Converges or diverges?

Question

Studying the behavior of the series $$ \sum _{n=1}^{\infty } \left(1-\frac{\log (n)}{n}\right)^{2 n}$$, I try in 12.3.1 on Windows 10

SumConvergence[(1 - Log[n]/n)^(2 n), n]

False

, but

SumConvergence[(1 - Log[n]/n)^(2 n), n, Method -> "RaabeTest"]

True

The result of

NSum[(1 - Log[n]/n)^(2 n), {n, 1, Infinity}]

1.33193*10^244

confirms the divergence and the results of

NSum[(1 - Log[n]/n)^(2 n), {n, 1, Infinity}, WorkingPrecision -> 30]

1.407104427435176587354

and

Series[(1 - Log[n]/n)^(2 n), {n, Infinity, 3}]

(1/n)^2-Log[n]^2/n^3+O(1/n)^4

stand for the convergence.

What should I trust?

Answer 1

Do the Raabe test by hand:

b[n_] = (1 - Log[n]/n)^(2 n);

Limit[n (b[n]/b[n + 1] - 1), n -> Infinity]
(* 2 *)

converges.

Answer 2

This looks like a horrible bug present in 12.3.1 (still happens on 13.0.1) (indeed due to "SumConvergence uses pre V11.2 Limit" and it is used for DivergenceTest). First of all even if on this limit Raabe test Method does work there is worse example here: Why doesn't Mathematica provide an answer while Wolfram|Alpha does, concerning a series convergence? I suppose DivergenceTest has higher priority (remember it is only meaningful for False, for True it means nothing). Or the problem there is that Raabe is done on nonnegative series and it somehow fails to check it. So:

a[n_] := (1 - Log[n]/n)^(2 n);
SumConvergence[a[n], n, Method -> Automatic] (* False, bug: does not TRY Raabe *)
SumConvergence[a[n], n, Method -> "RaabeTest"] (* True, NICE. It is 2. *)
SumConvergence[a[n], n, Method -> "RatioTest"] (* Error *)
SumConvergence[a[n], n, Method -> "RootTest"] (* Error *)
SumConvergence[a[n], n, Method -> "DivergenceTest"] (* False, a bug! *)
SumConvergence[a[n], n, Method -> "IntegralTest"] (* Fails, infinite loop  bug!! *)

For DivergenceTest see https://mathematica.stackexchange.com/a/163389/82985 Now, they broke IntegralTest long time ago, see SumConvergence difficulty first workaround on 1- Cos[Pi/n] there does not help this case though, it errors out after some time but FreeQ idea and myLCT do work, WOW. FreeQ does not work on (1 - Log[n]/n)^(2 n) though by myLCT does. So...

Now, indeed both Raabe test by hand and next level of it (in the series of Kummer's tests), Bertrand test (strange it is not one of Methods, WTF, one can even use Extended Betrand that is continuation further of that):

b[n_] = (1 - Log[n]/n)^(2 n);

Limit[Log[n] (n (b[n]/b[n + 1] - 1) - 1), n -> Infinity] (* prints Infinity, so convergent *)

Other example is worse. It is actually quite interesting, see: https://math.stackexchange.com/questions/2830362/sum-limitsn-1-infty1-cos-frac-pin-convergence-proof?noredirect=1&lq=1 and related questions.

a[n_] := 1 - Cos[Pi/n];
SumConvergence[a[n], n, Method -> Automatic] (* error *)
SumConvergence[a[n], n, Method -> "RaabeTest"] (* error *)
SumConvergence[a[n], n, Method -> "RatioTest"] (* error *)
SumConvergence[a[n], n, Method -> "RootTest"] (* error *)
SumConvergence[a[n], n, Method -> "DivergenceTest"] (* true, so useless *)
SumConvergence[a[n], n, Method -> "IntegralTest"] (* error *)

Raabe by hand again prints 2 while Betrand's prints Infinity. So converges.

b[n_] = 1 - Cos[Pi/n];

Limit[Log[n] (n (b[n]/b[n + 1] - 1) - 1), n -> Infinity] (* Infinity *)

Now I would not say this Raabe approach is always perfect. Try:

b[n_] = Abs[Sin[n]]^n/n;

Limit[n (b[n]/b[n + 1] - 1), n -> Infinity] (* Infinite loop *)

Answer 3

SumConvergence uses pre V11.2 Limit, which evaluates to the wrong answer below. This makes SumConvergence think the summand fails the divergence test and hence returns False.

Compare:

Asymptotics`ClassicLimit[(1 - Log[n]/n)^(2 n), n -> ∞, Assumptions -> n ∈ Integers]

1

Limit[(1 - Log[n]/n)^(2 n), n -> ∞, Assumptions -> n ∈ Integers]

0

Swapping definitions fixes the issue:

Block[{Asymptotics`ClassicLimit = Limit},
  SumConvergence[u[n], n]
]

True

Answer 4

Not a real answer, but rather a recommendation to ask a mathematician in this case (e.g., at https://math.stackexchange.com/). Mathematica does not seem to provide a definite answer. In the methods you have several options

a[n_] := (1 - Log[n]/n)^(2 n)
SumConvergence[a[n], n, Method -> Automatic]
SumConvergence[a[n], n, Method -> "RaabeTest"]
SumConvergence[a[n], n, Method -> "RatioTest"]
SumConvergence[a[n], n, Method -> "RootTest"]

False

True

I suppose the answers depend on the considered test. RatioTest and RootTest return nothing. Manual investigation of the ratio and root test yields

temp = FullSimplify[Abs[a[n + 1]/a[n]], n > 0];
Limit[temp, n -> Infinity]
temp = FullSimplify[Abs[a[n]]^(1/n), n > 0];
Limit[temp, n -> Infinity]

1

1

meaning that both of these test can NOT give an answer, see, e.g., Wikipedia. Try asking in https://math.stackexchange.com/. This question could also be moved there.