Problem 11932
(American Mathematical Monthly, Vol.123, October 2016) Proposed by H. Ohtsuka (Japan).
Letr be an integer. Prove that
∞
X
n=−∞
arctan sinh r cosh n
= πr.
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 arctan sinh r
cosh n
= arctan er−e−r en+ e−n
= arctan e−(n−r)−e−(n+r) 1 + e−2n
= arctan x − y 1 + xy
= arctan(x) − arctan(y)
= arctan(e−(n−r)) − arctan(e−(n+r)) because xy = e−2n> −1.
Without loss of generality, we may assume that r ≥ 0 (otherwise we use sinh(r) = − sinh(−r)).
Hence, since cosh(n) = cosh(−n), we have that
∞
X
n=−∞
arctan sinh r cosh n
= 2
∞
X
n=1
arctan sinh r cosh n
+ arctan(sinh r)
= 2
∞
X
n=1
harctan(e−(n−r)) − arctan(e−(n+r))i
+ arctan(er) − arctan(e−r).
NowP∞
n=1arctan(e−(n±r)) is convergent and we can split the series,
∞
X
n=−∞
arctan sinh r cosh n
= 2
∞
X
n=1
arctan(e−(n−r)) − 2
∞
X
n=1
arctan(e−(n+r)) + arctan(er) − arctan(e−r)
= 2 X
m≥1−r
arctan(e−m) − 2
∞
X
m≥r+1
arctan(e−m) + arctan(er) − arctan(e−r)
= 2 X
1−r≤m≤r
arctan(e−m) + arctan(er) − arctan(e−r)
= 2 X
−r≤m≤r
arctan(e−m) − arctan(er) − arctan(e−r)
= 2 X
1≤m≤r
arctan(em) + arctan(e−m) + 2 arctan(1) − arctan(er) − arctan(e−r)
= 2 X
1≤m≤r
π 2 + 2 ·π
4 −π 2 = πr
where, at the end, we used the identity arctan(x) + arctan(1/x) = π/2 for x > 0.
