• if å_{n = 1}^¥ b_n converges, so does å_{n = 1}^¥ a_n.
• if å_{n = 1}^¥ a_n diverges, so does å_{n = 1}^¥ b_n.

Proof. Let t_n be the n-th partial sums of å_{n = 1}^¥ b_n and s_n be the n-th partial sums of å_{n = 1}^¥ a_n. Then eventually |s_n-s_m| £ |t_n-t_m|.   ^[¯]

Defn.  A series å_{n = 1}^¥ x_n in a normed linear space X is said to converge absolutely if å_{n = 1}^¥ ||x_n||_X converges. Of course, the real and complex number systems are special cases.

Theorem. (Absolute Convergence Test) If a series converges absolutely, then it converges.

Proof. Let s_n be the sequence of partial sums of {x_n} and {t_n} be that for {||x_n||}, then for m > n the triangle inequality implies ||s_n-s_m|| = ||å_{k = n+1}^m x_n || £ å_{k = n+1}^m ||x_n|| = t_m-t_n.   ^[¯]

Theorem. (Limit Comparison Test) Suppose we have two nonnegative sequences which satisfy lim_n®¥ a_n/b_n = a with 0 < a < ¥. Then å_{n = 1}^¥ a_n and å_{n = 1}^¥ b_n converge and diverge together.

Proof. Without loss of generality, we can assume that all terms are nonnegative. By the hypothesis we see that eventually r a_n < b_n < R a_n where r = a/2 and R = 2 a.   ^[¯]

Theorem. (Alternating Series Test) Suppose that a_n decreases to 0, then the series å_{n = 1}^¥ (-1)ⁿ⁺¹ a_n converges (to s say). Furthermore, the error estimate holds:

Proof. Consider the sequence of partial sums {s_2n}, then these are monotone since 0 £ a_2k-1-a_2k and s_2n = s_2n-2+ (a_2n-1-a_2n). This sequence is also bounded since we can rewrite and estimate it as s_2n = a₁ -a_2n - å_{k = 1}^n-1 (a_2k-a_2k+1) £ a₁. Hence the sequence of s_2n's converge. Since the odd terms of the sequence of partial sums satisfy s_2n+1 = s_2n + a_2n+1 and a_2n+1® 0, we see that the sequence s_n also converges to s. For the error estimate, we may estimate the difference of partial sums by

| s_n - s_n+2k+1 | = | a_n+1 - å_{j = 1}^k(a_n+2j-a_n+2j+1) | £ |a_n+1 |.


Since the absolute value function is continuous and s_n+2k+1® s as k®¥, the error estimate follows in the limit.   ^[¯]

Example.   å_{n = 1}^¥ (-1)ⁿ⁺¹ 1/n converges but does not converge absolutely.

Details: Convergence of the series follows directly from the alternating series test. To show the series is not absolutely convergent, consider the function f(x) = 1/x and the partition P = {1,2,...,n+1} of [1,n+1], then log(n) £ ò₁ⁿ⁺¹ f(x) dx £ U(P,f) = å_{k = 1}ⁿ 1/k. But we know that log(n)®¥, which shows that the series å_{n = 1}^¥ 1/n does not converge.

Theorem.  (Integral Test) Suppose that f is nonnegative and monotone decreasing on [1,¥), then å_{n = 1}^¥ f(n) converges if and only if lim_n®¥ ò₁ⁿ f(x) dx is finite.

Proof. Simply note that

å_{n = 1}^¥ f(n)   ~   lim_n®¥ ò₁ⁿ f(x) dx.

by considering upper and lower Riemann sums.    ^[¯]