r] where 0 < r < 1 Essential for term-by-term integration and differentiation Without uniform convergence, interchanging limit operations and infinite sums is invalid.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Symmetric Logarithmic Expansion for \ln\left(\frac{1+x}{1-x}\right)

$$\ln\left(\frac{1+x}{1-x}\right) = 2 \left( x + \frac{x^3}{3} + \frac{x^5}{5} + \frac{x^7}{7} + \dots \right) \quad \text{for } |x| < 1$$

Rule: Only odd powers of $x$ appear with positive coefficients, eliminating even powers instantly and halving computation time.

Application: Frequently asked in UP PGT to sum alternating series or evaluate limits involving logarithmic combinations.
02

Summation of Series with Arithmetic-Geometric Coefficients

$$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = \ln 2$$

Rule: Recognize the alternating harmonic series directly as $\ln 2$ without testing convergence manually.

03

Evaluation of \ln 2 using x = 1/3 in Symmetric Form

$$\ln 2 = 2 \left( \frac{1}{3} + \frac{1}{3 \cdot 3^3} + \frac{1}{5 \cdot 3^5} + \dots \right)$$

Rule: Substituting $x = \frac{1}{3}$ into the symmetric logarithmic expansion yields a rapidly converging series for $\ln 2$.

Application: Used in numerical approximation questions in DSSSB PGT.
04

Coefficient Extraction via Differentiation

$$nx^n \text{ or } \frac{1}{n} \text{ patterns from } \sum \frac{x^n}{n}$$

Rule: To find the sum of $\sum_{n=1}^{\infty} \frac{x^n}{n(n+1)}$ or similar, use partial fractions combined with $\ln(1-x)$ integration.

Application: Directly tested in UP PGT series summation problems.
05

Logarithmic Expansion of \ln(1-x)

$$\ln(1-x) = -\left( x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots \right) \quad \text{for } -1 \le x < 1$$

Rule: All terms are negative when expanding $\ln(1-x)$ for positive $x$ in the domain.

06

Rapid Elimination via Domain Testing

$$x \in (-1, 1]$$

Rule: Check if the given $x$ lies outside the radius of convergence; options with divergent values can be eliminated instantly.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Divergence at x = -1

Common Pitfall: Candidates substitute $x = -1$ into $\ln(1+x) = x - \frac{x^2}{2} + \dots$ and incorrectly equate $\ln 0$ to an alternating harmonic series sum.

Counterexample: $x = -1 \implies \ln(0)$ is undefined, and series becomes $-(1 + \frac{1}{2} + \frac{1}{3} + \dots) = -\infty$.

Why it fails: The left boundary $x = -1$ is strictly excluded from the domain of convergence.

🛡️ Exam Defense: Never include $x = -1$ in logarithmic series evaluations; remember the interval is $(-1, 1]$.
⚠️ COMMON ERROR #02

Sign Confusion in \ln(1-x) vs \ln(1+x)

Common Pitfall: Forgetting that $\ln(1-x)$ has all negative terms: $-\sum \frac{x^n}{n}$.

Counterexample: Assuming $\ln(1-x) = \sum \frac{x^n}{n}$.

Why it fails: Leads to sign reversals in coefficient matching questions.

🛡️ Exam Defense: Always write out the first three terms explicitly: $\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \dots$.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Mercator Series (Logarithmic)-1 < x \le 1\ln(1+x) = \sum \frac{(-1)^{n-1}x^n}{n}Area under hyperbolaAlternating signs, denominator equals power
Symmetric Logarithmic Series|x| < 1\ln\left(\frac{1+x}{1-x}\right) = 2\sum \frac{x^{2n-1}}{2n-1}Area ratio under reciprocal curveOnly odd powers, all positive terms
Exponential Seriesx \in \mathbb{R}e^x = \sum \frac{x^n}{n!}Growth curve approximationFactorial denominators
Binomial Series-1 < x < 1(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dotsGeneral polynomial power expansionFractional or negative index

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the sum of the infinite series: $1 - \frac{1}{2}(1/2) + \frac{1}{3}(1/4) - \frac{1}{4}(1/8) + \dots$

(A) $\ln\left(\frac{3}{2}\right)$(B) $\ln 2$(C) $\ln\left(\frac{4}{3}\right)$(D) $\frac{1}{2} \ln 2$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Identify the general term as $\frac{(-1)^{n-1} (1/2)^n}{n}$ for $n = 1, 2, 3, \dots$. Step 2: Recognize this matches the logarithmic series expansion for $\ln(1+x)$ with $x = 1/2$. Step 3: Compute $\ln(1 + 1/2) = \ln(3/2)$.
⚡ Method B: 25-Second Exam Shortcut
Compare directly with $\ln(1+x) = x - x^2/2 + \dots$ setting $x = 1/2$, yielding $\ln(3/2)$ instantly.
💡 Tip: Look out for fractional powers multiplying the alternating denominators.
CANONICAL PROBLEM 02 Dual-Method Breakdown

If $x$ is small so that $x^2$ and higher powers can be neglected, then the value of $\ln\left(\frac{1+3x}{1-2x}\right)$ is approximately:

(A) $x$(B) $5x$(C) $-x$(D) $6x$
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Step 1: Use logarithm properties: $\ln(1+3x) - \ln(1-2x)$. Step 2: Apply the linear approximation $\ln(1+u) \approx u$ for small $u$. Step 3: Substitute $3x$ and $-2x$: $(3x) - (-2x) = 3x + 2x = 5x$.
⚡ Method B: 25-Second Exam Shortcut
For small $x$, $\ln(1+ax) - \ln(1-bx) \approx ax - (-bx) = (a+b)x$. Here $a=3, b=2 \implies 5x$.
💡 Tip: A favorite question type in UP TGT/PGT for linear approximation.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: The interval of convergence for the logarithmic series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n}$ is:
A $[-1, 1]$
B $(-1, 1]$
C $(-1, 1)$
D $[0, 1)$
Derivation & Analysis:

By D'Alembert's Ratio Test, the series converges for $|x| < 1$. At $x = 1$, it converges conditionally (alternating harmonic series). At $x = -1$, it diverges to $-\infty$. Thus, the exact interval is $(-1, 1]$.

Q2
MCQ 2: The sum of the series $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$ is equal to:
A $1$
B $\ln 2$
C $\frac{\pi}{4}$
D $e$
Derivation & Analysis:

This is the standard Mercator expansion of $\ln(1+x)$ evaluated at $x = 1$, which gives $\ln(1+1) = \ln 2$.

Q3
MCQ 3: If $|x| < 1$, then $\ln(1-x) =$ ?
A $x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$
B $-\left( x + \frac{x^2}{2} + \frac{x^3}{3} + \dots \right)$
C $x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$
D $-\left( x - \frac{x^2}{2} - \frac{x^3}{3} - \dots \right)$
Derivation & Analysis:

Substituting $-x$ in place of $x$ in $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$ gives $\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \dots = -\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \dots\right)$.

Q4
MCQ 4: The coefficient of $x^n$ in the expansion of $\ln(1 + 3x)$ is:
A $\frac{3^n}{n}$
B $\frac{(-1)^{n-1} 3^n}{n}$
C $\frac{(-1)^n 3^n}{n}$
D $\frac{3^n}{n!}$
Derivation & Analysis:

Replacing $x$ by $3x$ in $\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n}$ yields $\sum_{n=1}^{\infty} \frac{(-1)^{n-1} (3x)^n}{n} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} 3^n x^n}{n}$. Thus the coefficient of $x^n$ is $\frac{(-1)^{n-1} 3^n}{n}$.

Q5
MCQ 5: The series $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots$ sums to:
A $1$
B $\ln 2$
C $\frac{1}{2}$
D $2$
Derivation & Analysis:

Using partial fractions, $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. Telescoping sum gives $(1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots = 1$.

Q6
MCQ 6: For $0 < x < 1$, the series $x - \frac{x^3}{3} + \frac{x^5}{5} - \dots$ represents:
A $\tan^{-1} x$
B $\tanh^{-1} x$
C $\sin^{-1} x$
D $\ln(1+x)$
Derivation & Analysis:

This is the standard Gregory's series expansion for $\tan^{-1} x$ when $|x| \le 1$.

Q7
MCQ 7: If $x$ is very small, the approximate value of $\ln\left(\frac{1+x}{1-x}\right)$ is:
A $x$
B $2x$
C $\frac{1}{2}x$
D $x^2$
Derivation & Analysis:

Using the symmetric expansion $\ln\left(\frac{1+x}{1-x}\right) = 2\left(x + \frac{x^3}{3} + \dots\right)$, neglecting higher powers yields $2x$.

Q8
MCQ 8: Which of the following series is convergent?
A $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$
B $\sum_{n=1}^{\infty} \frac{1}{n}$
C $\sum_{n=1}^{\infty} (-1)^{n-1}$
D $\sum_{n=1}^{\infty} n$
Derivation & Analysis:

The alternating harmonic series $\sum \frac{(-1)^{n-1}}{n}$ converges conditionally by Leibniz's Test, whereas the others diverge.

Q9
MCQ 9: The value of $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n 2^n}$ is:
A $\ln 2$
B $\ln\left(\frac{3}{2}\right)$
C $\frac{1}{2} \ln 2$
D $\ln\left(\frac{4}{3}\right)$
Derivation & Analysis:

Substitute $x = 1/2$ into $\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n} = \ln(1+x)$, giving $\ln(1 + 1/2) = \ln(3/2)$.

Q10
MCQ 10: If $x = 1/3$, then $2 \left( \frac{1}{3} + \frac{1}{3(3^3)} + \frac{1}{5(3^5)} + \dots \right)$ evaluates to:
A $\ln 2$
B $\ln 3$
C $\frac{1}{2} \ln 3$
D $\ln\left(\frac{3}{2}\right)$
Derivation & Analysis:

This is the symmetric logarithmic expansion for $\ln\left(\frac{1+x}{1-x}\right)$ evaluated at $x = 1/3$, giving $\ln\left(\frac{1+1/3}{1-1/3}\right) = \ln 2$.

Common questions

At $x = -1$, the series $\sum \frac{(-1)^{n-1}(-1)^n}{n}$ transforms into $-\sum \frac{1}{n}$, which is the divergent harmonic series. Thus, $-1$ lies outside the valid domain of convergence.

Mercator series is specifically the Taylor series expansion of $\ln(1+x)$ about $x = 0$. All Mercator series are Taylor series, but Taylor series encompass expansions of arbitrary differentiable functions.

Always write out the first three terms explicitly: $\ln(1+x) = x - x^2/2 + x^3/3$ and $\ln(1-x) = -x - x^2/2 - x^3/3$. Memorize that $\ln(1-x)$ has all negative terms.

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