​
​
I asked GPT to give an integral form for the MRB constant from its table 1 and got the following.
CMRB=
1
∫
0
∞
∑
n=1
n
(-1)
log(n)exp
tlog(n)
n

n
t
∞
∑
n=1
n
(-1)
log(n)exp
tlog(n)
n

n
=
∞
∑
n=1
n
(-1)
-1+
t
n
n
Log[n]
Write a notebook to justify reversing the int and sum.
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Rigorous Abel-Regularized Derivation for an Integral Form of the MRB Constant

Goal

We want a rigorous version of the formal derivation starting from
C
MRB
:
∞
∑
n1
n
(-1)
(
1/n
n
-1).
The issue is that the manipulations
◼
  • exchanging the sum over
    • ◼
  • expanding
    • ◼
  • and reversing the order of the resulting sums,
    • are not justified directly at
    x1
    because absolute convergence fails.

    Abel regularization

    Introduce an Abel parameter
    x
    with
    0≤x<1
    and define
    C(x):
    ∞
    ∑
    n1
    n
    (-1)
    (
    1/n
    n
    -1)
    n
    x
    .
    Then the MRB constant is recovered, in the Abel sense, by
    C
    MRB
    
    lim
    x
    -
    1
    C(x),
    provided this limit exists.

    Step 1: Integral identity for each term

    Use
    u
    e
    -1u
    1
    ∫
    0
    su
    e
    s.
    With
    u
    logn
    n
    ,
    1/n
    n
    -1
    (logn)/n
    e
    -1
    logn
    n
    1
    ∫
    0
    slogn/n
    e
    s
    logn
    n
    1
    ∫
    0
    s/n
    n
    s.
    Hence
    C(x)
    ∞
    ∑
    n1
    n
    (-1)
    n
    x
    logn
    n
    1
    ∫
    0
    s/n
    n
    s.

    Step 2: Why sum and integral may be interchanged for
    0≤x<1

    For
    0≤s≤1
    ,
    s/n
    n
    
    slogn/n
    e
    ≤
    logn/n
    e
    
    1/n
    n
    .
    Also, since the function
    y↦
    logy
    y
    attains its maximum
    1/e
    at
    ye
    ,
    1/n
    n
    
    (logn)/n
    e
    ≤
    1/e
    e
    .
    Therefore
    n
    (-1)
    n
    x
    logn
    n
    s/n
    n
    ≤
    1/e
    e
    x
    n
    |
    logn
    n
    .
    Because
    ∞
    ∑
    n1
    x
    n
    |
    logn
    n
    <∞ (|x|<1),
    we get
    ∞
    ∑
    n1
    1
    ∫
    0
    n
    (-1)
    n
    x
    logn
    n
    s/n
    n
    ds<∞.
    So Tonelli/Fubini applies, giving
    C(x)
    1
    ∫
    0
    ∞
    ∑
    n1
    n
    (-1)
    n
    x
    logn
    n
    s/n
    n
    s.

    Step 3: Expand the exponential

    Since
    s/n
    n
    
    slogn/n
    e
    
    ∞
    ∑
    m0
    m
    s
    m!
    m
    logn
    n
    ,
    we obtain formally
    n
    (-1)
    n
    x
    logn
    n
    s/n
    n
    
    ∞
    ∑
    m0
    n
    (-1)
    n
    x
    m
    s
    m!
    m+1
    (logn)
    m+1
    n
    .

    Step 4: Why the order of summation may be reversed for
    0≤x<1

    For fixed
    n
    ,
    ∞
    ∑
    m0
    m
    s
    m!
    m
    logn
    n
    
    s/n
    n
    ≤
    1/n
    n
    ≤
    1/e
    e
    ,
    so
    ∞
    ∑
    m0
    n
    (-1)
    n
    x
    m
    s
    m!
    m+1
    (logn)
    m+1
    n
    ≤
    1/e
    e
    x
    n
    |
    logn
    n
    .
    Summing over
    n
    gives an absolutely convergent majorant. Hence we may interchange the sums and the integral:
    C(x)
    1
    ∫
    0
    ∞
    ∑
    m0
    ∞
    ∑
    n1
    n
    (-1)
    n
    x
    m
    s
    m!
    m+1
    (logn)
    m+1
    n
    s.
    Now integrate term by term:
    1
    ∫
    0
    m
    s
    s
    1
    m+1
    .
    Thus
    C(x)
    ∞
    ∑
    m0
    1
    (m+1)!
    ∞
    ∑
    n1
    n
    (-1)
    n
    x
    m+1
    (logn)
    m+1
    n
    .
    Equivalently, after reindexing
    km+1
    ,
    C(x)
    ∞
    ∑
    k1
    1
    k!
    ∞
    ∑
    n1
    n
    (-1)
    n
    x
    k
    (logn)
    k
    n
    .

    Step 5: Abel limit

    So the correct statement is:

    Clean final integral representation

    Remarks

    ◼
  • Without the factor
    • ◼
  • The bound
    • ◼
  • This is an instance of Abel regularization: establish identities for

    • Related documentation

    Mathematica confirms that this transformation gives over 200 correct digits :