In the case where all the terms are positive, If we stop evaluating a convergent positive series after adding n terms, the difference between the true infinite sum and our partial sum is called the error or remainder. 𝑅 𝑛 =𝑆− 𝑆 𝑛 In the case where all the terms are positive, the error (or remainder) will be positive. Here’s a representation of the series (Riemann Sum) and it’s related integral for n large. Drawn as a lower sum, the partial sum stops at the last yellow box. The error (shown in red) is less than 𝑛 ∞ 𝑓 𝑥 𝑑𝑥 . Shifted over, drawn as an upper sum, we see that the error is greater than 𝑛+1 ∞ 𝑓 𝑥 𝑑𝑥

Integral Test Remainder Estimate Given 𝑎 𝑛 is a decreasing positive series convergent series and 𝑓 𝑥 is a continuous function where 𝑓 𝑛 = 𝑎 𝑛 then if 𝑅 𝑛 =𝑆− 𝑆 𝑛 𝑛+1 ∞ 𝑓 𝑥 𝑑𝑥 ≤ 𝑅 𝑛 ≤ 𝑛 ∞ 𝑓 𝑥 𝑑𝑥

Example: Approximate the sum of the convergent series using the indicated number of terms. Include an estimate of the maximum error for your approximation.

The minimum number of terms needed is 32. 𝑛=1 ∞ 1 𝑛 3 , 𝑘=3 Example for #43 (read the directions, p. 582) We want 𝑅 𝑛 <.0005 It is sufficient is to find 𝑛 such that 𝑛 ∞ 1 𝑥 3 𝑑𝑥 <.0005 ∞ 𝑛 𝑛 ∞ 1 𝑥 3 𝑑𝑥 =− 1 2 𝑥 −2 = 1 ∞ − −1 2 𝑛 2 = 1 2 𝑛 2 1 2 𝑛 2 <.0005 →1<.001 𝑛 2 →1000< 𝑛 2 →𝑛>31.623 The minimum number of terms needed is 32. 𝑆≈𝑆 32 =1+ 1 8 + 1 27 + 1 64 +…+ 1 32768 ≈1.201583642

Remainder of Alternating Series

𝑛=1 ∞ −1 𝑛+1 𝑛 =1− 1 2 + 1 3 − 1 4 +…+ −1 𝑛+1 𝑛 +… 𝑛=1 ∞ −1 𝑛+1 𝑛 =1− 1 2 + 1 3 − 1 4 +…+ −1 𝑛+1 𝑛 +… Let’s look at some partial sums graphically. Clearly, the Partial Sums are headed for some limit in the middle of these points. Each “jump” is smaller than the previous one.

We know that the infinite sum 𝑆 is between What if we stopped at 𝑆 5 ? We know that the infinite sum 𝑆 is between 𝑆 5 =1− 1 2 + 1 3 − 1 4 + 1 5 = 47 60 and the next partial sum 𝑆 6 = 𝑆 5 − 1 6 = 37 60 So if I use 𝑆 5 = 47 60 as an estimate for 𝑆, my error is less than | 𝑎 6 |= 1 6 . How far off can my estimate be if I use 𝑆 11 for my estimate? 𝑆 100 ? 𝑎 12 = 1 12 , and |𝑎 101 |= 1 101 respectively.

Example

LaGrange Remainder Theorem The difference between 𝑓 𝑥 and 𝑝 𝑛 (𝑥) is 𝑅 𝑛 𝑥 = 𝑓 𝑛+1 (𝑧) 𝑛+1 ! 𝑥−𝑎 𝑛+1 where 𝑧 is some unknowable number between 𝑥 𝑎𝑛𝑑 𝑎. 𝑎 is the 𝑥−𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 at the point of tangency.

𝑓 𝑥 = 1 1−𝑥 , 𝑛=4 EX: #4 𝑓 𝑥 = 1 1−𝑥 →𝑓 0 =1 𝑓 ′ 𝑥 =−(1−𝑥 ) −2 −1 =(1−𝑥 ) −2 → 𝑓 ′ 0 =1 𝑓 ′′ 𝑥 =−2(1−𝑥 ) −3 −1 =2(1−𝑥 ) −3 → 𝑓 ′ ′ 0 =2 𝑓 ′ ′′ 𝑥 =−6(1−𝑥 ) −4 −1 =6(1−𝑥 ) −4 → 𝑓 ′ ′′ 0 =6 𝑓 (4) 𝑥 =−24(1−𝑥 ) −5 −1 =24(1−𝑥 ) −5 → 𝑓 4 0 =24 𝑝 4 𝑥 =𝑓 0 + 𝑓 ′ 0 𝑥+ 𝑓 ′′ 0 2! 𝑥 2 + 𝑓 ′′′ 0 3! 𝑥 3 + 𝑓 4 0 4! 𝑥 4 𝑝 4 𝑥 =1+1𝑥+ 2 2! 𝑥 2 + 6 3! 𝑥 3 + 24 4! 𝑥 4 𝑝 4 𝑥 =1+𝑥+ 𝑥 2 + 𝑥 3 + 𝑥 4 Look familiar? 1+r+ 𝑟 2 +𝑟 3 +…+ 𝑟 𝑛 +…= 1 1−𝑟 , 𝑖𝑓 𝑟 <1

𝑅 𝑛 𝑥 = 𝑓 𝑛+1 (𝑧) 𝑛+1 ! 𝑥−𝑎 𝑛+1 𝑎=0, 𝑛=4, 𝑓 5 𝑧 =5!(1−𝑧 ) −6 𝑅 4 𝑥 = 𝑓 5 (𝑧) 5 ! 𝑥−0 5 = 5!(1−𝑧 ) −6 5 ! 𝑥 5 =(1−𝑧 ) −6 𝑥 5

We will need four derivatives all evaluated at 𝑥=0. 𝑓 𝑥 =𝑠𝑖𝑛𝑥, 𝑛=3 We will need four derivatives all evaluated at 𝑥=0. 𝑓 𝑥 =𝑠𝑖𝑛𝑥→𝑓 0 =0 𝑓 ′ 𝑥 =𝑐𝑜𝑠𝑥→ 𝑓 ′ 0 =1 𝑓 ′′ 𝑥 =−𝑠𝑖𝑛𝑥→ 𝑓 ′′ 0 =0 𝑓 ′′′ 𝑥 =−𝑐𝑜𝑠𝑥→ 𝑓 ′′′ 0 =−1 𝑓 4 𝑥 =𝑠𝑖𝑛𝑥→ 𝑓 4 𝑥 =0 𝑝 3 𝑥 =𝑓 0 + 𝑓 ′ 0 𝑥+ 𝑓 ′′ 0 2! 𝑥 2 + 𝑓 ′′′ 0 3! 𝑥 3 =0+1𝑥+ 0 2! 𝑥 2 + −1 3! 𝑥 3 𝑓 𝑥 =𝑠𝑖𝑛𝑥 ≈ 𝑝 3 𝑥 =𝑥− 1 3! 𝑥 3 𝑅 𝑛 𝑥 = 𝑓 𝑛+1 (𝑧) 𝑛+1 ! 𝑥−𝑎 𝑛+1 𝑅 3 𝑥 = 𝑓 4 (𝑧) 4 ! 𝑥−0 4 = sin 𝑧 4 ! 𝑥 4