http://www.ck12.org Chapter 8. Infinite Series
Suppose an alternating series satisfies the conditions of the Alternating Series Test and has the sumS. Letsnbe the
nth partial sum of the series. Then|S−sn|≤un+ 1.
The main idea of the theorem is that the remainder|S−sn|cannot get larger than then+1st term in the series,un+ 1.
This is the term whose index is one more than the index of the partial sum used in the difference.
Example 2
Computes 3 for the series∑∞k= 1 (− 1 )k+^1 kk 3 ++^5 kand determine the bound on the remainder.
Solution
First we compute the third partial sum to approximate the sumSof the series:
s 3 = (− 1 )^2113 ++^51 +(− 1 )^3223 ++^52 +(− 1 )^4333 ++^53
=^64 − 107 + 308
=^90 −^4260 +^16 =^6460The theorem tells us to use the next term in the series,u 4 , to calculate the bound on the difference or remainder.
Remember that the part(− 1 )k+^1 just gives the sign of the term and, so we just use the partkk 3 ++^5 kto calculateu 4.
Thusu 4 = 443 ++^54 = 689. Then|S−sn|=
∣∣
S−^6460
∣∣
< 689 ≈ 0 .13. This tells us that the absolute value of the error or
remainder is less than 0.13.
Absolute and Conditional Convergence
There are other types of convergence for infinite series:absolute convergenceandconditional convergence.
Absolute Convergence
Let∑∞k= 1 uk=u 1 +u 2 +u 3 ...+uk+...be an infinite series. Then the series isabsolutely convergentif∑∞k= 1 |uk|=
|u 1 |+|u 2 |+|u 3 |+...+|uk|+...converges.
The infinite series∑∞k= 1 |uk|=|u 1 |+|u 2 |+|u 3 |+...+|uk|+...is the series made by taking the absolute values of the
terms of the series∑∞k= 1 |uk|=|u 1 |+|u 2 |+|u 3 |+...+|uk|+....
The convergence of the series of absolute values tells us something about the convergence of the series.
Theorem
If∑∞k= 1 |uk|=|u 1 |+|u 2 |+|u 3 |+...+|uk|+...converges, then∑∞k= 1 uk=u 1 +u 2 +u 3 +...+uk+...also converges.
This tells us that if you can show absolute convergence, then the series converges.
If the series of absolute value diverges, we cannot conclude anything about the series.
Example 3
Determine if the series∑k∞= 1 ( 3 −k^14 )−k+k^1 converges absolutely.
Solution
We find the series of absolute values:∑∞k= 1
∣∣
∣(− 3 k^14 )−k+k^1∣∣
∣=∑∞k= (^13) k (^41) −k, which behaves like the series∑∞k= (^131) k 4. This is a
p−series withp= 4 .The series∑∞k= 1 ( 3 −k^14 )−k+k^1 converges absolutely and hence converges.
Example 4