Problem 11384
(American Mathematical Monthly, Vol.115, October 2008) Proposed by M. Omarjee (France).
Let pn denote the nth prime. Show that
∞
X
n=1
(−1)⌊√n⌋
pn
converges.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Since pn= n log n + O(n log(log n)), then for M, N sufficiently large there is a positive constant C such that
M
X
n=N
(−1)⌊√n⌋
pn −
M
X
n=N
(−1)⌊√n⌋
nlog n
≤
M
X
n=N
1
pn − 1 nlog n
≤ C
M
X
n=N
log(log n) n(log n)2 which tends to zero as M, N goes to infinity because the series
∞
X
n=2
log(log n) n(log n)2 converges. Hence it suffices to prove that the series
X∞ n=2
(−1)⌊√n⌋
nlog n converges. For any M > N > 2
M
X
n=N
(−1)⌊√n⌋
nlog n =
M
X
n=N
An− An−1
nlog n =
M
X
n=N
An
nlog n−
M
X
n=N
An−1 nlog n
=
M
X
n=N
An
nlog n−
M−1
X
n=N −1
An
(n + 1) log(n + 1)
=
M
X
n=N
An
1
nlog n− 1
(n + 1) log(n + 1)
+ AM
Mlog M − AN −1 Nlog N. Let a = ⌊√
n⌋ then
An =
n
X
k=1
(−1)⌊√k⌋=
a−1
X
j=1
(j+1)2−1
X
k=j2
(−1)j+
n
X
k=a2
(−1)a
=
a−1
X
j=1
(−1)j(2j + 1) +
n
X
k=a2
(−1)a= −(1 + (−1)aa) + (−1)a(n − (a2− 1)).
Hence
|An| ≤√
n+ 2 = O(√ n).
Moreover
1
nlog n− 1
(n + 1) log(n + 1)= O
1
n2log n
.
Therefore for M, N sufficiently large there is a positive constant C such that
M
X
n=N
(−1)⌊√n⌋
nlog n
≤ C
M
X
n=N
1
n3/2log n+ 1
√Mlog M + 1
√Nlog N
!
which tends to zero as M, N goes to infinity and therefore the series converges.
