346 CHAPTER 6 Inferential Statistics
thatE must be infinite.
(b) Here we’ll give a nuts and bolts direct approach.^10 Note first
that the expectationE is given byE=
∑∞
k=0(2k+ 1)C(k)
22 k+1, whereC(k) =1
k+ 1Ñ
2 k
ké
, k= 0, 1 , 2 ,....(i) Show thatC(k) =
2 · 6 · 10 ···(4k−2)
(k+ 1)!, k≥1 (This is a
simple induction).^11(ii) Conclude thatC(k) =1
k+ 1∏k
m=1(
4 −2
m)
.(iii) Conclude thatC(k)2−(2k−1)=2
k+ 1∏k
m=1(
1 −1
2 m)
.(iv) Using the fact that lnx > x−1, show that ln(
1 − 21 m)
>
−m^1 , m= 1, 2 ,...
(v) Conclude that∏k
l=1(
1 −1
2 l)
= eln
∏k
l=1(
1 − 21 l)= e∑k
l=1
ln( 1 − 21 l)> e−
∑k
l=1(^1) l
e−(1+lnn)=
1
ne(see Exercise 5 on page 269)(vi) Finish the proof thatE=∞by showing that the series
for E is asymptotically a multiple of the divergent har-
monic series.- Here are two more simple problems where Catalan numbers ap-
pear.
(^10) I am indebted to my lifelong friend and colleague Robert Burckel for fleshing out most of the
details. 11
This still makes sense ifk= 0 for then the numerator is the “empty product,” and hence is 1.