Special Series

Overview

In the realm of advanced algebra and calculus, special series serve as indispensable tools for analyzing and approximating functions. This chapter delves into two cornerstone series – the Exponential and Logarithmic Series – which are not only elegant mathematical constructs but also powerful weapons in your problem-solving arsenal for the ISI MSQMS entrance examination. A firm understanding of these series is paramount for tackling a variety of questions, ranging from evaluating complex limits and summing intricate infinite series to approximating function values.

The ISI examination frequently features problems that directly or indirectly test your proficiency in manipulating and applying these expansions. Mastering the derivations, conditions of convergence, and various applications of exponential and logarithmic series will significantly enhance your ability to identify patterns, simplify expressions, and devise efficient solutions under exam conditions. This chapter provides the essential foundation for confidently approaching such problems, ensuring you're well-equipped to secure crucial marks.

Chapter Contents

| # | Topic | What You'll Learn |
|---|-------|-------------------|
| 1 | Exponential Series | Derive & apply series for $e^x$ and related functions. |
| 2 | Logarithmic Series | Derive & apply series for $\ln(1+x)$ and related functions. |

Learning Objectives

---

Now let's begin with Exponential Series...

Part 1: Exponential Series

Introduction

The exponential series is a fundamental concept in mathematics, particularly in calculus and analysis. It provides a powerful way to represent the exponential function $e^x$ as an infinite sum of terms. Understanding this series is crucial for solving various problems in sequences and series, limits, and approximations, which are frequently encountered in the ISI MSQMS examination.

This topic explores the definition of the mathematical constant $e$, the Maclaurin series expansion of $e^x$, and various derived series. We will also delve into techniques for manipulating these series to evaluate complex sums, equipping you with the necessary tools to tackle ISI-level questions.

---

Key Concepts

1. The Number $e$ as a Limit

The mathematical constant $e$ is a fundamental irrational and transcendental number, approximately equal to $2.71828$. It can be defined in several ways, one of the most important for ISI is its definition as a limit.

More generally, for any real number $a$, we have:

And similarly, for a variable $x$ approaching $0$:

n
f(n)

2

e ≈ 2.718

3

1

2

5

10

20

e

Worked Example:

Problem: Evaluate the limit:

Solution:

Step 1: Rewrite the expression to match the standard limit form.

We know:

The given expression is $\left(1 + \frac{3}{n}\right)^{2n}$.

Step 2: Apply the limit property.

We can rewrite $2n$ as $2 \times n$.

Step 3: Evaluate the limit.

Using the property:

Therefore,

Answer: $\boxed{e^6}$

---

---

2. Maclaurin Series Expansion of $e^x$

The Maclaurin series is a special case of the Taylor series, where the expansion is centered at $x = 0$. For a function $f(x)$, its Maclaurin series is given by:

For the exponential function $f(x) = e^x$:

• $f(x) = e^x \implies f(0) = e^0 = 1$


• $f'(x) = e^x \implies f'(0) = e^0 = 1$


• $f''(x) = e^x \implies f''(0) = e^0 = 1$


• In general, $f^{(n)}(x) = e^x \implies f^{(n)}(0) = e^0 = 1$

Substituting these values into the Maclaurin series formula gives us the exponential series:

Understanding the coefficients of this series is crucial. If $e^x = a_0 + a_1x + a_2x^2 + \dots$, then $a_n = \frac{1}{n!}$.

Worked Example:

Problem: If $e^{3x} = c_0 + c_1x + c_2x^2 + \dots$, find the value of $c_3$.

Solution:

Step 1: Recall the Maclaurin series for $e^u$.

The Maclaurin series for $e^u$ is $\sum_{n=0}^{\infty} \frac{u^n}{n!}$.

Step 2: Substitute $u=3x$ into the series.

Expanding the series, we get:

Step 3: Identify the coefficient $c_3$.

Comparing this with $e^{3x} = c_0 + c_1x + c_2x^2 + c_3x^3 + \dots$, the coefficient of $x^3$ is $c_3$.

Answer: $\boxed{\frac{9}{2}}$

---

3. Derived Exponential Series

By substituting specific values for $x$ in the general exponential series, we can obtain several important derived series.

a) Series for $e^{-x}$

Replace $x$ with $-x$ in the series for $e^x$:

b) Series for $e$

Substitute $x=1$ into the series for $e^x$:

c) Series for $e^{-1}$

Substitute $x=-1$ into the series for $e^x$, or $x=1$ into the series for $e^{-x}$:

Worked Example:

Problem: Find the sum of the infinite series $\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots$

Solution:

Step 1: Identify the pattern of the series.

The series has alternating signs and terms are reciprocals of factorials, starting from $2!$.

Step 2: Compare with known exponential series.

Recall the series for $e^{-1}$:

The first two terms, $1-1$, cancel out.

Step 3: Conclude the sum.

The given series is exactly the series for $e^{-1}$ after the cancellation of the initial terms.

Answer: $\boxed{e^{-1}}$

---

---

#
## 4. Series Involving Even and Odd Factorials

By combining the series for $e^x$ and $e^{-x}$, we can derive series that involve only even powers of $x$ (and even factorials) or only odd powers of $x$ (and odd factorials). These are closely related to hyperbolic functions.

#
### a) Series for Hyperbolic Cosine ($\cosh(x)$)

Add the series for $e^x$ and $e^{-x}$:

Summing them:

Dividing by 2 gives the series for $\cosh(x)$:

#
### b) Series for Hyperbolic Sine ($\sinh(x)$)

Subtract the series for $e^{-x}$ from $e^x$:

Subtracting:

Dividing by 2 gives the series for $\sinh(x)$:

Worked Example:

Problem: Find the sum of the infinite series $1 + \frac{1}{4 \times 2!} + \frac{1}{16 \times 4!} + \frac{1}{64 \times 6!} + \dots$

Solution:

Step 1: Write the general term of the series.

The given series is $1 + \frac{1}{4 \times 2!} + \frac{1}{16 \times 4!} + \frac{1}{64 \times 6!} + \dots$
We can observe the pattern:
For $n=0$, the term is $1 = \frac{1}{4^0 (0)!}$.
For $n=1$, the term is $\frac{1}{4 \times 2!} = \frac{1}{4^1 (2)!}$.
For $n=2$, the term is $\frac{1}{16 \times 4!} = \frac{1}{4^2 (4)!}$.
For $n=3$, the term is $\frac{1}{64 \times 6!} = \frac{1}{4^3 (6)!}$.

Thus, the general term of the series is $\frac{1}{4^n (2n)!}$ for $n=0, 1, 2, \dots$.

Step 2: Compare with the series for $\cosh(x)$.

The series for $\cosh(x)$ is $1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \dots = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$.
We can rewrite the general term of our series as $\frac{1}{(2n)!} \left(\frac{1}{4}\right)^n = \frac{1}{(2n)!} \left(\frac{1}{2^2}\right)^n = \frac{1}{(2n)!} \left(\frac{1}{2}\right)^{2n}$.

Step 3: Identify the value of $x$.

Comparing $\frac{x^{2n}}{(2n)!}$ with $\frac{(1/2)^{2n}}{(2n)!}$, we see that $x = \frac{1}{2}$.

Step 4: Evaluate the sum.

The given series is $\cosh\left(\frac{1}{2}\right)$.

Answer: $\frac{e+1}{2\sqrt{e}}$

---

#
## 5. Techniques for Series Manipulation

Many problems require manipulating the terms of a series to match a known exponential form.

#
### a) Splitting the Numerator

If the numerator of the general term contains a linear expression, it can often be split.
For example, if the general term is $\frac{n}{(n+k)!}$, we can write $n = (n+k) - k$.

#
### b) Adjusting the Index

Sometimes, a series might start from an index other than $0$, or its terms might be shifted. Adjusting the summation index can help match a standard form. For example, $\sum_{n=1}^\infty \frac{x^n}{n!} = e^x - 1$.

#
### c) Recognizing Patterns

Look for alternating signs, factorials in the denominator, and powers of $x$. These are strong indicators of exponential series.

Worked Example:

Problem: Find the sum of the infinite series $\frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \dots$

Solution:

Step 1: Write the general term of the series.

The terms are of the form $\frac{2n}{(2n+1)!}$ for $n=1, 2, 3, \dots$.
Let's verify:
For $n=1$: $\frac{2(1)}{(2(1)+1)!} = \frac{2}{3!}$
For $n=2$: $\frac{2(2)}{(2(2)+1)!} = \frac{4}{5!}$
For $n=3$: $\frac{2(3)}{(2(3)+1)!} = \frac{6}{7!}$

Step 2: Manipulate the general term using numerator splitting.

We want to simplify $\frac{2n}{(2n+1)!}$.
We can write $2n$ as $(2n+1) - 1$.

Step 3: Write out the sum with the manipulated general term.

Step 4: Relate the sum to known exponential series.

Recall the series for $e^{-1}$:

The terms $1-1$ cancel out. So,

This is exactly the series we are trying to sum.

Answer: $e^{-1}$

---

---

Problem-Solving Strategies

---

Common Mistakes

---

---

Practice Questions

:::question type="MCQ" question="The value of $\lim_{n \to \infty} \left(1 + \frac{5}{n}\right)^{n/2}$ is:" options=["$e^5$","$e^{5/2}$","$\sqrt{e}$","$e^{10}$"] answer="$e^{5/2}$" hint="Use the limit definition $e^a = \lim_{n \to \infty} (1 + \frac{a}{n})^n$." solution="Step 1: Rewrite the exponent to match the standard form.

Step 2: Apply the limit property.
We know that $\lim_{n \to \infty} \left(1 + \frac{a}{n}\right)^n = e^a$.
Here, $a=5$.

Step 3: Simplify the expression.

Answer: \boxed{e^{5/2}}" :::

:::question type="NAT" question="If $f(x) = e^{4x}$, and its Maclaurin series is $f(x) = \sum_{n=0}^\infty a_n x^n$, then the value of $a_2 \times 2!$ is:" answer="16" hint="Recall the general formula for Maclaurin coefficients $a_n = \frac{f^{(n)}(0)}{n!}$ or substitute into the series expansion." solution="Step 1: Write the Maclaurin series for $e^x$.

Step 2: Substitute $4x$ for $x$.

Step 3: Identify the coefficient $a_n$.

Comparing $e^{4x} = \sum_{n=0}^\infty a_n x^n$ with $\sum_{n=0}^\infty \frac{4^n x^n}{n!}$, we find that $a_n = \frac{4^n}{n!}$.

Step 4: Calculate $a_2$.

For $n=2$, $a_2 = \frac{4^2}{2!} = \frac{16}{2!} = \frac{16}{2} = 8$.

Step 5: Calculate $a_2 \times 2!$.

Answer: \boxed{16}" :::

:::question type="MCQ" question="The sum of the series $1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$ is:" options=["$e$","$e-1$","$e+1$","$\frac{1}{e}$"] answer="$e$" hint="This is a direct application of a standard exponential series." solution="Step 1: Recognize the series pattern.
The given series is $1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$.

Step 2: Compare with the standard series for $e$.
The series for $e$ is defined as:

Since $0! = 1$, the first term is $1/1 = 1$.

So, $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$.

Step 3: Conclude the sum.
The given series is exactly the series expansion of $e$.

The sum is $e$.
Answer: \boxed{e}"
:::

:::question type="SUB" question="Prove that the sum of the series $\frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \dots$ is $\frac{e - e^{-1}}{2}$." answer="Proof shows that the sum is $\sinh(1) = \frac{e-e^{-1}}{2}$." hint="Consider the series expansions of $e^x$ and $e^{-x}$ and how they combine." solution="Step 1: Write down the series for $e^x$ and $e^{-x}$.
The series for $e^x$ is:

The series for $e^{-x}$ is:

Step 2: Subtract the series for $e^{-x}$ from $e^x$.

Step 3: Divide by 2 to get the series for $\sinh(x)$.

This is the series expansion for $\sinh(x)$.

Step 4: Substitute $x=1$ into the series.
The given series is $\frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \dots$.
This corresponds to the series for $\sinh(x)$ with $x=1$.

Since $1 = \frac{1}{1!}$, we can write the series as:

Step 5: Conclude the proof.
Therefore, the sum of the series is $\sinh(1) = \frac{e^1 - e^{-1}}{2} = \frac{e - e^{-1}}{2}$.
Hence proved.
Answer: \boxed{\frac{e - e^{-1}}{2}}"
:::

:::question type="MCQ" question="The value of the series $\sum_{n=0}^{\infty} \frac{(-1)^n 2^n}{n!}$ is:" options=["$e^2$","$e^{-2}$","$e$","$e^{-1}$"] answer="$e^{-2}$" hint="Identify the form of the exponential series and the value of $x$." solution="Step 1: Recall the series for $e^x$.

Step 2: Compare the given series with the standard form.
The given series is $\sum_{n=0}^{\infty} \frac{(-1)^n 2^n}{n!} = \sum_{n=0}^{\infty} \frac{(-2)^n}{n!}$.

Step 3: Identify the value of $x$.
By comparing $\sum_{n=0}^{\infty} \frac{x^n}{n!}$ with $\sum_{n=0}^{\infty} \frac{(-2)^n}{n!}$, we see that $x = -2$.

Step 4: Evaluate the sum.
Therefore, the sum of the series is $e^{-2}$.
Answer: \boxed{e^{-2}}"
:::

:::question type="MSQ" question="Select ALL correct statements regarding the series $S = \frac{1}{0!} + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots$:" options=["A. $S = \cosh(1)$","B. $S = \frac{e+e^{-1}}{2}$","C. The series involves only odd factorials.","D. The series converges to a value less than 2."] answer="A,B,D" hint="Consider the series for $e^x$, $e^{-x}$, and $\cosh(x)$." solution="Let's analyze each option:

A. $S = \cosh(1)$
The series for $\cosh(x)$ is $\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots$.
If we set $x=1$, we get $\cosh(1) = 1 + \frac{1^2}{2!} + \frac{1^4}{4!} + \dots = \frac{1}{0!} + \frac{1}{2!} + \frac{1}{4!} + \dots$.
This matches the given series $S$. So, statement A is correct.

B. $S = \frac{e+e^{-1}}{2}$
By definition, $\cosh(x) = \frac{e^x + e^{-x}}{2}$.
Since $S = \cosh(1)$, we have $S = \frac{e^1 + e^{-1}}{2} = \frac{e+e^{-1}}{2}$.
So, statement B is correct.

C. The series involves only odd factorials.
The series terms are $\frac{1}{0!}, \frac{1}{2!}, \frac{1}{4!}, \dots$. The denominators are $0, 2, 4, \dots$, which are all even non-negative integers. Therefore, the series involves only even factorials, not odd factorials. So, statement C is incorrect.

D. The series converges to a value less than 2.
We know $e \approx 2.718$ and $e^{-1} \approx 0.368$.
So, $S = \frac{e+e^{-1}}{2} \approx \frac{2.718 + 0.368}{2} = \frac{3.086}{2} = 1.543$.
This value $1.543$ is indeed less than 2. So, statement D is correct.
Answer: \boxed{A,B,D}"
:::

:::question type="NAT" question="What is the sum of the series $\sum_{n=0}^{\infty} \frac{1}{n!} - \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}$?" answer="2.3504" hint="Recall the series expansions for $e$ and $e^{-1}$." solution="Step 1: Identify the first sum.
The first sum is $\sum_{n=0}^{\infty} \frac{1}{n!} = e$.

Step 2: Identify the second sum.
The second sum is $\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} = e^{-1}$.

Step 3: Perform the subtraction.
The required sum is $e - e^{-1}$.

Step 4: Calculate the numerical value.
Using $e \approx 2.71828$ and $e^{-1} \approx 0.36788$,

Rounding to four decimal places, the sum is $2.3504$.
Answer: \boxed{2.3504}"
:::

:::question type="MCQ" question="The sum of the series $1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} + \dots$ is:" options=["$\sin(1)$","$\cos(1)$","$\sinh(1)$","$\cosh(1)$"] answer="$\sin(1)$" hint="Recall the Maclaurin series for trigonometric functions and hyperbolic functions." solution="Step 1: Write out the Maclaurin series for common functions.

Step 2: Compare the given series with these expansions.
The given series is $1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} + \dots$.
The Maclaurin series for $\sin(x)$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$.
If we substitute $x=1$ into the series for $\sin(x)$, we get:

This exactly matches the given series.

Step 3: Conclude the sum.
The sum of the series is $\sin(1)$.
Answer: \boxed{\sin(1)}"
:::

---

Summary

---

What's Next?

---

---

---

Part 2: Logarithmic Series

Introduction

Logarithmic series are a special type of infinite series that represent logarithmic functions as polynomials. These series are fundamental in calculus and various branches of mathematics, providing a powerful tool for approximating logarithmic values, analyzing functions, and solving complex problems in sequences and series. For the ISI MSQMS examination, a strong understanding of logarithmic series, their derivations, properties, and applications is crucial. This topic frequently appears in questions involving series expansion, summation of specific series, and simplification of logarithmic expressions. Mastering the standard forms and manipulation techniques will significantly aid in tackling such problems efficiently.

---

Key Concepts

1. Fundamentals of Logarithms

Understanding the basic definition and properties of logarithms is essential before delving into logarithmic series. These properties allow for simplification and transformation of complex expressions into forms suitable for series expansion.

Worked Example:

Problem: Simplify the expression $2 \log_3 9 + \log_5 \left(\frac{1}{25}\right) + \log_2 (\log_2 16)$.

Solution:

Step 1: Evaluate each term using logarithm properties.

For the first term, $2 \log_3 9$:

Apply the power rule $\log_a (M^k) = k \log_a M$:

Apply the identity property $\log_a a = 1$:

For the second term, $\log_5 \left(\frac{1}{25}\right)$:

Apply the power rule:

Apply the identity property:

For the third term, $\log_2 (\log_2 16)$:

First, evaluate the inner logarithm $\log_2 16$:

Now substitute this back into the expression:

Step 2: Sum the simplified terms.

Answer: $4$

---

---

2. Standard Logarithmic Series Expansions

The core of logarithmic series involves expressing logarithmic functions as infinite sums of powers of $x$. These expansions are derived from the geometric series and calculus.

Worked Example:

Problem: Find the coefficient of $x^3$ in the expansion of $\log_e\left(\frac{1}{1+2x}\right)$ for $|2x| < 1$.

Solution:

Step 1: Rewrite the expression using logarithm properties.

Step 2: Apply the standard series expansion for $\log_e(1+y)$.

Let $y = 2x$. The series expansion for $\log_e(1+y)$ is:

Substitute $y = 2x$:

Step 3: Find the expansion for $-\log_e(1+2x)$.

Step 4: Identify the coefficient of $x^3$.

The term containing $x^3$ is $-\frac{8}{3}x^3$.

Therefore, the coefficient of $x^3$ is $-\frac{8}{3}$.

Answer: $\boxed{-\frac{8}{3}}$

---

3. Summation of Special Series

Sometimes, logarithmic expressions involve sums of powers. Knowing the formulas for these sums is crucial for simplification. The sum of cubes is particularly relevant as it appeared in a PYQ.

Other useful sums (though not directly tested in the provided PYQs, they are standard knowledge for ISI):
* Sum of first $n$ natural numbers: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
* Sum of first $n$ squares: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

Worked Example:

Problem: Simplify the expression $\log_e \left(\sum_{k=1}^{n} k^3\right) - 2 \log_e \left(\frac{n(n+1)}{2}\right)$.

Solution:

Step 1: Apply the formula for the sum of cubes.

We know that $\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$.

Substitute this into the expression:

Step 2: Use the power rule of logarithms.

Apply $\log_a (M^k) = k \log_a M$ to the first term:

Step 3: Simplify the expression.

The two terms are identical and one is subtracted from the other, resulting in zero.

Answer: $\boxed{0}$

---

4. Techniques for Series Manipulation and Coefficient Extraction

Many ISI problems require algebraic manipulation to transform a given series into one of the standard logarithmic forms. This often involves factorization, substitution, or splitting terms. Coefficient extraction is about finding the multiplier of a specific power of $x$ in an expanded series.

Algebraic Factorization and Simplification:
Before applying series expansions, it's often necessary to simplify the argument of the logarithm. For example, expressions like $1-x-x^2+x^3$ need to be factorized.
$1-x-x^2+x^3 = (1-x) - x^2(1-x) = (1-x)(1-x^2) = (1-x)(1-x)(1+x) = (1-x)^2(1+x)$.

Then, using logarithm properties:
$\log_e \left\{\frac{1}{(1-x)^2(1+x)}\right\} = -\log_e \left\{(1-x)^2(1+x)\right\}$
$= -[\log_e (1-x)^2 + \log_e (1+x)]$
$= -[2\log_e (1-x) + \log_e (1+x)]$

Now, the standard series for $\log_e(1-x)$ and $\log_e(1+x)$ can be applied.

Splitting Series Terms:
Sometimes a series can be broken down into two or more standard series. For example, a series of the form $\sum \frac{1}{n} \left(A^n - B^n\right)$ can be split into $\sum \frac{1}{n} A^n - \sum \frac{1}{n} B^n$. Each of these may then correspond to a standard logarithmic series.

Coefficient Extraction:
To find the coefficient of $x^k$ in an expanded series:

Expand the function into a power series using known formulas.

Identify the term containing $x^k$.

The numerical factor multiplying $x^k$ is the coefficient.

Worked Example:

Problem: Determine the coefficient of $x^2$ in the expansion of $\log_e\left(\frac{1+x}{1-3x}\right)$ for sufficiently small $x$.

Solution:

Step 1: Rewrite the expression using logarithm properties.

Step 2: Apply the standard series expansions.

For $\log_e(1+x)$:

For $\log_e(1-3x)$, let $y = 3x$. The series for $\log_e(1-y)$ is $-y - \frac{y^2}{2} - \frac{y^3}{3} - \dots$.
Substitute $y=3x$:

Step 3: Combine the expansions.

Step 4: Collect terms with $x^2$.

The terms containing $x^2$ are $-\frac{x^2}{2}$ and $+\frac{9x^2}{2}$.

Coefficient of $x^2 = -\frac{1}{2} + \frac{9}{2}$

Answer: $\boxed{4}$

---

Problem-Solving Strategies

---

Common Mistakes

---

Practice Questions

:::question type="MCQ" question="The coefficient of $x^2$ in the expansion of $\log_e \left(\frac{1-x}{1+3x}\right)$ for sufficiently small $x$ is:" options=["$2$","$4$","$6$","$8$"] answer="$4$" hint="Use the property $\log_e(A/B) = \log_e A - \log_e B$ and then apply the standard series expansions for $\log_e(1-y)$ and $\log_e(1+y)$." solution="Step 1: Rewrite the expression using logarithm properties.

Step 2: Expand $\log_e(1-x)$ using the series $-\sum_{n=1}^{\infty} \frac{x^n}{n}$.

The coefficient of $x^2$ in $\log_e(1-x)$ is $-\frac{1}{2}$.

Step 3: Expand $\log_e(1+3x)$ using the series $\sum_{n=1}^{\infty} (-1)^{n-1} \frac{y^n}{n}$ with $y=3x$.

The coefficient of $x^2$ in $\log_e(1+3x)$ is $-\frac{9}{2}$.

Step 4: Combine the coefficients for $\log_e(1-x) - \log_e(1+3x)$.
The coefficient of $x^2$ in the full expansion is the coefficient from $\log_e(1-x)$ minus the coefficient from $\log_e(1+3x)$.

Answer: $\boxed{4}$"
:::

---

Chapter Summary

---

Chapter Review Questions

:::question type="MCQ" question="Evaluate the limit:

Given that the limit exists and is equal to $4$, find the possible value(s) of $a$ if $a=b$." options=["A) $a = 2$", "B) $a = \pm 2$", "C) $a = 4$", "D) $a = \pm 4$"] answer="B" hint="Use the Maclaurin series expansions for $e^x$ and $\ln(1+x)$ up to the $x^2$ term." solution="
We use the Maclaurin series expansions for $e^{ax}$ and $\ln(1+bx)$:
$e^{ax} = 1 + (ax) + \frac{(ax)^2}{2!} + O(x^3) = 1 + ax + \frac{a^2x^2}{2} + O(x^3)$
$\ln(1+bx) = (bx) - \frac{(bx)^2}{2} + O(x^3) = bx - \frac{b^2x^2}{2} + O(x^3)$

Substitute these into the limit expression:

For the limit to exist and be finite, the coefficient of $x$ in the numerator must be zero.
So, $a-b = 0 \implies a=b$.

Given that $a=b$, the limit simplifies to:

We are given that the limit is $4$.
Therefore, $a^2 = 4 \implies a = \pm 2$.

The final answer is $\boxed{B}$"
:::

:::question type="NAT" question="Let $S = \sum_{n=1}^{\infty} \frac{1}{n \cdot 2^n}$. Find the value of $e^S$." answer="2" hint="Recall the logarithmic series expansion for $\ln(1-x)$ or $\ln(1+x)$." solution="
The logarithmic series expansion is given by:
$\ln(1-x) = -\sum_{n=1}^{\infty} \frac{x^n}{n}$ for $|x|<1$.

Comparing this with the given series $S = \sum_{n=1}^{\infty} \frac{1}{n \cdot 2^n}$, we can rewrite $S$ as:
$S = \sum_{n=1}^{\infty} \frac{(1/2)^n}{n}$

This matches the form of $-\ln(1-x)$ if we set $x = 1/2$.
So, $S = -\ln\left(1 - \frac{1}{2}\right)$
$S = -\ln\left(\frac{1}{2}\right)$
$S = -(\ln(1) - \ln(2))$
$S = -(0 - \ln(2))$
$S = \ln(2)$

Now, we need to find $e^S$:
$e^S = e^{\ln(2)}$
Using the property $e^{\ln(y)} = y$:
$e^S = 2$

The final answer is $\boxed{2}$"
:::

:::question type="MCQ" question="Evaluate the sum of the series:

" options=["A) $e$", "B) $2e$", "C) $e+1$", "D) $e^2$"] answer="B" hint="Rewrite the general term $\frac{n^2}{n!}$ by splitting $n^2$ as $n(n-1) + n$ to relate it to terms of $e^x$ series." solution="
Let the given sum be $S$.

For $n=0$, $\frac{0^2}{0!} = 0$.
For $n=1$, $\frac{1^2}{1!} = 1$.
For $n \ge 2$, we can simplify $\frac{n^2}{n!}$:

We can further split $n$ in the numerator as $(n-1)+1$:

For $n \ge 2$, $\frac{n-1}{(n-1)!} = \frac{1}{(n-2)!}$.
So, for $n \ge 2$:

Now, let's write out the sum:


Let $k=n-2$ for the first sum and $m=n-1$ for the second sum.
When $n=2$, $k=0$ and $m=1$.

We know that $e = \sum_{k=0}^{\infty} \frac{1}{k!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \dots$
And $\sum_{m=1}^{\infty} \frac{1}{m!} = \frac{1}{1!} + \frac{1}{2!} + \dots = e - \frac{1}{0!} = e-1$.

Substitute these values back into the expression for $S$:

The final answer is $\boxed{B}$"
:::

:::question type="MCQ" question="Find the sum of the infinite series:

" options=["A) $\ln(2)$", "B) $\ln(3)$", "C) $\ln(3/2)$", "D) $\ln(1/2)$"] answer="C" hint="Identify the pattern and relate it to the logarithmic series for $\ln(1+x)$." solution="
The given series is:

This can be written in summation notation as:

We know the Maclaurin series expansion for $\ln(1+x)$:

Comparing the given series with the expansion of $\ln(1+x)$, we can see that if we substitute $x = \frac{1}{2}$, we get:


This is exactly the series we need to sum.
Therefore, the sum of the series is:

The final answer is $\boxed{C}$"
:::

---

What's Next?