Assume that Theorem 2.4 (the root test) is true and prove the ratio test (Theorem 2.5).
Solution
1. Recall the root test (Theorem 2.4). Proof: if \(\limsup_{n\to\infty}|a_n|^{1/n} < 1\) then \(\sum a_n\) converges absolutely; if \(\limsup |a_n|^{1/n} > 1\) then \(\sum a_n\) diverges.
2. Relate the root-test quantity to the ratio-test quantity.Proof: for a sequence of positive numbers \(c_n\), \[\liminf_{n\to\infty}\frac{c_{n+1}}{c_n} \le \liminf_{n\to\infty} c_n^{1/n} \le \limsup_{n\to\infty} c_n^{1/n} \le \limsup_{n\to\infty}\frac{c_{n+1}}{c_n}.\] The upper bound: write \(q = \limsup c_{n+1}/c_n\); for \(\epsilon > 0\) there is \(N\) with \(c_{n+1} \le (q+\epsilon)c_n\) for \(n \ge N\), hence \(c_n \le C(q+\epsilon)^n\), so \(c_n^{1/n} \le C^{1/n}(q+\epsilon) \to q+\epsilon\).The lower bound is analogous (or apply the upper bound to \(1/c_n\)).
3. Apply to \(c_n = |a_n|\) and invoke the root test.Proof: if \(\limsup |a_{n+1}|/|a_n| = r < 1\), then by 2, \(\limsup |a_n|^{1/n} \le r < 1\), so the root test gives absolute convergence.If \(\liminf |a_{n+1}|/|a_n| > 1\) (the ratio test’s divergence hypothesis), then \(\liminf |a_n|^{1/n} > 1\), so in particular \(\limsup |a_n|^{1/n} > 1\) and the root test gives divergence.
4. Q.E.D.