Problem 11954
(American Mathematical Monthly, Vol.124, January 2017) Proposed by P. Bracken (USA).
Determine the largest constantc and the smallest constant d such that, for all positive integers n, 1
n − c≤
∞
X
k=n
1 k2 ≤ 1
n − d.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. For x > 0, let
Ψ1(x) =
∞
X
k=0
1 (k + x)2 the so-called trigamma function. We show that the function
F (x) := x − 1 Ψ1(x)
is strictly increasing in (0, +∞). Then, the required constants are easy to be determined:
c = inf
n≥1
n − 1 Ψ1(n)
= 1 − 1
Ψ1(1) = 1 − 6 π2, and
d = sup
n≥1
n − 1 Ψ1(n)
= lim
n→+∞
n − 1 Ψ1(n)
= 1 2 where in the last step we used the fact that
Ψ1(n) =
∞
X
k=n
1 k2 = 1
n + 1
2n2 + o(1/n2) which follows from the Euler-Maclaurin formula.
Now, by differentiating F we obtain F′(x) = 1 +Ψ′1(x)
Ψ21(x) =Ψ21(x) + Ψ′1(x)
Ψ21(x) where Ψ′1(x) = −2
∞
X
k=0
1 (k + x)3. Hence, it suffices to show that for x > 0, G(x) := Ψ21(x) + Ψ′1(x) > 0.
Since limx→+∞G(x) = 0, it remains to prove that for x > 0, G(x + 1) − G(x) < 0, that is Ψ21(x + 1) + Ψ′1(x + 1) − Ψ21(x) − Ψ′1(x) =
Ψ1(x) − 1 x2
2 +
Ψ′1(x) + 2 x3
−Ψ21(x) − Ψ′1(x)
= −2Ψ1(x) x2 + 1
x4 + 2 x3 = − 2
x2
Ψ1(x) − 1 x− 1
2x2
< 0 which holds if and only if H(x) := Ψ1(x) −x1−2x12 > 0.
Since limx→+∞H(x) = 0, we still have to show that for x > 0, H(x + 1) − H(x) < 0, that is Ψ1(x + 1) − 1
x + 1− 1
2(x + 1)2−Ψ1(x) + 1 x+ 1
2x2 =
Ψ1(x) − 1 x2
− 2x + 3
2(x + 1)2 −Ψ1(x) +2x + 1 2x2
= − 1
2x2(x + 1)2 < 0
which is true.