Problem 11956
(American Mathematical Monthly, Vol.124, January 2017) Proposed by H. Ohtsuka (Japan).
Show that
∞
X
n=1
arctan(sinh n) · arctan sinh 1 cosh n
converges, and find the sum.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. For non-negative real numbers a, b, we have that arctan sinh a
cosh b
= arctan ea−e−a eb+ e−b
= arctan e−(b−a)−e−(b+a) 1 + e−2b
= arctan x − y 1 + xy
= arctan(x) − arctan(y)
= arctan(e−(b−a)) − arctan(e−(b+a)).
Then
arctan sinh 1 cosh n
= arctan(e−n+1) − arctan(e−n−1), and
arctan(sinh n) = arctan sinh n cosh 0
= arctan(en) − arctan(e−n) = π
2 −2 arctan(e−n) where we used the identity arctan(x) + arctan(1/x) = π/2 for x > 0. Hence
N
X
n=1
arctan(sinh n) · arctan sinh 1 cosh n
=π
2SN −2TN. where
SN =
N
X
n=1
arctan(e−n+1) − arctan(e−n−1)
=
N −1
X
n=0
arctan(e−n) −
N+1
X
n=2
arctan(e−n)
= arctan(1) + arctan(1/e) − arctan(e−N) − arctan(e−N −1) → π
4 + arctan(1/e), and
TN =
N
X
n=1
arctan(e−n) arctan(e−n+1) − arctan(e−n−1)
=
N −1
X
n=0
arctan(e−n) arctan(e−n−1) −
N
X
n=1
arctan(e−n) arctan(e−n−1)
= arctan(1) arctan(1/e) − arctan(e−N) arctan(e−N −1) → π
4arctan(1/e).
Therefore the desired series is convergent and
∞
X
n=1
arctan(sinh n) · arctan sinh 1 cosh n
=π 2
π
4 + arctan(1/e)
−2π
4 arctan(1/e)
= π2 8 .
