CORRIGENDUM:
ON A LIMIT PROBLEM ASSOCIATED WITH THE
ARITHMETIC-GEOMETRIC MEAN INEQUALITY
[Journal London Math. Soc, 42 (1967), 712-718.]
W. N. EVERITT
Professor P. H. Diananda, University of Singapore, in writing a review of this
paper for Mathematical Reviews [36, (1969), 528] has noticed an error which
appeared in one part of the proof of Theorem 1. The proof given of Case 3 in §6
is correct for the " sufficiency " half of Theorem 1 but is not correct for the " necessity "
half. In lines 8 and 9 on page 716 the terms
(
1
N
should be replaced respectively by
1
(•s&'S
N
This implies that (6.6) on that page should be replaced by
\
I N
N
\
IN
/
\p = l
N(AN-GN) = - £ <x/ + O £ |ap|3 + 0 £
2<Xp = l
v
\p = l
P
and the line displayed 7 lines below (6.6) is not valid.
It is clear that the revised form of (6.6) still proves that the derived sequence is
00
bounded when £ a 2 < oo. However, the revised form of the line displayed 7
lines below (6.6) will not now suffice to prove the result in the opposite direction.
Fortunately, the results stated in Theorem 1 remain true. Professor P. H.
Diananda has shown that this is the case and his proof is given below.
THEOREM
(P. H. Diananda).
With the notation of Theorem 1
lim N(AN — GN) = oo
if (i) ap > 0 for all p ^ 1 and (ii) there is a real number a such that 0 < a < oo and
oo
lim ap = a and £ (ap—a) = oo.
p-»oo
p=l
Proof. Define TN> P for N ^ 1 and p ^ 1 by
x
NtP = aP-AN;
(1)
then a calculation shows that (recall that ap = ap — a)
N
N
£ xN = 0 and
P»Z = i
'
2 2
£ ao22=N(A
= N(A
++
N-ot)
N-oc)
p=i
N
2
£ xx2NtP
<p..
p=i
Received 5 July 1968.
[J. LONDON M A T H . SOC. (2), 1 (1969), 428-430]
(2)
CORRIGENDUM: ON A LIMIT PROBLEM
429
From (2) it follows that
oo
N
X oD2 = oo if and only if lim £ *$ . = oo.
p=l
(3)
JY->oop = l
JV
JV
For clearly J ffp2> £ rjv>p for all JV > 1 which gives the result in one direction.
P =i
00
P =i
Suppose £ <7p = oo; then given K > 0 there is an integer P = P(K) such that
p=1
P
E <r p 2 >K. IfnowJV^ P
P=
i
£ T^,P ^ f T2N>P = f ((rp + a - ^ N ) 2 -> £ ff,2 as iV ^ oo
p=i
P=I
p=i
P=I
AT
since ^ -> a as JV -> oo. Thus lim inf E TN,P ^ -^
completes the proof of (3).
-»» p -i
With the hypothesis of the Theorem we now have
an
d since X is arbitrary this
lim £ < p = o o .
N>
(4)
=l
From the definition (1) of xN
p
1/JV
Now rN p = Gp + (ct — AN) and ap -> 0 as p -*• oo, and a — A^ -> 0 as JV -*• oo; in
particular given s > 0 there is an integer P = P(e) such that \ap\ < e for all p ^ P.
Hence, if JV > P
p
i.e.
N
P=I
N
p=p+i
'
'
p=i
and so, in view of (4),
E T L
=1
i.e.
[Y
J
\?N P\ I <o(l) + e + o(l) asJV->-oo
•
\P = I
/
/
N
N
\
3
Z l^jv P | = o S tl
P=I
•
\p=i
as JV -> oo.
P
'
(6)
/
Then from the first result of (2) and since AN2 -* a2 > 0 as JV -> oo the equation
(5) gives, as JV -> oo,
The second result of (2) gives
430
CORRIGENDUM: ON A LIMIT PROBLEM
since ap -* 0 as p -» oo. Thus on expanding the exponential in (7)
asJV >0
" °
2a
ftS.," a s * . * ,
p=i
The result stated in the theorem now follows on use of (4) above.
A similar proof is required in establishing the result given in Theorem 2.
I am grateful to Professor P. H. Diananda for his correction.
Minor
misprints
In (6.5) replace "p -> oo " by " N -* oo ".
In (*) of " Note added in proof" replace £ by II.