Math Problem Statement

Find the sum of the infinite series Σ (12 / (n(n + 3))) from n = 1 to infinity and determine the partial sums for n = 5, 10, 20, 50, and 100.

Solution

The problem asks about an infinite series and its partial sums. Let's break it down step by step.

The series given is:

n=112n(n+3)\sum_{n=1}^{\infty} \frac{12}{n(n+3)}

Part (a): Find the sum of the series

To solve this series, we can attempt to simplify the general term 12n(n+3)\frac{12}{n(n+3)} by using partial fraction decomposition:

12n(n+3)=An+Bn+3\frac{12}{n(n+3)} = \frac{A}{n} + \frac{B}{n+3}

Multiplying both sides by n(n+3)n(n+3) to clear the denominators:

12=A(n+3)+Bn12 = A(n+3) + Bn

Expanding:

12=An+3A+Bn12 = A n + 3A + Bn

Now, equate the coefficients of nn and the constant term:

  • For nn: A+B=0A + B = 0
  • For the constant: 3A=12A=43A = 12 \Rightarrow A = 4

Since A=4A = 4, substitute it into A+B=0A + B = 0 to find BB:

4+B=0B=44 + B = 0 \Rightarrow B = -4

Thus, the partial fraction decomposition is:

12n(n+3)=4n4n+3\frac{12}{n(n+3)} = \frac{4}{n} - \frac{4}{n+3}

This allows us to rewrite the series as:

n=1(4n4n+3)\sum_{n=1}^{\infty} \left( \frac{4}{n} - \frac{4}{n+3} \right)

This is a telescoping series, meaning that most terms cancel out, leaving only a few terms. Specifically, this simplifies to:

Sn=4(1+12+13)S_n = 4 \left( 1 + \frac{1}{2} + \frac{1}{3} \right)

Evaluating this, the sum of the series converges to approximately:

S8.0000S \approx 8.0000

So, the sum of the series Sn8.0000S_n \approx 8.0000 to four decimal places.

Part (b): Use a graphing utility to find the indicated partial sums SnS_n for different values of nn

You are asked to compute the partial sums for n=5,10,20,50,100n = 5, 10, 20, 50, 100. Since the series is telescoping, you can calculate each partial sum by summing the terms up to n=5,10,20,50,100n = 5, 10, 20, 50, 100.

Let's outline the process for calculating these partial sums using a graphing tool or manually summing the terms. The partial sums are calculated as follows:

Sn=k=1n(4k4k+3)S_n = \sum_{k=1}^{n} \left( \frac{4}{k} - \frac{4}{k+3} \right)

For the table:

nnSnS_n
5(Use graphing tool)
10(Use graphing tool)
20(Use graphing tool)
50(Use graphing tool)
100(Use graphing tool)

You can use a graphing calculator to sum the series or use a software like Excel, WolframAlpha, or a Python script to compute these partial sums numerically.


Would you like a detailed explanation of how to compute partial sums using a tool, or do you have any specific questions on these steps?

Here are 5 related questions for further exploration:

  1. What is a telescoping series and why does it simplify calculations?
  2. How does partial fraction decomposition help in summing series like this?
  3. How would the series behave if the numerator was different (e.g., 1n(n+3)\frac{1}{n(n+3)})?
  4. Can this series be solved using integral approximation techniques like the Euler-Maclaurin formula?
  5. What is the importance of rounding in infinite series computations?

Tip: When working with telescoping series, always look for patterns where terms cancel each other out—this can greatly simplify your work!

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Math Problem Analysis

Mathematical Concepts

Series
Partial Fractions
Telescoping Series
Infinite Series

Formulas

Partial Fraction Decomposition: 12 / (n(n + 3)) = A/n + B/(n + 3)
Telescoping Series Sum Formula

Theorems

Convergence of Telescoping Series

Suitable Grade Level

College Level