Calculus 2 Series and Sequences Review Sheet PDF

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Summary

Entire middle section of the course. Info on series and sequences including series convergence tests and taylor polynomials....

Description

Calculus 2: Series and Sequences Taylor Series Taylor-Lagrange Theorem •

𝐸𝑓𝑛 (𝑥; 𝑎) = |

𝑓 (𝑛+1) (𝑐) (𝑛+1)!

∗ (𝑥 − 𝑎)𝑛−1 |

for some 𝑐 ∈ (𝑎, 𝑥)

Sequences and Series Sequence: a list of numbers which is ordered • lim 𝑎𝑛 = 𝐿 𝑛→∞

• If the limit exists and is finite, the sequence is convergent • If the limit is infinite or DNE, the sequence is divergent Squeeze/Pinch/Sandwich Theorem • If lim 𝑓(𝑥) = lim 𝑔(𝑥) 𝑥→∞

𝑥→∞

And 𝑓(𝑛) ≤ 𝑎𝑛 ≤ 𝑔(𝑛) Then, lim 𝑎𝑛 = lim 𝑓(𝑥) = lim 𝑔(𝑥) 𝑛→∞ 𝑥→∞ 𝑥→∞ Rates of Growth Priorities • ln(𝑛) ≪ 𝑛𝑘 for any 𝑘 > 0 𝑘 𝑛 • 𝑛 ≪𝑐 for any 𝑘 > 0, 𝑐 > 1 • 𝑐 𝑛 ≪ 𝑛! for any 𝑐 > 1 • 𝑛! ≪ 𝑛𝑛 Series: a sum of a sequence • 𝑆𝑛 = ∑ 𝑛𝑘=1 𝑎𝑘 = 𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 ∞ • If {𝑆𝑛 } converges, ∑ 𝑘=1 𝑎𝑘= lim 𝑆𝑛 • •

𝑛→∞

• Otherwise, the series diverges Geometric Sequence/Series • 𝑎𝑛 = 𝑎 ∗ 𝑟 𝑛−1 𝑎 ∈ ℝ, 𝑎 ≠ 0, 𝑟 ∈ ℝ, 𝑟 ≠ 0 𝑎 𝑟=1 𝐷𝑁𝐸 𝑟 = −1 • lim 𝑎𝑛 = { |𝑟| < 1} 0 𝑛→∞ 𝐷𝑁𝐸 |𝑟| > 1 𝑎 |𝑟| < 1 1−𝑟

𝐷𝑁𝐸 |𝑟| > 1 ∑∞ 𝑛=1 𝑎𝑛 = 𝐷𝑁𝐸 𝑟 = 1 {𝐷𝑁𝐸 𝑟 = −1} ∑∞ 𝑛=1 (𝑎𝑛 + 𝑏𝑛 ) Convergence/Divergence ∞ ∞ • If ∑∞ 𝑛=1 𝑎𝑛 converges and ∑ 𝑛=1 𝑏𝑛 converges, ∑𝑛=1 (𝑎𝑛 + 𝑏𝑛 ) converges ∞ ∞ ∞ • If ∑𝑛=1 𝑎𝑛 converges but ∑ 𝑛=1 𝑏𝑛 diverges, ∑ 𝑛=1(𝑎𝑛 + 𝑏𝑛 ) diverges ∞ • If ∑∞ 𝑛=1 𝑎𝑛 diverges and ∑ 𝑛=1 𝑏𝑛 diverges, no information Term Test for Divergence • If lim 𝑎𝑘 ≠ 0, then ∑ ∞ 𝑘=1 𝑎𝑘 diverges •

•

𝑘→∞

If lim 𝑎𝑘 = 0, no information 𝑘→∞

Integral Test • Assuming 𝑎𝑛 ≥ 0, 𝑓(𝑛) = 𝑎𝑛 , and decreasing in 𝑛 ∞ • ∑∞ 𝑛=1 𝑎𝑛 converges if and only if ∫1 𝑓(𝑥)𝑑𝑥 converges P-Test

∞

• ∑ 𝑛=1 𝑛𝑝 for any 𝑝 ∈ ℝ • The series diverges if 𝑝 ≤ 1 • The series converges if 𝑝 > 1 Comparison Test • 𝑎 𝑛 ≤ 𝑏𝑛 ∞ • If ∑∞ 𝑛=1 𝑏𝑛 converges, then ∑ 𝑛=1 𝑎𝑛 also converges ∞ ∞ • If ∑𝑛=1 𝑎𝑛 diverges, then ∑ 𝑛=1 𝑏𝑛 also diverges Limit Comparison Test 𝑎 0 1, then ∑𝑛=1 𝑎𝑛 diverges • If 𝐿 = 1, no information Alternating Series Absolute Convergence ∞ • ∑∞ 𝑛=1 𝑎𝑛 converges absolutely if ∑𝑛=1 |𝑎𝑛 | converges Alternating Series Test • If |𝑎𝑘+1 | ≤ |𝑎𝑘 | • And lim |𝑎𝑘 | = 0 𝑘→∞ • Then ∑ ∞ 𝑘=1 𝑎𝑘 converges Alternating Series Conditional Convergence ∞ ∞ • If ∑∞ 𝑘=1 𝑎𝑘 converges, but ∑ 𝑘=1|𝑎𝑘 | diverges, ∑ 𝑘=1 𝑎𝑘 converges conditionally...