7 Note that both bounds have sign opposite to bn ; moreover, by the bound an = Ω(n(k−1)/k ), both bounds have absolutely value strictly less than that of bn for n sufficiently large. Consequently, for n large,
But turning this around, the fact that an+1 − an
= a−1/k n −1/(k+1) k+1 ≤ n−1/(k+1) (1 + o(1)), k
|e1 | ≤ |bn |. We now work on e2 . By Taylor’s theorem with remainder applied to (1 + x)m for x > 0 and 0 < m < 1,
where o(1) denotes a function tending to 0 as n → ∞, yields
1 + mx ≥ (1 + x)m ≥ 1 + mx +
an −1/(k+1) X n k+1 i−1/(k+1) (1 + o(1)) k i=1 −1/(k+1) k+1 k+1 nk/(k+1) (1 + o(1)) = k k k/(k+1) k+1 = nk/(k+1) (1 + o(1)), k ≤
so lim sup n→∞
an nk/(k+1)
≤
k+1 k
m(m − 1) 2 x . 2
The “main term” of L((n + 1)k/(k+1) − nk/(k+1) ) k n−1/(k+1) . To make this coincide with is L k+1 −1/k −1/(k+1) L n , we take L=
k+1 k
k/(k+1)
.
We then find that |e2 | = O(n−2 ),
k/(k+1)
and this completes the proof. Third solution: We argue that an → ∞ as in the first solution. Write bn = an − Lnk/(k+1) , for a value of L to be determined later. We have bn+1
and because bn+1 = e1 + e2 , we have |bn+1 | ≤ |bn | + |e2 |. Hence ! n X −2 i = O(1), |bn | = O i=1
and so a−1/k n
= bn + = e1 + e2 ,
k/(k+1)
− L((n + 1)
−n
k/(k+1)
)
where − L−1/k n−1/(k+1) e1 = bn + a−1/k n
e2 = L((n + 1)k/(k+1) − nk/(k+1) ) − L−1/k n−1/(k+1) .
We first estimate e1 . For −1 < m < 0, by the convexity of (1 + x)m and (1 + x)1−m , we have 1 + mx ≤ (1 + x)m
≤ 1 + mx(1 + x)m−1 .
ak+1 n = Lk+1 = n→∞ nk lim
k+1 k
k
.
Remark: The case k = 2 appeared on the 2004 Romanian Olympiad (district level). Remark: One can make a similar argument for any sequence given by an+1 = an + f (an ), when f is a decreasing function. Remark: Richard Stanley suggests a heuristic for determining the asymptotic behavior of sequences of this type: replace the given recursion an+1 − an = a−1/k n by the differential equation
Hence 1 − L−(k+1)/k n−1 bn ≤ e1 − bn k 1 ≤ − bn an−(k+1)/k . k
y ′ = y −1/k and determine the asymptotics of the latter.
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