Answer:
Step 1. If
I1(n)=∑1≤k≤n√(Γ(kn))−k
Then limn→∞I1(n)=0.
Proof. Indeed, since Γ is decreasing on (0,1] we have
I1(n)≤∑1≤k≤n√(Γ(1n−−√))−k≤∑k=1∞(Γ(1n−−√))−k=1Γ(1/n−−√)−1
and step 1. follows.
Step 2. If
I2(n)=∑n√<k≤n/2(Γ(kn))−k
Then limn→∞I2(n)=0.
Proof. Recall that Γ attains its minimum ≈0.8856, on [1,2], at some some point x0≈1.4616. In particular, Γ(x)≥2/3 for 1≤x≤2. So, for n−−√<k≤n/2 we have
knΓ(kn)=Γ(1+kn)≥23
Thus, for n−−√<k≤n/2, we have Γ(k/n)>4/3. It follows that
I2(n)≤∑k>n√(34)k=4(34)⌈n√⌉
and step 2. follows.
Step 3. If
I3(n)=∑n/2<k≤n(Γ(kn))−k
Then limn→∞I3(n)=eγeγ−1. where γ is the Euler-Mascheroni constant.
Proof. Note first that, with p=n−k,
I3(n)=∑0≤p<n/2(Γ(1−pn))p−n=∑p=0∞ap(n)
with
ap(n)={(Γ(1−pn))p−n0ifotherwise0≤p<n/2
Now, since Γ(1)=1 and Γ′(1)=−γ we have, for a fixed p and large n:
(p−n)lnΓ(1−pn)=(p−n)ln(1+γpn+O(1n2))=−γp+O(1n)
Thus
∀p≥0,limn→∞ap(n)=e−γp.(1)
desired conclusion follows:
limn→∞∑1≤k≤n(Γ(kn))−k=limn→∞(I1(n)+I2(n)+I3(n))=eγeγ−1.
Step-by-step explanation:
