Proof. Define bn := logan . Then bn+1 is the arithmetic mean of bn , · · · , bn−k+1 . We want
to check whether limit of bn exists. So, it is enough to show that bn is a Cauchy sequence.
Assume that b1 , · · · , bk is contained for some interval of length x. Then it is obvious that
bi is contained in that interval for all i. Also, for all i ≥ k + 1,
bi+1 − bi =
bi + · · · + bi−k+1
bi−1 + · · · + bi−k
bi − bi−k
−
=
k
k
k
Then for all i, j between k + 1 and 2k,
j−1 j−1
j−1
X b − b X
bl − bl−k X
x
k−1
l
l−k ≤
|bi − bj | = ≤
x
≤
k
k
k
k
l=i
l=i
l=i
Therefore, bk+1 , bk+2 , · · · , b2k is contained in some interval of length k−1
k x, and so is ai for
all i ≥ 2k + 1. In the same way, we can show that for any n, there is an interval of length
n
( k−1
k ) x which contains bi ‘s for all i > nk. Thus bi is a Cauchy sequence and its limit exists.
Then limit of an = ebn exists.
From definition, we can observe that
bn+1 + 2bn+2 + · · · + kbn+k = bn+1 + 2bn+2 + · · · + (k − 1)bn+k−1 + (bn + · · · + bn+k−1 )
= bn + 2bn+1 + · · · + kbn+k−1
= ······
= b1 + 2b2 + · · · + kbk
Therefore, by taking the limit for each side,
b = lim bn =
n→∞
2
(b1 + · · · + kbk )
k(k + 1)
So we obtain
2
a = lim an = elimn→∞ bn = e k(k+1) (b1 +···+kbk ) =
n→∞
1
q
k(k+1)
2
a1 a22 · · · akk