For p>=2
I need to calculate a limit
With[{p = 2}, Limit[E^(ProductLog[n p]/p)*(E^(-n^(-1 + 1/p) p^(-2 + 1/p) ProductLog[n p]^(1 - 1/p)) + 1/p - 1) + E^ProductLog[n p]/ p*(E^(-n^(-1 + 1/p) p^(-1 + 1/p) ProductLog[n p]^(1 - 1/p)) - 1), n -> Infinity]]
The limit is returned unevaluated.
Unfortunately, my efforts so far have stalled on finding a bug in Mathematica (versions 12 and 13), see Why do these identical limits give different results?
Then I found out that versions 10 and 11 don't have this bug. So I continued with them. However, I can't get a result.
My conjecture is that the limit is equal to -1/4
for p=2
and is equal to zero for p>=3
.
With[{p = 2}, Plot[E^(ProductLog[n p]/p)*(E^(-n^(-1 + 1/p) p^(-2 + 1/p) ProductLog[n p]^(1 - 1/p)) + 1/p - 1) + E^ProductLog[n p]/p*(E^(-n^(-1 + 1/p) p^(-1 + 1/p) ProductLog[n p]^(1 - 1/p)) - 1), {n, 1, 100000000000}]]
With[{p = 3}, Plot[E^(ProductLog[n p]/p)*(E^(-n^(-1 + 1/p) p^(-2 + 1/p) ProductLog[n p]^(1 - 1/p)) + 1/p - 1) + E^ProductLog[n p]/p*(E^(-n^(-1 + 1/p) p^(-1 + 1/p) ProductLog[n p]^(1 - 1/p)) - 1), {n, 1, 100000000000}]]
My question is: How to calculate these limits with Mathematica ?
Versions 10 and 11 are able to calculate a similar limit
Limit[E^(E^ProductLog[n] (-1 + E^(Sqrt[ProductLog[n]]/(2 Sqrt[n]))))*(E^(-(1/2) E^(ProductLog[n]/2))) , n -> Infinity]