Summary of Convergence Tests for ∞

Let ∑an be an infinite series of positive terms. n=1 ∞

The series ∑ an converges if and only if the of partial sums, S n = a1 + a2 + a3 +an , converges. n=1 ∞

NOTE: lim S n = an n→∞ ∑ n=1

∞

Divergence Test: If lim an ≠ 0 , the series an diverges. n→∞ ∑ n=1 ∞ n n 1 Example: The series is divergent since lim = lim = 1 ∑ 2 n→∞ 2 n→∞ n=1 n +1 n +1 1+ 1 n 2

This means that the terms of a must approach zero. That is, if ∑an converges, then

lim an = 0. However, lim an = 0 does not imply convergence. n→∞ n→∞ : THIS is our model series A geometric series a + ar + ar 2 +  + ar n−1 +  converges for −1< r <1. a a Note: r = n+1 If the series converges, the sum of the series is . an 1− r n ∞  7  35 7 Example: The series ∑5  converges with a = a1 = and r = . The sum of the series is 35. n=1  8  8 8

Integral Test: If f is a continuous, positive, decreasing function on [1,∞) with f (n) = an , then the series ∞ ∞ ∑an converges if and only if the improper ∫ f (x)dx converges. n=1 1

∞ 1 p-series: The series ∑ p is convergent for p > 1 and diverges otherwise. n=1 n ∞ 1 ∞ 1 Examples: The series ∑ 1.001 is convergent but the series ∑ is divergent. n=1 n n=1 n

∞ an+1 an+1 : (a) If lim < 1 then the series an converges; (b) if lim > 1 the series diverges. n→∞ ∑ n→∞ an n=1 an Otherwise, you must use a different test for convergence.

This says that if the series eventually behaves like a convergent (divergent) geometric series, it converges (diverges). If this limit is one, the test is inconclusive and a different test is required. Specifically, the Ratio Test does not work for p-series. ∞ ∞

Comparison Test: Suppose ∑an and ∑bn are series with positive terms. n=1 n=1 ∞ ∞

(a) If ∑bn is convergent and an ≤ bn for all n, then ∑an converges. n=1 n=1 ∞ ∞

(b) If ∑bn is divergent and an ≥ bn for all n, then ∑an diverges. n=1 n=1 The Comparison Test requires that you make one of two comparisons: • Compare an unknown series to a LARGER known convergent series (smaller than convergent is convergent) • Compare an unknown series to a SMALLER known divergent series (bigger than divergent is divergent) ∞ 3n ∞ 3n ∞ 1 > = Examples: ∑ 2 ∑ 2 3∑ which is a divergent harmonic series. Since the original series is larger n=2 n − 2 n=2 n n=2 n by comparison, it is divergent.

∞ 5n ∞ 5n 5 ∞ 1 < = We have ∑ 3 2 ∑ 3 ∑ 2 which is a convergent p-series. Since the original series is n=1 2n + n +1 n=1 2n 2 n=1 n smaller by comparison, it is convergent. ∞ ∞

Limit Comparison Test: Suppose ∑an and ∑bn are series with positive terms. If n=1 n=1 a lim n = c where 0 < c < ∞ , then either both series converge or both series diverge. (Useful for p-series) n→∞ bn

Rule of Thumb: To obtain a series for comparison, omit lower order terms in the numerator and the denominator and then simplify.

∞ n ∞ n ∞ 1 = Examples: For the series ∑ 2 , compare to ∑ 2 ∑ 3 which is a convergent p-series. n=1 n + n + 3 n=1 n n=1 n 2

n ∞ π n + n ∞ π n ∞  π  = For the series ∑ n 2 , compare to ∑ n ∑  which is a divergent geometric series. n=1 3 + n n=1 3 n=1  3 

∞ n−1 Test: If the alternating series ∑(−1) bn = b1 − b2 + b3 − b4 + b5 − b6 +  n=1

satisfies (a) bn > bn+1 and (b) limbn = 0 , then the series converges. n→∞

Remainder: Rn = s − sn ≤ bn+1

Absolute convergence simply means that the series converges without alternating (all signs and terms are positive). ∞ (−1)n Examples: The series ∑ is convergent but not absolutely convergent. n=0 n +1

∞ (−1)n Alternating p-series: The alternating p-series ∑ p converges for p > 0. n=1 n ∞ (−1) n ∞ (−1) n Examples: The series ∑ and the Alternating Harmonic series ∑ are convergent. n=1 n n=1 n