Root test

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Root test: a quick way to check if a series converges

What it uses
- For a series sum a_n, look at the nth roots of the absolute values: (|a_n|)^(1/n).

How it works
- Let L = limsup as n → ∞ of (|a_n|)^(1/n).
- If L < 1, the series sum a_n converges absolutely.
- If L > 1, the series diverges.
- If L = 1, the test is inconclusive: the series may converge or diverge.

Note
- If the limit lim (|a_n|)^(1/n) exists, you can use L = lim (|a_n|)^(1/n) in place of the limsup.

Examples
- sum 1/n^2: (|a_n|)^(1/n) = (1/n^2)^(1/n) = n^(-2/n) → 1, so L = 1 (inconclusive). The series actually converges by comparison to a p-series.
- sum 1/n: (1/n)^(1/n) → 1, so L = 1 (inconclusive). The series diverges.

Power series
- For sum c_n (z − p)^n, the root test gives the radius of convergence:
- R = 1 / limsup_{n→∞} (|c_n|)^(1/n).
- The series converges for |z − p| < R and diverges for |z − p| > R.
- What happens on the boundary |z − p| = R usually needs separate checks.

Intuition
- If eventually (|a_n|)^(1/n) ≤ k < 1, then |a_n| ≤ k^n and the tail behaves like a geometric series, so it converges.
- If infinitely many n have (|a_n|)^(1/n) > 1, then |a_n| does not go to 0, so the series diverges.

This gives a clear routine: compute L, compare to 1, and use the radius of convergence formula for power series.