A26 MAT Lecture 44
1
Then
Root Test for Convergence
(The Root Test) Suppose that lim |ak |1/k = r exists.
k k=1 k=1
proposition 1.1. r<1 r>1 r=1
ak converges absolutely ak diverges
nothing at all (test is inconclusive in this case)
proof. We the proof of the case |r| < 1 only. Since |r| < 1 we may choose s (r, 1). Note that 0 < r < 1 and then 0 < |ak |1/k < s for k N (for some N ). Thus |ak | < sk for k N and so
|ak | <
k=N k=N k=N
sk =
sk 1-s
So by the comparison test, k=1 ak .
ak converges absolutely, and hence so does
example 1.2. Does the root test apply to the series
k=1
xk k!
1/k
solution. We have |ak | The limit r = lim
1/k
xk = k!
=
|x| (k!)1/k
|x| can indeed be found as follows (Note: Our (k!)1/k k method here is by brute force; using the the ratio test is easier.) Assuming 1
k = 2 is even we have (k!)1/k = (1 2 3 ( + 1) k)1/k k = 2 factors (( + 1) k)1/k factors
/k
= ( 1 k) 2 2 On the other hand if k = 2 + 1 is odd, we have (k!)1/k = (1 2 3 ( + 1) k)1/k k = 2 + 1 factors (( + 1) k)1/k + 1 factors +1 ( + 1) k ( 1 k) 2 (since + 1 = 2
1
1
1
k+1 2
> 1k 2
1 +1 > ) k 2
1 In either case, (k!)1/k ( 2 k) 2 as k . Thus r = 0 and we have convergence.
example 1.3. 1 . 2k+(-1)k k=1
See whether or not the the root test applies to the series
solution. We have |ak |
1/k
=
1 2k+(-1)k
1/k
=
1 2(k+(-1)k )/k
=
1 2
1 1+ k (-1)k
1 as k 2
Since lim |ak |1/k =
k
1 < 1 the series converges. 2 2