Problem 11771
(American Mathematical Monthly, Vol.121, April 2014)
Proposed by D. M. Batinetu-Giurgiu (Romania) and Neculai Stanciu (Romania).
Find
n→∞lim pn
(2n − 1)!! tan πn+1p(n + 1)!
4√n n!
!
− 1
! .
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
By Stirling’s approximation formula
ln(n!) = n ln(n) − n +ln(2πn)
2 + O(1/n), we have that
xn:= e√n
n! = exp(ln(n!)/n + 1) = n +ln(2πn)
2 + O(1/n) and
n→∞lim(xn+1− xn) = 1 and lim
n→∞
xan xn
= a for any positive integer a. Now
pn
(2n − 1)!! = n r(2n)!
2nn! = x22n 2xn
and lim
x→1
tan(π4x) − 1 x − 1 = π
2. Therefore
pn
(2n − 1)!! tan πn+1p(n + 1)!
4√n n!
!
− 1
!
= x22n 2exn
tan(π4xn+1x
n ) − 1
xn+1 xn − 1
xn+1 xn
− 1
= 1 2e
x2n xn
2tan(π4xn+1x
n ) − 1
xn+1
xn − 1 (xn+1− xn) → π e.
