8.2. Infinite Series http://www.ck12.org
Convergence/Divergence of Series
Let∑∞k= 1 ukbe an infinite series and let{sn}be the sequence of partial sums for the series. If{sn}has a finite limitl
, then the infinite series converges and∑∞k= 1 uk=l.
If{sn}does not have a finite limit, then the infinite series diverges. The infinite series does not have a sum.
Example 3
Does the infinite series 1+ 0. 1 + 0. 01 + 0. 001 +...converge or diverge?
Solution
To make our work easier, write the infinite series 1+ 0. 1 + 0. 01 + 0. 001 +...as an infinite series of fractions:
1 + 101 + 1012 + 1013 +...
To solve for convergence or divergence of the infinite series, write the formula for thenth partial sumsn=∑nk= 1 uk:
sn= 1 + 101 + 1012 + 1013 +...+ 101 n− 1. Note that thenth partial sum ends with a power ofn−1 in the denominator
because 1 is the first term of the infinite series.
It is rather difficult to find limn→+∞sn=limn→+∞ 1 + 101 + 1012 + 1013 +...+ 101 n− 1 as it is written. We will “work” the
sum into a different form so that we can find the limit of the sequence of partial sums.
First, multiply both sides of the equationsn= 1 + 101 + 1012 + 1013 +...+ 101 n− 1 by 101 :
1
10 sn=1
10
(
1 + 101 + 1012 + 1013 +...+ 101 n− 1)
1
10 sn=1
10 +
1
102 +
1
103 +
1
104 +...+
1
10 nNow we have two equations:
sn= 1 + 101 + 1012 + 1013 +...+ 101 n− 1
1
10 sn=1
10 +
1
102 +
1
103 +
1
104 +...+
1
10 nSubtract the bottom equation from the top equation to cancel terms and simplifying:
sn= 1 + 101 + 1012 + 1013 + 1014 +...+ 101 n− 1
− 101 sn=−( 1
10 +
1
102 +
1
103 +
1
104 +...+
1
10 n)
sn− 101 sn= 1 − 101 n
9
10 sn=^1 −1
10 nSolve forsnby multiplying both sides of the last equation by^109 :