Problem 11941
(American Mathematical Monthly, Vol.123, November 2016) Proposed by O. Furdui (Romania).
Let
L= lim
n→∞
Z 1 0
pn
xn+ (1 − x)ndx.
(a) Find L.
(b) Find
n→∞lim n2
Z 1 0
pn
xn+ (1 − x)ndx − L
.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution.
(a) Let t = 1−xx , then x = 1+tt , dx = (1+t)dt 2, and Z 1
0
pn
xn+ (1 − x)ndx= 2 Z 1/2
0 (1 − x)n s
1 +
x 1 − x
n
dx= 2 Z 1
0
√n
1 + tn (1 + t)3 dt.
Since for all t ∈ [0, 1],
√n
1 + tn
(1 + t)3 → 1
(1 + t)3 and 0 ≤
√n
1 + tn
(1 + t)3 ≤ 2 (1 + t)3, by the Lebesgue’s Dominated Convergence Theorem,
L= lim
n→∞
Z 1 0
pn
xn+ (1 − x)ndx= 2 Z 1
0
1
(1 + t)3 dt=
− 1
(1 + t)2
1 0
=3 4. (b) By letting s = tn, we obtain t = s1/n, dt = n1s1/n−1ds, and
n2
Z 1 0
pn
xn+ (1 − x)ndx − L
= 2n2 Z 1
0
√n
1 + tn− 1 (1 + t)3 dt
= 2n Z 1
0
(1 + s)1/n− 1
(1 + s1/n)3 s1/n−1ds
= 2n Z 1
0
exp(n1ln(1 + s)) − exp(0)
(1 + s1/n)3 s1/n−1ds
= 2 Z 1
0
exp(hn(s)) ln(1 + s)
(1 + s1/n)3 s1/n−1ds
where 0 < hn(s) < n1ln(1 + s) is given by the Mean Value Theorem. Now for all s ∈ (0, 1], exp(hn(s)) ln(1 + s)
(1 + s1/n)3 s1/n−1→ ln(1 + s)
8s and 0 ≤ exp(hn(s)) ln(1 + s)
(1 + s1/n)3 s1/n−1≤ 2 ln(1 + s)
s .
Hence, by the Lebesgue’s Dominated Convergence Theorem,
n→∞lim n2
Z 1 0
pn
xn+ (1 − x)ndx − L
=1 4
Z 1 0
ln(1 + s) s ds= 1
4 Z 1
0
∞
X
k=1
(−1)k−1sk−1 k
=1 4
∞
X
k=1
(−1)k−1 k2 =π2
48.
