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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]]

Limit

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}]]

p=2

 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}]]

p=3

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]

similar limit

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2 Answers 2

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Using my first approach from Why do these identical limits give different results?

With[{p = 2}, 
 expr = 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) // FullSimplify]

Limit[expr /. n -> 1/n, n -> 0, Direction -> "FromAbove"]

(*  -(1/4)  *)

(Did you try it on this? You left no comment. With p = 3, I get 0, too.)

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  • $\begingroup$ This is a valuable result for me that confirms my conjectures. Thank you! But this only works under version 13. Under version 10 (which I have more confidence in after yesterday) I get an error message: "Limit::ldir: Value of Direction -> FromAbove should be a number or Automatic. >>" $\endgroup$ Commented Jul 9, 2022 at 15:46
  • $\begingroup$ @VaclavKotesovec Follow the hint in the error message, and, consulting the docs, replace "FromAbove" with -1, IIRC. Then see if it works. I can no longer test such an old version as V10. (BTW, p = 4 has been running for a half hour or more with no answer yet. It probably won't finish.) $\endgroup$
    – Michael E2
    Commented Jul 9, 2022 at 16:03
  • $\begingroup$ Yes, it works now even under version 10.. $\endgroup$ Commented Jul 9, 2022 at 16:11
  • $\begingroup$ @VaclavKotesovec Good. I was afraid there was a real change in what Limit could do. $\endgroup$
    – Michael E2
    Commented Jul 9, 2022 at 16:12
  • $\begingroup$ Interesting situation: cases p=5, p=6 and p=7 (all with a result of 0) ended in 2-3 minutes each. The p=4 and p=8 cases have been running for over 13 hours now and each consumes over 60 GB of RAM (on my 384 GB server). Only integer p are interesting to me, but I can also add that the cases p=5/2, p=7/2, p=9/2 and p=11/2 also ended with a result of zero in just a few minutes. $\endgroup$ Commented Jul 10, 2022 at 6:14
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After substitution $$t=\frac{n}{W(p n)}$$ we have $$W(p n)=\frac{n}{t}$$ and $$\exp(W(p n))=\frac{p n}{W(p n)}=p t$$ where W is the LambertW function. Now the expression transforms to

 (-1 + E^(-p^(-1 + 1/p) t^(-1 + 1/p))) t + (-1 + E^(-p^(-2 + 1/p) t^(-1 + 1/p)) + 1/p) p^(1/p) t^(1/p)

If n tends to infinity, t also tends to infinity and as a result we get

 Assuming[p > 2, Limit[(-1 + E^(-p^(-1 + 1/p) t^(-1 + 1/p))) t + (-1 + E^(-p^(-2 + 1/p) t^(-1 + 1/p)) + 1/p) p^(1/p) t^(1/p), t -> Infinity]]

limits

My conjecture is proven.

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  • 2
    $\begingroup$ Math beats Mathematica! :) $\endgroup$
    – Michael E2
    Commented Jul 11, 2022 at 15:17

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