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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 +Ψ1(x)

Ψ21(x) =Ψ21(x) + Ψ1(x)

Ψ21(x) where Ψ1(x) = −2

X

k=0

1 (k + x)3. Hence, it suffices to show that for x > 0, G(x) := Ψ21(x) + Ψ1(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) + Ψ1(x + 1) − Ψ21(x) − Ψ1(x) =



Ψ1(x) − 1 x2

2 +



Ψ1(x) + 2 x3



−Ψ21(x) − Ψ1(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) −x12x12 > 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. 

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