=+13. Continuing Problems 11 and 12, suppose a constant a ≥ 0 and positive integer p exist such that

|ck| ≤ a

|k|

p+1 for all k = 0. Integration by parts shows that this criterion holds if f(p+1)(x) is piecewise continuous. Verify the inequality

|

ˆbk − ck| ≤ a np+1 ∞

j=1 

1



j + k n

p+1 +

1



j − k n

p+1 

when |k| < n/2. To simplify this inequality, demonstrate that

∞

j=1 1

(j + α)p+1 <

 ∞

1 2

(x + α)

−p−1dx

= 1 p

1 2 + α

p for α > −1/2. Finally, conclude that

|

ˆbk − ck| <

a pnp+1 

1

1 2 + k n

p +

1

1 2 − k n

p



.