Using Partial Sums To Approximate Infinite Series

1) Find the 6th partial sum of the series: 1 + 1/2 + 1/4 + 1/8 + ...

63/32

31/16

63/64

127/64

This is a geometric series with a₁=1 and r=1/2. The nth partial sum is Sₙ = (1 - (½)ⁿ)/(1 - 1/2) = (1 - (½)ⁿ)/(½) = 2(1 - (½)ⁿ). For n=6, (½)⁶ = 1/64, so S₆ = 2(1 - 1/64) = 2(63/64) = 63/32.

Explanation

This is a geometric series with a₁=1 and r=1/2. The nth partial sum is Sₙ = (1 - (½)ⁿ)/(1 - 1/2) = (1 - (½)ⁿ)/(½) = 2(1 - (½)ⁿ). For n=6, (½)⁶ = 1/64, so S₆ = 2(1 - 1/64) = 2(63/64) = 63/32.

2) You are given the partial sum formula Sₙ = n² + 3n for a specific series. What is the value of the fourth term, a4, of this series?

28

18

10

7

The relationship between a specific term aₙ and the partial sums is given by aₙ = Sₙ - Sn-1. To find the fourth term a₄, we need to calculate S₄ and S_3. First, we calculate S₄ by substituting n=4 into the formula n² + 3n, which gives 4² + 3(4) = 16 + 12 = 28. Next, we calculate S_3 by substituting n=3 into the formula, which gives 3² + 3(3) = 9 + 9 = 18. Finally, we subtract S_3 from S₄ to find a₄, so 28 - 18 equals 10.

Explanation

The relationship between a specific term aₙ and the partial sums is given by aₙ = Sₙ - Sn-1. To find the fourth term a₄, we need to calculate S₄ and S_3. First, we calculate S₄ by substituting n=4 into the formula n² + 3n, which gives 4² + 3(4) = 16 + 12 = 28. Next, we calculate S_3 by substituting n=3 into the formula, which gives 3² + 3(3) = 9 + 9 = 18. Finally, we subtract S_3 from S₄ to find a₄, so 28 - 18 equals 10.

3) Which of the following statements about partial sums is FALSE?

The partial sums of a convergent series must approach a finite limit.

The partial sums of a divergent series always go to infinity.

The nth partial sum is the sum of the first n terms.

If the terms of a series are all positive, the partial sums are increasing.

Option B is false because there are divergent series whose partial sums do not go to infinity, such as the series 1 - 1 + 1 - 1 + ... which oscillates between 0 and 1. Divergent means the partial sums do not approach a finite limit, but they may oscillate or behave erratically without necessarily going to infinity.

Explanation

Option B is false because there are divergent series whose partial sums do not go to infinity, such as the series 1 - 1 + 1 - 1 + ... which oscillates between 0 and 1. Divergent means the partial sums do not approach a finite limit, but they may oscillate or behave erratically without necessarily going to infinity.

4) The alternating harmonic series Σ (-1)ⁿ⁺¹/n converges to ln(2) ≈ 0.6931. Using the alternating series error bound, what is the minimum number of terms needed to approximate ln(2) with an error less than 0.1?

5

10

20

100

The error bound for an alternating series is the absolute value of the next term. So we need to find the smallest n such that |a_{n+1}| < 0.1. For this series, a_{n+1} = 1/(n+1). So we need 1/(n+1) < 0.1, which means n+1 > 10, so n > 9. Thus, the minimum n is 10. That means if we use the partial sum S₁₀, the error is at most 1/11 ≈ 0.0909 < 0.1. So the answer is 10.

Explanation

The error bound for an alternating series is the absolute value of the next term. So we need to find the smallest n such that |a_{n+1}| < 0.1. For this series, a_{n+1} = 1/(n+1). So we need 1/(n+1) < 0.1, which means n+1 > 10, so n > 9. Thus, the minimum n is 10. That means if we use the partial sum S₁₀, the error is at most 1/11 ≈ 0.0909 < 0.1. So the answer is 10.

5) Consider the series defined by the sum from n=1 to infinity of (1/√(n) - 1/√(n+1)). What is the value of the third partial sum, S_3?

0.5

1

1 − 1/√3

1.5

This is a telescoping series involving square roots. We calculate the first three terms.

For n=1: 1/1 - 1/√(2) = 1 - 1/√(2).

For n=2: 1/√(2) - 1/√(3).

For n=3: 1/√(3) - 1/√(4) = 1/√(3) - 1/2.

When we add these terms to find S_3, the -1/√(2) cancels with +1/√(2), and the -1/√(3) cancels with +1/√(3). We are left with the first part of the first term and the last part of the third term: 1 - 1/2 = 0.5.

Explanation

This is a telescoping series involving square roots. We calculate the first three terms.

For n=1: 1/1 - 1/√(2) = 1 - 1/√(2).

For n=2: 1/√(2) - 1/√(3).

For n=3: 1/√(3) - 1/√(4) = 1/√(3) - 1/2.

When we add these terms to find S_3, the -1/√(2) cancels with +1/√(2), and the -1/√(3) cancels with +1/√(3). We are left with the first part of the first term and the last part of the third term: 1 - 1/2 = 0.5.

6) The alternating series Σ (-1)ⁿ⁻¹ / (2n-1) from n=1 to infinity converges to π/4 ≈ 0.7854. If you use the partial sum S₅ to approximate the sum, what is the maximum possible error?

1/9

1/11

1/13

1/15

The terms are aₙ = (-1)ⁿ⁻¹/(2n-1). S₅ uses terms for n=1 to 5. The next term is for n=6: a₆ = (-1)⁵/(2*6-1) = (-1)/(11) = -1/11. The absolute value is 1/11. So the maximum error is 1/11.

Explanation

The terms are aₙ = (-1)ⁿ⁻¹/(2n-1). S₅ uses terms for n=1 to 5. The next term is for n=6: a₆ = (-1)⁵/(2*6-1) = (-1)/(11) = -1/11. The absolute value is 1/11. So the maximum error is 1/11.

7) The partial sums of a series are given by Sₙ = 4 - 2/(n+1). What is the value of the infinite series?

0

2

4

The series diverges.

The value of the infinite series is the limit of Sₙ as n→∞. lim (n→∞) [4 - 2/(n+1)] = 4 - 0 = 4.

Explanation

The value of the infinite series is the limit of Sₙ as n→∞. lim (n→∞) [4 - 2/(n+1)] = 4 - 0 = 4.

8) Find the partial sum Sₙ for the series sum from k=1 to n of ln(k / (k+1)).

Ln(n)

−ln(n+1)

Ln(n) − ln(n+1)

0

We use the property of logarithms that ln(a/b) = ln(a) - ln(b). This allows us to rewrite the general term as ln(k) - ln(k+1). This transforms the series into a telescoping sum. Writing out the terms, we get: (ln(1) - ln(2)) + (ln(2) - ln(3)) + (ln(3) - ln(4)) + ... + (ln(n) - ln(n+1)). The -ln(2) cancels with +ln(2), -ln(3) cancels with +ln(3), and so on. We are left with the first term ln(1) and the last term -ln(n+1). Since ln(1) is 0, the final expression for the partial sum is simply -ln(n+1).

Explanation

We use the property of logarithms that ln(a/b) = ln(a) - ln(b). This allows us to rewrite the general term as ln(k) - ln(k+1). This transforms the series into a telescoping sum. Writing out the terms, we get: (ln(1) - ln(2)) + (ln(2) - ln(3)) + (ln(3) - ln(4)) + ... + (ln(n) - ln(n+1)). The -ln(2) cancels with +ln(2), -ln(3) cancels with +ln(3), and so on. We are left with the first term ln(1) and the last term -ln(n+1). Since ln(1) is 0, the final expression for the partial sum is simply -ln(n+1).

9) Consider the series sum from n=1 to infinity of 1 / (n² + 3n + 2). Using partial fraction decomposition, determine the closed form for the n-th partial sum Sₙ.

1/2 − 1/(n+2)

1/2 − 1/(2(n+2))

1 − 1/(n+1)

1/2 − 1/(n+1) + 1/(n+2)

First, we factor the denominator n² + 3n + 2 into (n+1)(n+2). We then use partial fractions to write 1/((n+1)(n+2)) as A/(n+1) + B/(n+2). Solving for A and B, we find that A=1 and B=-1, so the term is 1/(n+1) - 1/(n+2). This creates a telescoping series. We write out the first few terms: (1/2 -⅓) + (1/3 - 1/4) + (1/4 - 1/5) + ... + (1/(n+1) - 1/(n+2)). We observe that -1/3 cancels with +1/3, -1/4 cancels with +1/4, and so on. The only terms that survive are the very first positive term, 1/2, and the very last negative term, -1/(n+2). Thus, Sₙ = 1/2 - 1/(n+2).

Explanation

First, we factor the denominator n² + 3n + 2 into (n+1)(n+2). We then use partial fractions to write 1/((n+1)(n+2)) as A/(n+1) + B/(n+2). Solving for A and B, we find that A=1 and B=-1, so the term is 1/(n+1) - 1/(n+2). This creates a telescoping series. We write out the first few terms: (1/2 -⅓) + (1/3 - 1/4) + (1/4 - 1/5) + ... + (1/(n+1) - 1/(n+2)). We observe that -1/3 cancels with +1/3, -1/4 cancels with +1/4, and so on. The only terms that survive are the very first positive term, 1/2, and the very last negative term, -1/(n+2). Thus, Sₙ = 1/2 - 1/(n+2).

10) A sequence of partial sums is defined by Sₙ = 3n/(n+1). What is the 100th term a₁₀₀ of the original series?

3/10100

3/10000

3/101

3/10001

The nth term aₙ = Sₙ - Sₙ₋₁. So a₁₀₀ = S₁₀₀ - S₉₉. Compute S₁₀₀ = 3×100/(100+1) = 300/101. Compute S₉₉ = 3×99/(99+1) = 297/100. Then a₁₀₀ = 300/101 - 297/100 = (300×100 - 297×101)/(101×100) = (30000 - 29997)/(10100) = 3/10100.

Explanation

The nth term aₙ = Sₙ - Sₙ₋₁. So a₁₀₀ = S₁₀₀ - S₉₉. Compute S₁₀₀ = 3×100/(100+1) = 300/101. Compute S₉₉ = 3×99/(99+1) = 297/100. Then a₁₀₀ = 300/101 - 297/100 = (300×100 - 297×101)/(101×100) = (30000 - 29997)/(10100) = 3/10100.

11) The partial sums of a series are given by Sₙ = (n² + 3n)/(n² + 1). What is the sum of the first 10 terms?

130/101

13/10

1.3

1.287

The sum of the first 10 terms is S₁₀. Substitute n=10 into the formula: S₁₀ = (10² + 3×10)/(10² + 1) = (100 + 30)/(100 + 1) = 130/101, an exact value. Although the sum is approximately ≈1.287, Option D should be understood as a decimal approximation and not the exact sum.

Explanation

The sum of the first 10 terms is S₁₀. Substitute n=10 into the formula: S₁₀ = (10² + 3×10)/(10² + 1) = (100 + 30)/(100 + 1) = 130/101, an exact value. Although the sum is approximately ≈1.287, Option D should be understood as a decimal approximation and not the exact sum.

12) Let aₙ = 3 for all n. What is the partial sum Sₙ?

3

3n

3ⁿ

N³

The partial sum Sₙ is the sum of the first n terms.

Since every term is 3, we are summing 3 + 3 + 3 + ... + 3 (n times).

Multiplication is defined as repeated addition.

Therefore, 3 added to itself n times is 3n.

Explanation

The partial sum Sₙ is the sum of the first n terms.

Since every term is 3, we are summing 3 + 3 + 3 + ... + 3 (n times).

Multiplication is defined as repeated addition.

Therefore, 3 added to itself n times is 3n.

13) Which of the following expressions represents the n-th partial sum, Sₙ, for the series sum from k=1 to infinity of (2^k - 2^(k-1))?

2ⁿ

2ⁿ⁻¹

2ⁿ − 1

2 − 2ⁿ

We write out the terms to identify the telescoping pattern.

k=1: 2^1 - 2^0

k=2: 2² - 2^1

k=3: 2³ - 2²

...

k=n: 2ⁿ - 2ⁿ⁻¹

Summing these up, the positive 2^1 cancels with the negative 2^1 from the next term. This cancellation continues up the chain. The only terms that remain are the negative part of the first term (-2⁰which is -1) and the positive part of the last term (2ⁿ). Therefore, Sₙ = 2ⁿ - 1.

Explanation

We write out the terms to identify the telescoping pattern.

k=1: 2^1 - 2^0

k=2: 2² - 2^1

k=3: 2³ - 2²

...

k=n: 2ⁿ - 2ⁿ⁻¹

Summing these up, the positive 2^1 cancels with the negative 2^1 from the next term. This cancellation continues up the chain. The only terms that remain are the negative part of the first term (-2⁰which is -1) and the positive part of the last term (2ⁿ). Therefore, Sₙ = 2ⁿ - 1.

14) You are observing a sequence of partial sums Sₙ. If the sequence oscillates between two values, such as 2, 4, 2, 4..., which of the following statements is TRUE?

Series converges to 3

Series converges to 0

Series diverges

Terms are positive

For an infinite series to converge, the sequence of its partial sums Sₙ must approach a single, finite limit. If the partial sums oscillate between two different values, they do not approach a single limit. Therefore, by definition, the series diverges.

Explanation

For an infinite series to converge, the sequence of its partial sums Sₙ must approach a single, finite limit. If the partial sums oscillate between two different values, they do not approach a single limit. Therefore, by definition, the series diverges.