Problem 12078
(American Mathematical Monthly, Vol.125, December 2018) Proposed by H. Ohtsuka (Japan).
Let nk
q be the q-binomial coefficient defined by
n k
q
=
k−1
Y
i=0
1 − qn−i 1 − qi+1.
For a positive integer s and for0 < q < 1, prove
∞
X
n=1
qsn
s+n s+1
q
=qs(1 − qs+1) 1 − qs .
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. We have that qs(n−1)
s+n−1 s
q
− qsn
s+n s
q
= qs(n−1)
s+n s+1
q
·
1 − qs+n 1 − qs+1 −
qsn
s+n s+1
q
·
1 − qn 1 − qs+1
= 1
(1 − qs+1) s+ns+1
q
qs(n−1)(1 − qs+n) − qsn(1 − qn)
= qs(n−1)− qsn (1 − qs+1) s+ns+1
q
= qsn(1 − qs) qs(1 − qs+1) s+ns+1
q
.
Hence, for any positive integer N , the partial sum is telescopic,
N
X
n=1
qsn
s+n s+1
q
= qs(1 − qs+1) 1 − qs
N
X
n=1
qs(n−1)
s+n−1 s
q
− qsn
s+n s
q
!
= qs(1 − qs+1)
1 − qs 1 − qsN
s+N s
q
! .
It follows that
∞
X
n=1
qsn
s+n s+1
q
= lim
N →∞
N
X
n=1
qsn
s+n s+1
q
=qs(1 − qs+1) 1 − qs because for 0 < q < 1,
N →∞lim qsN
s+N s
q
= lim
N →∞qsN Qs
i=1(1 − qi) Qs
i=1(1 − qN+i)= 0.
