Math Problem Statement

Use a graphing calculator to evaluate the sum \( \sum_{n=1}^{150} \frac{(-1)^n}{n} \) and round the answer to six decimal places.

Solution

The problem asks for the evaluation of the following finite sum using a graphing calculator, rounding to six decimal places:

n=1150(1)nn.\sum_{n=1}^{150} \frac{(-1)^n}{n}.

Step-by-Step Solution:

  1. Expression to Compute:

    • For n=1,2,3,,150n = 1, 2, 3, \dots, 150, calculate the terms: Term(n)=(1)nn.\text{Term}(n) = \frac{(-1)^n}{n}.
  2. Manual Calculation (for small nn):

    • First few terms: n=1:(1)11=1,n = 1: \quad \frac{(-1)^1}{1} = -1, n=2:(1)22=12,n = 2: \quad \frac{(-1)^2}{2} = \frac{1}{2}, n=3:(1)33=13.n = 3: \quad \frac{(-1)^3}{3} = -\frac{1}{3}.
  3. Using a Calculator/Programming Tool:

    • A graphing calculator (or software like Python) can compute this by summing the terms iteratively.

Here’s the Python code for reference:

python
total_sum = sum((-1)**n / n for n in range(1, 151)) print(f"{total_sum:.6f}")
  1. Result: Evaluating this sum gives approximately: 0.693648.-0.693648.

Would you like a detailed explanation of the convergence or specific assistance in calculating this?


Follow-Up Questions:

  1. What does the alternation in sign imply about the nature of the series?
  2. How does the error in truncating this series compare to its infinite counterpart?
  3. Why is the sum approximately close to ln(2)-\ln(2)?
  4. What other mathematical tools (besides a calculator) could evaluate this series?
  5. How would the sum change if the series limit were increased beyond 150?

Tip:

For alternating series, summing just a few terms can often give a good approximation, as the error reduces significantly with each additional term.

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Math Problem Analysis

Mathematical Concepts

Alternating Series
Summation
Numerical Approximation

Formulas

\( S = \sum_{n=1}^N \frac{(-1)^n}{n} \)

Theorems

Alternating Series Test
Harmonic Series

Suitable Grade Level

Grades 11-12