SUMMARY of INFINITE SERIES TESTS 1. Definition of the Sum of A

SUMMARY of INFINITE SERIES TESTS 1. Definition of the Sum of A

SUMMARY OF INFINITE SERIES TESTS 1. Definition of the sum of a series. P∞ Let sn = a1 + a2 + ··· + an. If limn→∞ sn exists and equals s then n=1 an is convergent and its sum is s. 2. Well understood series. • (geometric series) The series a + ar + ar2 + ar3 + ··· is convergent if |r| < 1 and the sum is a/(1 − r) and the series is divergent with no sum if |r| ≥ 1. P∞ 1 • (p series) The series n=1 np is convergent if p > 1 and divergent if p ≤ 1. P∞ • (Telescoping series) For some series n=1 an, a partial fraction decomposition of an results in a telescoping sum of sn making it easy to determine the sum of the series. P∞ P∞ 3. Tests for positive term series only. Suppose n=1 an and n=1 bn are positive term series. P∞ P∞ • (limit comparison test) If an ∼ bn for large n, then n=1 an and n=1 bn have the same behavior. • (comparison test ) Suppose an ≤ bn for all n. P∞ P∞ (a) If n=1 an is divergent then n=1 bn is divergent, P∞ P∞ (b) If n=1 bn is convergent then n=1 an is convergent. 4. Tests for any series. P∞ • (nth term test ) If limn→∞ an 6= 0 or does not exist then n=1 an is divergent. • (ratio test) Suppose lim an+1 = L. If L < 1 then P∞ a is convergent; if L > 1 n→∞ an n=1 n the series is divergent; if L = 1 then try another test. STRATEGY FOR USING THE TESTS The following steps will work fairly frequently. 1. If it is a p series or a geometric series or a telescoping series then we know its behavior. 2. If it is a positive term series then use the limit comparison test to simplify the series - the series may already be in the simplest possible form. Now analyze the simpler series using steps 1, 3, 4 and 5. 3. If the series has a factorial or a power of a number (such as 3n) then try the ratio test. 4. If the series is a positive term series then try the ordinary comparison test. 5. If the above failed, try computing limn→∞ an; • If the limit is non-zero or does not exist then the series is divergent because of the nth term test. • If the limit is zero or too hard to compute then try ratio test..

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