Find ∞ X n=1 sinh−1  1 √2n+2+ 2

Problem 11930

(American Mathematical Monthly, Vol.123, October 2016) Proposed by C. I. V˘alean (Romania).

Find

∞

X

n=1

sinh⁻¹

 1

√2ⁿ⁺²+ 2 +√

2ⁿ⁺¹+ 2

 .

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Solution. Let

aⁿ= 1

√2ⁿ⁺²+ 2 +√

2ⁿ⁺¹+ 2 and bⁿ=

√2ⁿ+ 1 −√ 3 2ⁿ⁺¹² . Then, it can be easily verified that

bn+1p1 + b²n− bⁿ q

1 + b²n+1= aⁿ. By the known identity,

sinh⁻¹(xp1 + y²− y

p1 + x²) = sinh⁻¹(x) − sinh⁻¹(y), it follows that

N

X

n=1

sinh⁻¹(an) =

N

X

n=1

sinh⁻¹(bn+1) − sinh⁻¹(bn)
= sinh⁻¹(bN +1) − sinh⁻¹(b1).

Now b1= 0, bN +1→ 1/√

2, and therefore

∞

X

n=1

sinh⁻¹

 1

√2ⁿ⁺²+ 2 +√

2ⁿ⁺¹+ 2



= lim

N →∞sinh⁻¹(bN +1) = sinh⁻¹

 1

√2



=ln(2 +√ 3)

2 .