Math Problem Statement

Show that the sum of the first 2n term of the series 1 minus one half plus one half minus one third plus one third minus one fourth plus one fourth minus one fifth plus one fifth minus one sixth plus ... is the same as the sum of the first n terms of the series StartFraction 1 Over 1 times 2 EndFraction plus StartFraction 1 Over 2 times 3 EndFraction plus StartFraction 1 Over 3 times 4 EndFraction plus StartFraction 1 Over 4 times 5 EndFraction plus StartFraction 1 Over 5 times 6 EndFraction plus ... . Do these series​ converge? What is the sum of the first 2nplus1 terms of the first​ series? If the series​ converge, what is their​ sum? Question content area bottom Part 1 Given ngreater than or equals​1, write expressions for the 2nminus1th and 2nth terms in the first series. The 2nminus1th term in the first series is

enter your response here. ​(Simplify your answer. Use integers or fractions for any numbers in the​ expression.)

Solution

To analyze the first series, let's break it down:

Part 1: Terms of the First Series

The first series is:

S=112+1213+1314+1415+1516+S = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6} + \ldots

The pattern alternates between positive and negative terms, with each positive term followed by a negative term that matches its value but decreases by the next integer.

  • 2n-1th term: This corresponds to the last positive term in the sequence up to nn pairs, which is:

    a2n1=1na_{2n-1} = \frac{1}{n}

  • 2nth term: This corresponds to the last negative term in the sequence up to nn pairs, which is:

    a2n=1n+1a_{2n} = -\frac{1}{n+1}

Summary of Results

  • 2n-1th term: 1n\frac{1}{n}

  • 2nth term: 1n+1-\frac{1}{n+1}

Part 2: Convergence of the Series

  1. First Series: The first series is an alternating series. By the Alternating Series Test, it converges since the absolute values of the terms decrease and approach zero.

  2. Second Series: The second series is:

T=112+123+134+145+T = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \ldots

This series can be expressed in the form:

Tn=k=1n1k(k+1)=k=1n(1k1k+1)T_n = \sum_{k=1}^{n} \frac{1}{k(k+1)} = \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right)

By telescoping:

Tn=11n+1=nn+1T_n = 1 - \frac{1}{n+1} = \frac{n}{n+1}

This series also converges as nn \to \infty, converging to 1.

Part 3: Sum of the First 2n+12n+1 Terms of the First Series

To find the sum of the first 2n+12n+1 terms, we add the first nn positive terms and the first nn negative terms, and then add the additional positive term:

S2n+1=(1+12+13++1n)(12+13++1n)+last termS_{2n+1} = \left(1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n}\right) - \left(\frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n}\right) + \text{last term}

Thus, the final sum can be expressed as:

S2n+1=11n+1S_{2n+1} = 1 - \frac{1}{n+1}

Conclusion

  1. The series converge.
  2. The sum of the first 2n+12n+1 terms is 11n+11 - \frac{1}{n+1}.

Follow-up Questions

  1. How do we derive the general term for both series?
  2. What is the explicit form of the sum for the first nn terms of the second series?
  3. Can you illustrate the convergence behavior of both series graphically?
  4. What happens if we extend the first series to infinity?
  5. How do the two series relate to the harmonic series?

Tip

For alternating series, always check the convergence using the Alternating Series Test, which provides a useful criterion for convergence.

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Alternating Series
Telescoping Series

Formulas

S_{2n+1} = 1 - 1/(n+1)
T_n = sum(1/(k(k+1)))

Theorems

Alternating Series Test
Telescoping Series Theorem

Suitable Grade Level

Grades 11-12