Defn. If {a_n} is a sequence of real numbers, then define the limit inferior and limit superior, respectively, by

Note. If we define a_k : = inf_{n ³ k} a_n, then it is clear that the {a_n} form a nondecreasing sequence and will converge in the extended sense. Similarly, b_k : = sup_{n ³ k} a_n form a nonincreasing sequence and will converge in the extended sense to its infimum.

Proposition.  Suppose a = lim inf_n®¥ a_n, then for each e > 0, eventually a-e < a_n and infinitely often a_n < a+e. Similarly, if lim sup_n®¥ a_n = b, then for each e > 0 eventually a_n < b+ e and infinitely often b < a_n + e.

Proof. We prove the first statement and leave the other for the student. By the definition of the limit inferior, we see that if a_k : = inf_{n ³ k} a_n, then for e > 0 there is an n such that a-e < a_k £ a. The statement of the theorem follows directly from the definition of infimum applied to a_k.   ^[¯]

Corollary.  If lim inf_n®¥ a_n > a, then eventually a_n > a. Similarly, if lim sup_n®¥ a_n < b, then eventually a_n < b.

Proof. For the first statement, let a: = lim inf_n®¥ a_n, set e = a-a, and apply the previous proposition. For the second, set e = b-b.   ^[¯]

Corollary.  A sequence {a_n} converges if and only if lim inf_n®¥ a_n = lim sup_n®¥ a_n . The common value is the limit of the sequence.

Proof. Apply the previous proposition.   ^[¯]

Theorem.  (Ratio Test) For a sequence of nonnegative numbers, define

then for the series å_{n = 1}^¥ a_n

• R < 1 implies convergence,
• r > 1 implies divergence,
• R = 1, r = 1 the test is inconclusive.

Proof. To prove the first statement is true, we observe as shown above that R < 1 implies that eventually a_n+1/a_n £ T where T is strictly between R and 1. So there exists N such that 0 £ a_n+1 £ T a_n for all N £ n. By induction we then see that eventually (i.e. for N £ n), we have a_n £ C Tⁿ where C: = a_N/T^N. Applying the comparison test and the fact that 0 < T < 1, we see that the series converges. On the other hand, if r > 1, then a similiar argument shows that eventually a_n > C tⁿ> C, where r>t>1 and C = a_N /t^N for some N. Hence by the n-th term test, the series must diverge. The last statement of the theorem follows since å_{n = 1}^¥ 1/n diverges, å_{n = 1}^¥ 1/n² converges, but R = r = 1 for both series.   ^[¯]

Note. The special limit lim_n®¥  n^1/n  =  1 will be useful in what follows.

Details:  By taking logarithms and using the continuity of the log function at x=0, we see that it will suffice to show that lim_n®¥ log(n)/n = 0. This follows however by an application of L'Hospital's rule.   ^[¯]

Theorem. (n-th Root Test) For a sequence of nonnegative numbers, define

then for the series å_{n = 1}^¥ a_n

• R < 1 implies convergence,
• R > 1 implies divergence,
• R = 1 implies the test is inconclusive.

Proof. To prove the first statement is true, we observe as shown above that R < 1 implies that eventually (a_n)^1/n £ T where T is strictly between R and 1. So there exists N such that 0 £ a_n £ Tⁿ for all N £ n. Applying the comparison test and the fact that 0 < T < 1, we see that the series converges. On the other hand, if R > 1, then infinitely often a_n > ((R+1)/2)ⁿ > 1, so by the n-th term test, the series must diverge. The third part of the theorem follows since å_{n = 1}^¥ 1/n diverges, å_{n = 1}^¥ 1/n² converges, but lim_n®¥ n^1/n = 1 and so R = 1 for both series.   ^[¯]