158 CHAPTER 8. POWER SERIES
Using the binomial theorem,
f (x+h)− f (x)h
=1h
∞
∑k=0
ak
((x+h−a)k− (x−a)k
)=
1h
∞
∑k=1
ak
(k
∑j=0
(kj
)(x−a) j hk− j− (x−a)k
)
=∞
∑k=1
ak
(k−1
∑j=0
(kj
)(x−a) j h(k−1)− j
)
Then ∣∣∣∣∣ f (x+h)− f (x)h
−∞
∑k=1
akk (x−a)k−1
∣∣∣∣∣=
∣∣∣∣∣ ∞
∑k=1
ak
(k−1
∑j=0
(kj
)(x−a) j h(k−1)− j− k (x−a)k−1
)∣∣∣∣∣=
∣∣∣∣∣ ∞
∑k=2
ak
(k−1
∑j=0
(kj
)(x−a) j h(k−1)− j− k (x−a)k−1
)∣∣∣∣∣=
∣∣∣∣∣ ∞
∑k=2
ak
(k−2
∑j=0
(kj
)(x−a) j h(k−1)− j
)∣∣∣∣∣Therefore,∣∣∣∣∣ f (x+h)− f (x)
h−
∞
∑k=1
akk (x−a)k−1
∣∣∣∣∣≤ ∞
∑k=2|ak|(
k−2
∑j=0
(kj
)|x−a| j |h|(k−1)− j
)
Now it is clear that k (k−1)(k−2
j
)≥(k
j
)and so
= |h|∞
∑k=2|ak|(
k−2
∑j=0
(kj
)|x−a| j |h|(k−2)− j
)
≤ |h|∞
∑k=2|ak|k (k−1)
k−2
∑j=0
(k−2
j
)|x−a| j |h|(k−2)− j
= |h|∞
∑k=2|ak|k (k−1)(|x−a|+ |h|)k−2 < |h|
∞
∑k=2|ak|k (k−1)rk−2 (8.4)
By assumption and what was just observed about limk→∞ k1/k,
lim supk→∞
(|ak|k (k−1)rk−2
)1/k< 1
and so the series on the right in 8.4 converges. Therefore, assuming |h| is small enough,∣∣∣∣∣ f (x+h)− f (x)h
−∞
∑k=1
akk (x−a)k−1
∣∣∣∣∣<C |h|