Proofs of Divergence of the Harmonic Series - Prairie State College

elsewhere. For example,
⎡
⎤
0 0 0 0 0 0 ···
⎢
⎥
⎢
0 0 0 0 0 0 ··· ⎥
⎢
⎥
⎢ 0 0 0 1 0 0 ··· ⎥
J3 = ⎢
⎥
⎢ 0 0 0 0 1 0 ··· ⎥.
⎥
⎢
⎥
⎢ 0 0 0 0 0 1 ··· ⎥
⎣
⎦
.
. . . . . . ..
It follows that M = J1+J1+J2+J3+J4+··· (in the sense that, for any n, the equality holds for the n×n
leading, principal submatrices).
By expanding h T J1h, we find that
h T J1h =
∞
n=1
From here it is easy to show (perhaps by induction) that
1
= 1.
n(n+1)
h T Jkh = 1
k .
Now suppose that the harmonic series converges with sum S. Then we have
S = h T Mh = h T (J1 +J1 +J2 +J3 +J4 +···)h
= h T J1h+h T J1h+h T J2h+h T J3h+h T J4h+···
= 1+1+ 1 1 1
2 + 3 + 4 +···
= 1+S.
The conclusion, S = 1+S, is impossible unless S = ∞.
Proof 39 (Cesàro summability)
It is not difficult to prove the following result (see for example [10, pages 128–129]), which was studied in
detail by the Italian mathematician Ernesto Cesàro.
Suppose that ∞ k=1ak converges with sum S and let Sn = n k=1ak. Then the sequence of
average partial sums, { 1 n n k=1Sk} ∞ n=1, also converges to S.
In order to use this result, let
and notice that
1
n
k=1
Hn = 1+ 1 1 1 1
+ + +···+
2 3 4 n ,
n
Hk = 1
n n−1
+
n 1 2
= 1
n
n
k=1
n−k +1
k
(A Proof Without Words for this last fact is given in [17].)
14
+ n−2
3
1
+···+
n
= Hn −1+ Hn
n
.