Summary of Series Convergence Tests 1. Harmonic Series 1. The harmonic series is the series ∞ ∑ 1 π π=1 = 1 1 1 + + + β―. 1 2 3 2. The harmonic series is divergent. 2. Geometric Series 1. A geometric series is any series of the form ∞ ∑ πππ−1 = π + ππ + ππ2 + β― , π=1 where π, π ∈ β. 2. A geometric series converges to π if |π| < 1, and is divergent if |π| ≥ 1. 1−π 3. Divergence Test ∑∞ Let π=1 ππ be a series. 1. If lim ππ ≠ 0, then the series π→∞ ∞ ∑ ππ is divergent. π=1 2. Caution: If lim ππ = 0, you CANNOT conclude that the series ∞ ∑ π→∞ ππ is convergent. π=1 4. Integral Test ∑∞ Let π=1 ππ be a series, and let π βΆ [1, ∞) → β be function that satisfies the following four properties: (1) π (π) = ππ for all π. (2) π (π₯) is continuous on [1, ∞). (3) π (π₯) > 0 on [1, ∞). (4) π (π₯) is decreasing on [1, ∞). Then ∑∞ π=1 ∞ ππ is convergent if and only if ∫1 1 π (π₯) ππ₯ is convergent. 5. Remainder Estimate for the Integral Test ∑∞ Let π=1 ππ be a series. ∑∞ 1. Suppose that π=1 ππ is convergent by the Integral Test. Let π ∈ β. π ∑ ∞ ππ + π=1 ∫π+1 π (π₯) ππ₯ ≤ ∞ ∑ π=1 ππ ≤ π ∑ ∞ ππ + π=1 ∫π π (π₯) ππ₯. 2. An approximate value for the sum of the series is the average of the upper bound and the lower bound in the above inequalities. The difference between this approximate value and the actual sum of the series is at most half the distance between the upper bound and the lower bound. 6. π-Series 1. A π-series is any series of the form ∞ ∑ 1 1 1 1 = π + π + π + β―, π π 1 2 3 π=1 where π ∈ β. 2. A π-series is convergent if π > 1, and is divergent if π ≤ 1. 7. Comparison Test ∑∞ ∑∞ Let π=1 ππ and π=1 ππ be series. Suppose that ππ ≥ 0 and ππ ≥ 0 for all π ∈ β. Suppose that ππ ≤ ππ for all π ∈ β. 1. If ∞ ∑ ππ is convergent, then 2. If ππ is convergent. π=1 π=1 ∞ ∑ ∞ ∑ ππ is divergent, then π=1 3. Caution: If ∞ ∑ ππ is divergent. π=1 ∞ ∑ π=1 ππ is convergent or if ∞ ∑ ππ is divergent, you CANNOT conclude anything about π=1 the other series by the Comparison Test. 2 8. Limit Comparison Test ∑∞ ∑∞ Let π=1 ππ and π=1 ππ be series. Suppose that ππ ≥ 0 and ππ ≥ 0 for all π ∈ β. Suppose that ππ = πΏ, π→∞ π π lim for some πΏ ∈ β or πΏ = ∞. 1. Suppose that 0 < πΏ < ∞. Then either both ∑∞ and π=1 ππ are divergent. 2. Suppose that πΏ = 0. If ∞ ∑ ∑∞ π=1 ππ is convergent, then π=1 3. Suppose that πΏ = ∞. If ∞ ∑ ππ and ∞ ∑ ∑∞ π=1 ππ are convergent, or both ∑∞ π=1 ππ ππ is convergent. π=1 ππ is divergent, then π=1 ∞ ∑ ππ is divergent. π=1 9. Alternating Series Test ∑∞ Let π=1 (−1)π−1 ππ be an alternating series, where ππ > 0 for all π ∈ β. 1. Suppose that the alternating series satisfies the following two properties: { }∞ (a) the sequence ππ π=1 is decreasing. (b) lim ππ = 0. π→∞ Then the alternating series is convergent. 1. The same result holds for alternating series of the form ∑∞ π=1 (−1)π ππ . 10. Remainder Estimate for the Alternating Series Test ∑∞ Let π=1 (−1)π−1 ππ be an alternating series, where ππ > 0 for all π ∈ β. Let π ∈ β. 1. The πth remainder of the alternating series, denoted π π , is defined by π π = ∞ ∞ ∑ ∑ (−1)π−1 ππ − π π = (−1)π ππ . π=1 π=π+1 2. Suppose that the alternating series satisfies the hypotheses of the Alternating Series Test, and hence is convergent. Then |π π | ≤ ππ+1 . ∑∞ 3. The same result holds for alternating series of the form π=1 (−1)π ππ . 3 11. Ratio Test ∑∞ Let π=1 ππ be a series. Suppose that ππ ≠ 0 for all π ∈ β. Suppose that | ππ+1 | | = πΏ, lim || π→∞ | π || π for some πΏ ∈ β or πΏ = ∞. ∑∞ 1. If πΏ < 1, then π=1 ππ is absolutely convergent. ∑∞ 2. If πΏ > 1, then π=1 ππ is divergent. 3. Caution: If πΏ = 1, you CANNOT conclude conclude that divergent by the Ratio Test. 4 ∑∞ π=1 ππ is either convergent or