Lecture Notes in Differential Equations - Bruce E. Shapiro

265
By the triangle inequality,
|(k + 2)(k + 1)c k+2 | ≤
k∑
[(j + 1) |c j+1 | |p k−j | + |c j | |q k−j |] (28.84)
j=0
Choose any r such that 0 < r < R, where R is the radius of convergence
of (28.76). Then since the two series for p and q converge there is some
number M such that
|p j | r j ≤ M, |q j | r j ≤ M (28.85)
Otherwise there would be a point within the radius of convergence at which
the two series would diverge. Hence
|(k + 2)(k + 1)c k+2 | ≤ M r k ∑ k
j=0 rj [(j + 1) |c j+1 | + |c j |] (28.86)
where in the second line we are merely adding a positive number to the
right hand side of the inequality. Define C 0 = |c 0 | , C 1 = |c 1 |, and define
C 3 , C 4 , ... by
(k + 2)(k + 1)C k+2 = M k∑
r k r j [(j + 1)C j+1 + C j ] + MC k+1 r (28.87)
j=0
Then since |(k + 2)(k + 1)c k+2 | ≤ (k+2)(k+1)C k+2 we know that |c k | ≤ C k
for all k. Thus if the series ∑ ∞
k=0 C kt k converges then the series ∑ ∞
k=0 c kt k
must also converge by the comparison test. We will do this by the ratio
test. From (28.87)
k(k + 1)C k+1 =
M
k−1
∑
r k−1 r j [(j + 1)C j+1 + C j ] + MC k r (28.88)
j=0
k(k − 1)C k =
M
k−2
∑
r k−2 r j [(j + 1)C j+1 + C j ] + MC k−1 r (28.89)
j=0
Multiplying (28.88) by r, and writing the last term of the sum explicitly,
k(k + 1)C k+1 r =
M ∑ k−1
r k−2 j=0 rj [(j + 1)C j+1 + C j ] + MC k r 2
= M ∑ k−2
r k−2 j=0 rj [(j + 1)C j+1 + C j ] + Mr [kC k + C k−1 ] + MC k r 2
(28.90)