# 2.3 MCQs-Sequences and Series

## Sequences and Series

### Basic Concepts and Definitions

1\. A sequence is defined as:

1. A sum of numbers
2. A list of numbers arranged in a specific order
3. An unordered collection of numbers
4. The limit of a function

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**Answer:** 2. A list of numbers arranged in a specific order

**Explanation:** A sequence is an ordered list of numbers: $$a\_1, a\_2, a\_3, \ldots, a\_n, \ldots$$ Each number is called a term of the sequence. The position of each term is important, with $$a\_n$$ representing the nth term. A sequence can be finite or infinite.

Example: $$1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots$$ is an infinite sequence.

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2\. Which of the following represents an arithmetic sequence?

1. $$2, 4, 8, 16, 32$$
2. $$3, 6, 9, 12, 15$$
3. $$1, 4, 9, 16, 25$$
4. $$1, 1, 2, 3, 5, 8$$

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**Answer:** 2. $$3, 6, 9, 12, 15$$

**Explanation:** An arithmetic sequence has a constant difference between consecutive terms.

For the sequence $$3, 6, 9, 12, 15$$: $$6 - 3 = 3$$ $$9 - 6 = 3$$ $$12 - 9 = 3$$ $$15 - 12 = 3$$

The common difference is constant: $$d = 3$$.

General form: $$a\_n = a\_1 + (n-1)d$$

Option 1 is a geometric sequence, option 3 represents squares of natural numbers, and option 4 is the Fibonacci sequence.

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3\. Which of the following represents a geometric sequence?

1. $$5, 10, 15, 20, 25$$
2. $$2, 4, 6, 8, 10$$
3. $$3, 9, 27, 81, 243$$
4. $$1, 3, 6, 10, 15$$

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**Answer:** 3. $$3, 9, 27, 81, 243$$

**Explanation:** A geometric sequence has a constant ratio between consecutive terms.

For the sequence $$3, 9, 27, 81, 243$$: $$\frac{9}{3} = 3$$ $$\frac{27}{9} = 3$$ $$\frac{81}{27} = 3$$ $$\frac{243}{81} = 3$$

The common ratio is constant: $$r = 3$$.

General form: $$a\_n = a\_1 \cdot r^{n-1}$$

Options 1 and 2 are arithmetic sequences, option 4 represents triangular numbers.

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4\. The nth term of the sequence $$2, 5, 8, 11, \ldots$$ is:

1. $$3n - 1$$
2. $$2n + 1$$
3. $$3n + 2$$
4. $$2n + 3$$

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**Answer:** 1. $$3n - 1$$

**Explanation:** This is an arithmetic sequence with: First term: $$a\_1 = 2$$ Common difference: $$d = 3$$

General formula for arithmetic sequence: $$a\_n = a\_1 + (n-1)d$$

Substitute values: $$a\_n = 2 + (n-1) \cdot 3$$ $$a\_n = 2 + 3n - 3$$ $$a\_n = 3n - 1$$

Verification: $$n=1: 3 \cdot 1 - 1 = 2$$ $$n=2: 3 \cdot 2 - 1 = 5$$ $$n=3: 3 \cdot 3 - 1 = 8$$ $$n=4: 3 \cdot 4 - 1 = 11$$

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### Arithmetic Progressions (AP)

5\. The sum of first n terms of an AP is given by:

1. $$S\_n = \frac{n}{2} \[2a + (n-1)d]$$
2. $$S\_n = \frac{a(r^n - 1)}{r - 1}$$
3. $$S\_n = \frac{n(n+1)}{2}$$
4. $$S\_n = \frac{a}{1 - r}$$

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**Answer:** 1. $$S\_n = \frac{n}{2} \[2a + (n-1)d]$$

**Explanation:** For an AP with first term $$a$$ and common difference $$d$$, the sum of first n terms is:

$$S\_n = \frac{n}{2} \[2a + (n-1)d]$$

Derivation: Write the sum forwards and backwards, then add:

Alternative form: $$S\_n = \frac{n}{2} \[\text{first term} + \text{last term}] = \frac{n}{2} (a + a\_n)$$

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6\. If the 5th term of an AP is 17 and the 9th term is 33, the common difference is:

1. $$2$$
2. $$3$$
3. $$4$$
4. $$5$$

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**Answer:** 3. $$4$$

**Explanation:** For an AP: $$a\_n = a + (n-1)d$$

Given: $$a\_5 = a + 4d = 17 \quad \text{(1)}$$ $$a\_9 = a + 8d = 33 \quad \text{(2)}$$

Subtract equation (1) from equation (2): $$(a + 8d) - (a + 4d) = 33 - 17$$ $$4d = 16$$ $$d = 4$$

To find $$a$$: From equation (1): $$a + 4 \cdot 4 = 17$$ $$a + 16 = 17$$ $$a = 1$$

The AP is: $$1, 5, 9, 13, 17, 21, 25, 29, 33, \ldots$$

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7\. The arithmetic mean between 7 and 19 is:

1. $$11$$
2. $$12$$
3. $$13$$
4. $$14$$

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**Answer:** 3. $$13$$

**Explanation:** The arithmetic mean (AM) between two numbers $$a$$ and $$b$$ is:

$$\text{AM} = \frac{a + b}{2}$$

Substitute $$a = 7$$, $$b = 19$$: $$\text{AM} = \frac{7 + 19}{2} = \frac{26}{2} = 13$$

Alternatively, consider an AP of three terms: $$7, x, 19$$ The common difference must be constant: $$x - 7 = 19 - x$$ $$2x = 26$$ $$x = 13$$

The arithmetic mean is also called the average of two numbers.

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### Geometric Progressions (GP)

8\. The sum of first n terms of a GP is given by:

1. $$S\_n = \frac{n}{2} \[2a + (n-1)d]$$
2. $$S\_n = \frac{a(1 - r^n)}{1 - r} \quad \text{for } r \neq 1$$
3. $$S\_n = \frac{a}{1 - r}$$
4. $$S\_n = \frac{n(n+1)}{2}$$

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**Answer:** 2. $$S\_n = \frac{a(1 - r^n)}{1 - r} \quad \text{for } r \neq 1$$

**Explanation:** For a GP with first term $$a$$ and common ratio $$r$$:

For $$r = 1$$: $$S\_n = na$$

Alternative form: $$S\_n = \frac{a(r^n - 1)}{r - 1} \quad \text{for } r \neq 1$$

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9\. The geometric mean between 4 and 16 is:

1. $$6$$
2. $$8$$
3. $$10$$
4. $$12$$

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**Answer:** 2. $$8$$

**Explanation:** The geometric mean (GM) between two positive numbers $$a$$ and $$b$$ is:

$$\text{GM} = \sqrt{ab}$$

Substitute $$a = 4$$, $$b = 16$$: $$\text{GM} = \sqrt{4 \times 16} = \sqrt{64} = 8$$

Alternatively, consider a GP of three terms: $$4, x, 16$$ The common ratio must be constant: $$\frac{x}{4} = \frac{16}{x}$$ $$x^2 = 64$$ $$x = 8$$ (positive value)

For n numbers, $$\text{GM} = (a\_1 \times a\_2 \times \cdots \times a\_n)^{1/n}$$

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10\. If the 3rd term of a GP is 12 and the 6th term is 96, the common ratio is:

1. $$2$$
2. $$3$$
3. $$4$$
4. $$6$$

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**Answer:** 1. $$2$$

**Explanation:** For a GP: $$a\_n = ar^{n-1}$$

Given: $$a\_3 = ar^2 = 12 \quad \text{(1)}$$ $$a\_6 = ar^5 = 96 \quad \text{(2)}$$

Divide equation (2) by equation (1): $$\frac{ar^5}{ar^2} = \frac{96}{12}$$ $$r^3 = 8$$ $$r = 2$$ (since $$2^3 = 8$$)

To find $$a$$: From equation (1): $$a \cdot 2^2 = 12$$ $$a \cdot 4 = 12$$ $$a = 3$$

The GP is: $$3, 6, 12, 24, 48, 96, \ldots$$

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### Infinite Geometric Series

11\. The sum to infinity of a geometric series exists when:

1. $$|r| < 1$$
2. $$|r| > 1$$
3. $$r = 1$$
4. $$r = -1$$

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**Answer:** 1. $$|r| < 1$$

**Explanation:** For an infinite geometric series: $$S = a + ar + ar^2 + ar^3 + \cdots$$

The sum to infinity exists when $$|r| < 1$$ and is given by: $$S\_{\infty} = \frac{a}{1 - r}$$

Reason: As $$n \to \infty$$, $$r^n \to 0$$ when $$|r| < 1$$.

If $$|r| \geq 1$$, the series diverges (the sum goes to infinity or oscillates).

Example: $$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = \frac{1}{1 - \frac{1}{2}} = 2$$

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12\. The sum to infinity of the series $$1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots$$ is:

1. $$\frac{3}{2}$$
2. $$2$$
3. $$3$$
4. $$\infty$$

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**Answer:** 1. $$\frac{3}{2}$$

**Explanation:** This is an infinite geometric series with: First term: $$a = 1$$ Common ratio: $$r = \frac{1}{3}$$

Since $$|r| = \frac{1}{3} < 1$$, the sum to infinity exists: $$S\_{\infty} = \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}$$

Verification using partial sums: $$S\_1 = 1 = 1$$ $$S\_2 = 1 + \frac{1}{3} = \frac{4}{3} \approx 1.333$$ $$S\_3 = 1 + \frac{1}{3} + \frac{1}{9} = \frac{13}{9} \approx 1.444$$ $$S\_4 = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} = \frac{40}{27} \approx 1.481$$ Approaching $$1.5 = \frac{3}{2}$$

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### Harmonic Progression (HP)

13\. A sequence is said to be in harmonic progression if:

1. The reciprocals of its terms are in arithmetic progression
2. The terms have a constant difference
3. The terms have a constant ratio
4. The terms are reciprocals of natural numbers

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**Answer:** 1. The reciprocals of its terms are in arithmetic progression

**Explanation:** A sequence $$h\_1, h\_2, h\_3, \ldots$$ is in harmonic progression if the sequence of reciprocals $$\frac{1}{h\_1}, \frac{1}{h\_2}, \frac{1}{h\_3}, \ldots$$ is in arithmetic progression.

Example: $$\frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \ldots$$ is in HP because the reciprocals $$2, 4, 6, 8, \ldots$$ are in AP.

The nth term of HP: $$h\_n = \frac{1}{a + (n-1)d}$$ where $$a = \frac{1}{h\_1}$$ and $$d = \frac{1}{h\_2} - \frac{1}{h\_1}$$.

Harmonic mean between a and b: $$\text{HM} = \frac{2ab}{a+b}$$

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14\. The harmonic mean between 2 and 8 is:

1. $$3.2$$
2. $$4$$
3. $$5$$
4. $$6.4$$

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**Answer:** 1. $$3.2$$

**Explanation:** The harmonic mean between two numbers $$a$$ and $$b$$ is:

$$\text{HM} = \frac{2ab}{a+b}$$

Substitute $$a = 2$$, $$b = 8$$: $$\text{HM} = \frac{2 \times 2 \times 8}{2 + 8} = \frac{32}{10} = 3.2$$

Verification: The HP with three terms would be: $$2, 3.2, 8$$ Reciprocals: $$\frac{1}{2} = 0.5$$, $$\frac{1}{3.2} = 0.3125$$, $$\frac{1}{8} = 0.125$$

Differences: $$0.5 - 0.3125 = 0.1875$$ $$0.3125 - 0.125 = 0.1875$$

The differences are constant, so the reciprocals are in AP, hence the original numbers are in HP.

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### Special Series

15\. The sum of first n natural numbers is:

1. $$\frac{n(n+1)}{2}$$
2. $$\frac{n(n-1)}{2}$$
3. $$n^2$$
4. $$\frac{n(n+1)(2n+1)}{6}$$

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**Answer:** 1. $$\frac{n(n+1)}{2}$$

**Explanation:** The sum of first n natural numbers:

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

Derivation (Gauss method): Let $$S = 1 + 2 + 3 + \cdots + n$$ Write backwards: $$S = n + (n-1) + (n-2) + \cdots + 1$$ Add the two equations:

Example: Sum of first 100 natural numbers = $$\frac{100 \times 101}{2} = 5050$$

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16\. The sum of squares of first n natural numbers is:

1. $$\frac{n(n+1)}{2}$$
2. $$\frac{n(n+1)(2n+1)}{6}$$
3. $$\left\[\frac{n(n+1)}{2}\right]^2$$
4. $$\frac{n(n+1)(n+2)}{6}$$

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**Answer:** 2. $$\frac{n(n+1)(2n+1)}{6}$$

**Explanation:** The sum of squares of first n natural numbers:

$$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$

Derivation can be done using mathematical induction or telescoping sums.

Verification: $$n=1: 1^2 = 1, \quad \text{formula: } \frac{1 \times 2 \times 3}{6} = 1$$ $$n=2: 1^2+2^2 = 1+4=5, \quad \text{formula: } \frac{2 \times 3 \times 5}{6} = 5$$ $$n=3: 1^2+2^2+3^2 = 1+4+9=14, \quad \text{formula: } \frac{3 \times 4 \times 7}{6} = 14$$

This formula is useful in calculus for Riemann sums and calculating areas under curves.

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17\. The sum of cubes of first n natural numbers is:

1. $$\frac{n(n+1)(2n+1)}{6}$$
2. $$\left\[\frac{n(n+1)}{2}\right]^2$$
3. $$\frac{n^2(n+1)^2}{4}$$
4. Both 2 and 3

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**Answer:** 4. Both 2 and 3

**Explanation:** The sum of cubes of first n natural numbers:

$$1^3 + 2^3 + 3^3 + \cdots + n^3 = \left\[\frac{n(n+1)}{2}\right]^2 = \frac{n^2(n+1)^2}{4}$$

Note that $$\left\[\frac{n(n+1)}{2}\right]^2 = \frac{n^2(n+1)^2}{4}$$, so options 2 and 3 are equivalent.

Remarkable fact: $$(1 + 2 + 3 + \cdots + n)^2 = 1^3 + 2^3 + 3^3 + \cdots + n^3$$

Example for n=3: $$(1+2+3)^2 = 6^2 = 36$$ $$1^3+2^3+3^3 = 1+8+27 = 36$$

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### Convergence and Divergence

18\. A sequence $${a\_n}$$ is said to converge to L if:

1. $$a\_n$$ approaches L as n increases
2. $$a\_n = L$$ for all n
3. $$a\_n$$ is always close to L
4. $$|a\_n - L| < \epsilon$$ for all $$n > N$$, for any $$\epsilon > 0$$

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**Answer:** 4. $$|a\_n - L| < \epsilon$$ for all $$n > N$$, for any $$\epsilon > 0$$

**Explanation:** Formal definition of convergence: A sequence $${a\_n}$$ converges to L if for every $$\epsilon > 0$$, there exists a natural number N such that:

$$|a\_n - L| < \epsilon \quad \text{for all } n > N$$

Intuitively: $$a\_n$$ gets arbitrarily close to L as n increases.

Notation: $$\lim\_{n \to \infty} a\_n = L$$

Example: $$a\_n = \frac{1}{n}$$ converges to 0.

If a sequence does not converge, it diverges.

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19\. The sequence $$a\_n = \frac{n^2 + 1}{2n^2 - 3}$$ converges to:

1. $$0$$
2. $$\frac{1}{2}$$
3. $$1$$
4. $$\infty$$

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**Answer:** 2. $$\frac{1}{2}$$

**Explanation:** To find the limit:

Divide numerator and denominator by the highest power of n (which is $$n^2$$):

As $$n \to \infty$$: $$\frac{1}{n^2} \to 0$$ and $$\frac{3}{n^2} \to 0$$

Therefore:

Alternative method: For rational functions (ratio of polynomials), the limit as $$n \to \infty$$ is the ratio of leading coefficients: $$\frac{1}{2}$$.

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### Tests for Convergence of Series

20\. The p-series $$\sum \frac{1}{n^p}$$ converges if:

1. $$p > 0$$
2. $$p > 1$$
3. $$p < 1$$
4. $$p = 1$$

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**Answer:** 2. $$p > 1$$

**Explanation:** The p-series:

$$\sum\_{n=1}^{\infty} \frac{1}{n^p}$$

Converges if $$p > 1$$ Diverges if $$p \leq 1$$

Important special cases:

* $$p = 1$$: Harmonic series $$\sum \frac{1}{n}$$ diverges
* $$p = 2$$: $$\sum \frac{1}{n^2}$$ converges (to $$\frac{\pi^2}{6}$$)
* $$p = \frac{1}{2}$$: $$\sum \frac{1}{\sqrt{n}}$$ diverges

The p-series test is a special case of the integral test.

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21\. A series $$\sum a\_n$$ converges absolutely if:

1. $$\sum |a\_n|$$ converges
2. $$\sum a\_n$$ converges
3. $$a\_n \to 0$$
4. $$a\_n$$ are all positive

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**Answer:** 1. $$\sum |a\_n|$$ converges

**Explanation:** Absolute convergence: A series $$\sum a\_n$$ converges absolutely if $$\sum |a\_n|$$ converges.

Important properties:

1. Absolute convergence implies convergence (but not vice versa)
2. Absolutely convergent series can be rearranged without changing the sum
3. Conditionally convergent series (converges but not absolutely) can be rearranged to sum to any real number (Riemann rearrangement theorem)

Example: $$\sum\_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$$ converges conditionally (alternating harmonic series), but $$\sum\_{n=1}^{\infty} \left|\frac{(-1)^{n+1}}{n}\right| = \sum\_{n=1}^{\infty} \frac{1}{n}$$ diverges.

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### Ratio Test

22\. For the series $$\sum a\_n$$, if $$\lim\_{n \to \infty} \left|\frac{a\_{n+1}}{a\_n}\right| = L$$, then the series converges absolutely if:

1. $$L < 1$$
2. $$L > 1$$
3. $$L = 1$$
4. $$L = 0$$

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**Answer:** 1. $$L < 1$$

**Explanation:** Ratio test: Let $$L = \lim\_{n \to \infty} \left|\frac{a\_{n+1}}{a\_n}\right|$$

* If $$L < 1$$: series converges absolutely
* If $$L > 1$$: series diverges
* If $$L = 1$$: test is inconclusive (need another test)

Useful for series with factorials or exponentials.

Example: For $$\sum \frac{x^n}{n!}$$:

Converges for all x.

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### Power Series

23\. A power series centered at $$x\_0$$ has the form:

1. $$\sum a\_n (x - x\_0)^n$$
2. $$\sum a\_n x^n$$
3. $$\sum \frac{(x - x\_0)^n}{n!}$$
4. $$\sum \frac{a\_n}{n!}$$

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**Answer:** 1. $$\sum a\_n (x - x\_0)^n$$

**Explanation:** A power series centered at $$x\_0$$:

$$\sum\_{n=0}^{\infty} a\_n (x - x\_0)^n$$

When $$x\_0 = 0$$, it becomes a Maclaurin series: $$\sum a\_n x^n$$

The set of x values for which the series converges is called the interval of convergence.

Within its interval of convergence, a power series defines a function that is infinitely differentiable.

Example: $$e^x = \sum\_{n=0}^{\infty} \frac{x^n}{n!}$$ centered at 0, converges for all x.

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### Taylor and Maclaurin Series

24\. The Maclaurin series expansion of $$e^x$$ is:

1. $$\sum \frac{(-1)^n x^n}{n!}$$
2. $$\sum \frac{x^n}{n!}$$
3. $$\sum x^n$$
4. $$\sum \frac{x^{2n}}{(2n)!}$$

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**Answer:** 2. $$\sum \frac{x^n}{n!}$$

**Explanation:** A Maclaurin series is a Taylor series centered at 0.

For $$f(x) = e^x$$, all derivatives are: $$f^{(n)}(x) = e^x$$ $$f^{(n)}(0) = 1$$ for all n

Therefore:

Expanded form: $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$

This series converges for all $$x \in \mathbb{R}$$.

Euler's formula: $$e^{ix} = \cos x + i \sin x$$ comes from this series.

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25\. The radius of convergence R for a power series $$\sum a\_n (x - x\_0)^n$$ satisfies:

1. $$R = \frac{1}{\limsup |a\_n|^{1/n}}$$
2. $$R = \lim \left|\frac{a\_n}{a\_{n+1}}\right|$$
3. Both 1 and 2 can be used
4. R is always 1

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**Answer:** 3. Both 1 and 2 can be used

**Explanation:** The radius of convergence R can be found using:

1. Cauchy-Hadamard theorem: $$R = \frac{1}{\limsup\_{n \to \infty} |a\_n|^{1/n}}$$
2. Ratio method (when the limit exists): $$R = \lim\_{n \to \infty} \left|\frac{a\_n}{a\_{n+1}}\right|$$

Within $$|x - x\_0| < R$$, the series converges absolutely. At $$|x - x\_0| = R$$, convergence needs to be checked separately.

Example: For $$\sum \frac{x^n}{n}$$:

Using ratio method: $$\lim\_{n \to \infty} \left|\frac{a\_n}{a\_{n+1}}\right| = \lim\_{n \to \infty} \frac{n+1}{n} = 1$$ So $$R = 1$$.

At $$x = 1$$: $$\sum \frac{1}{n}$$ diverges. At $$x = -1$$: $$\sum \frac{(-1)^n}{n}$$ converges conditionally.

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