Problem 11808
(American Mathematical Monthly, Vol.121, December 2014)
Proposed by D.M. Batinetu-Giurgiu (Romania) and N. Stanciu (Romania).
Compute
n→∞lim n2
Z ((n)!)−1/n
((n+1)!)−1/(n+1)
Γ(nx) dx.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
We will show that if f is a continuous real function in (a, b) and e ∈ (a, b) then
n→∞lim n2
Z ((n)!)−1/n
((n+1)!)−1/(n+1)
f (nx) dx = ef (e).
In our case Γ is a continuous real function in (0, +∞) and therefore the required limit is eΓ(e).
Let bn= n(n!)−1/n and an = n((n + 1)!)−1/(n+1), then by the Mean Value Theorem for integrals,
n2
Z ((n)!)−1/n
((n+1)!)−1/(n+1)
f (nx) dx = n Z bn
an
f (t) dt = n(bn− an)f (tn)
for some tn ∈ (an, bn). Now, by the Stirling approximation formula,
ln(n!) = n ln(n) − n +1
2ln(n) + ln(√
2π) + O(1/n).
Hence
bn= n exp(− ln(n!)/n) = e − e ln(n)
2n −e ln(√ 2π)
n + O(ln2(n)/n2), bn− an = bn−nbn+1
n + 1 = e
n+ O(ln(n)/n2), which imply that
n→∞lim bn= lim
n→∞an= lim
n→∞tn= e and, by the continuity of f at e,
n→∞lim n(bn− an)f (tn) = ef (e).
