Root Test Calculator E
The Root Test Calculator E helps determine whether an infinite series converges by analyzing the limit of the nth root of the absolute value of its terms. This test is particularly useful for series with terms that involve powers of n.
What is the Root Test?
The Root Test is a convergence test used to determine whether an infinite series converges or diverges. It's particularly effective for series where the terms involve powers of n, such as 1/n² or 2ⁿ/n³.
The test involves calculating the limit of the nth root of the absolute value of the series terms as n approaches infinity. The result of this limit determines whether the series converges:
- If the limit is less than 1, the series converges absolutely.
- If the limit is greater than 1, the series diverges.
- If the limit equals 1, the test is inconclusive.
The Root Test is often used in conjunction with other convergence tests, especially when the Ratio Test is inconclusive.
How to Use the Calculator
- Enter the general term of your series in the input field. For example, for the series 1/n², you would enter "1/n^2".
- Specify the starting index (usually 1 or 2).
- Click "Calculate" to determine the limit of the nth root of the absolute value of the terms.
- Interpret the result based on the limit value.
The calculator will display the limit value and provide an interpretation of what this means for your series.
Root Test Formula
For a series Σaₙ, the Root Test involves calculating:
L = lim (n→∞) |aₙ|^(1/n)
Where:
- L is the limit value
- aₙ is the nth term of the series
The series converges if L < 1, diverges if L > 1, and the test is inconclusive if L = 1.
Examples
Example 1: Convergent Series
Consider the series Σ(1/n²).
Using the Root Test:
L = lim (n→∞) (1/n²)^(1/n) = lim (n→∞) n^(-2/n) = 0 (since 2/n → 0 as n → ∞)
Since L = 0 < 1, the series converges.
Example 2: Divergent Series
Consider the series Σ(2ⁿ/n³).
Using the Root Test:
L = lim (n→∞) (2ⁿ/n³)^(1/n) = lim (n→∞) 2^(1/n) * n^(-3/n)
As n → ∞, 2^(1/n) → 1 and n^(-3/n) → 1 (since -3/n → 0)
Thus, L = 2 > 1, so the series diverges.