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# 2n−1 X k=1 (−1)kHk− 2n−1 X k=1 (−1)k− 2n−1 X k=1 (−1)kHk k = H2n−Hn 2 − 1 − 2n−1 X k=1 (−1)kHk k

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Problem 12026

(American Mathematical Monthly, Vol.125, February 2018) Proposed by M. Bataille (France).

Forn ∈ N, let Hn=Pn

k=11/k and Sn =

n

X

k=1

(−1)n−k k

k

X

j=1

Hj

Find lim

n→∞Sn/ ln(n) and lim

n→∞(S2n− S2n−1).

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

Solution. It easy to verify by induction thatPk

j=1Hj= k(Hk− 1) + Hk, hence S2n=

2n

X

k=1

(−1)kHk

2n

X

k=1

(−1)k+

2n

X

k=1

(−1)kHk

k = Hn

2 +

2n

X

k=1

(−1)kHk

k and

S2n−1= −

2n−1

X

k=1

(−1)kHk

2n−1

X

k=1

(−1)k

2n−1

X

k=1

(−1)kHk

k = H2n−Hn

2 − 1 −

2n−1

X

k=1

(−1)kHk

k . Therefore

S2n− S2n−1= Hn+ 1 − H2n−H2n

2n + 2

2n

X

k=1

(−1)kHk

k . Now, as N goes to infinity,

N

X

k=1

(−1)kHk

k =

N

X

k=1

(−1)k k2 +

N

X

k=2

(−1)kHk−1

k

=

N

X

k=1

(−1)k k2 +1

2

N

X

k=2 k−1

X

j=1

(−1)k k

 1 j + 1

k − j



=

N

X

k=1

(−1)k k2 +1

2

N −1

X

j=1 N

X

k=j+1

(−1)k j(k − j)

=

N

X

k=1

(−1)k k2 +1

2

N −1

X

j=1

(−1)j j

N −j

X

k=1

(−1)k k

=

N

X

k=1

(−1)k k2 +1

2

N −1

X

j=1

(−1)j j

2

−1 2

N −1

X

j=1

(−1)j j

N −1

X

k=N −j+1

(−1)k

k → −π2

12+ln2(2) 2

where

N −1

X

j=1

(−1)j j

N −1

X

k=N −j+1

(−1)k k

≤ 1

√N → 0.

Hence, since Hn = ln(n) + γ + o(1),

n→∞lim(S2n−S2n−1) = ln(n)+1−ln(2n)−π2

6 +ln2(2)+o(1) → ln2(2)−ln(2)+1−π2

6 ≈ −0.85760324596 and

n→∞lim Sn

ln(n) = lim

n→∞

Hn

2 ln(n) =1 2.



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