Along the lines of how normal the MRB constant is, I noticed the following about its original sum,
$\displaystyle{\sum_{n=1}^{\infty}(-1)^n(n^{1/n}-1)}.$
I wondered, (what would happen to the sum if you added smaller and smaller steps?)
ReImPlot[ NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/n}, WorkingPrecision -> 30, Method ->"AlternatingSigns"], {n, 1, Infinity}]
`
In[35]:= Re[
NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/10^20},
WorkingPrecision -> 70, Method -> "AlternatingSigns",
NSumTerms -> 2000]]
Out[35]= -1.\
999999749999999973333339999999998228052447182132376098872359604490555*\
10^-14
In[36]:= Re[
NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/10^30},
WorkingPrecision -> 70, Method -> "AlternatingSigns",
NSumTerms -> 2000]]
Out[36]= -1.\
999999749999999999999999997333333999999999999999999982280524444387501*\
10^-24
In[37]:= Re[
NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/10^40},
WorkingPrecision -> 70, Method -> "AlternatingSigns",
NSumTerms -> 2000]]
Out[37]= -1.\
999999749999999999999999999999999999733333399999999999999990562538808*\
10^-34
In[38]:= Re[
NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/10^50},
WorkingPrecision -> 70, Method -> "AlternatingSigns",
NSumTerms -> 2000]]
Out[38]= -1.\
999999749999999999999999999999999999999999999973243568040505517332039*\
10^-44
In this case, finding exactly where the error starts is difficult for me. But let's assume the following is true.
In[5]:= Re[ NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/10^30}, WorkingPrecision -> 40, Method -> "AlternatingSigns"]]
Out[5]= -1.12250000000000000000000000002637761741*10^-28
In[6]:= Re[NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 2/10^40}, WorkingPrecision -> 50, Method -> "AlternatingSigns"]]
Out[6]= -2.244999999999999999999999999999999999982964553611*10^-38
In[39]:= p = Re[NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/10^200}, WorkingPrecision -> 1000, Method -> "AlternatingSigns"]]
Out[39]= \
-1.1224999999999999999999999999999999999999999999999999999999999999999\
9999999999999999999999999999999999999999999999999999999999999999999999\
9999999999999999999999999999999999999999999999999999999999999887999999\
9999999999999999999999999999999999999999999999999999999999999999999999\
9999999999999999999999999999999999999999999999999999999999999999999999\
9999999999999999999999999999999999999999999999999951865743238994910613\
...*10^-198
In[41]:= Rationalize[N[Q100 = 10^98 p + 1, 200], 0]]
$$\frac{8908685968819599109131403118040089086859688195991091314031180400890868596881959910913140311804008907}{8908685968819599109131403118040089086859688195991091314031180400890868596881959910913140311804008908}$$
I think it's worth pointing out that the rational approximation consists, entirely, of repeating
$$89086859688195991091314031180400's$$ except for the last digit of the numerator.
Some steps of multiples of powers of 10 also give solutions with many repeating 9s.
In[59]:= N[ Re[NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/10^40}, WorkingPrecision -> 1000, Method -> "AlternatingSigns"]], 100]
Out[59]=-1.122499999999999999999999999999999999998879999999999999999999999999999999999995186574323899491061367*10^-38
In[54]:= N[Re[NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/20^40}, WorkingPrecision -> 1000, Method ->"AlternatingSigns"]], 55]
Out[54]= \
-1.020907802740111947059631347656249999999999999999999074*10^-50
In[69]:= N[Re[NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/50^40}, WorkingPrecision -> 1000, Method -> "AlternatingSigns"]], 150]
Out[69]= \
-1.2342018021785599999999999999999999999999999999999999999999999999998\
6460030820316153243290828799999999999999999999999999999999999999993601\
859835697267*10^-66
In[84]:= N[ Re[NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/200^40}, WorkingPrecision -> 1000, Method -> "AlternatingSigns"]], 200]
Out[84]= \
-1.0209078027401119470596313476562499999999999999999999999999999999999\
9999999999999999999999990735577139406090041440222648816416040062904357\
91015624999999999999999999999999999999999999637878099817017722*10^-90
I posted it at https://math.stackexchange.com/questions/4575530/sum-infty-textstylex-11-k-atopk-in-1-ldots-infty-1x-x1 to see if anyone can explain it.
I found the following that shows a little bit of what happens for smaller and smaller steps of powers 2 and 5, which are related to powers 10.
By a factor of pi, the imaginary part has at least as many repetitions of decimals.
In
680380635051101*10^-79, \
-1.0S4e9t9A9t9t9r9i9b9u9t9e9s9[9S9 ep9a9r9a9t9e9,9 9H9o99d9R9e9s9t9]99999988
9999999999S9e9p9a9r9a9t9e9[9d9a9t\a_
pat9t9_9]9 9:9=9 9M9o9d9u9l9e9[85
0680 3 8{0r6e3s5}0,5
1*10 ^ -r8e1s, =\ G
her-B1y.[0d1a4t9a9,9 9p9a9t9t9]9;99
9999 9 9I9f9[9p9a9t9t9@9r9e9s9[9[919,9 919]9]9,9 8r8e9s7,4 9R9e9v9e9r9s9e9@9r9e9s9]9
9999 9 9]\
999e9p9a9r9a9t9e9[9d9a9t9a9,9 8E5v3e8n0Q6]8
806{3{520,5 *41,0 ^6-,8 38,, \1
1, 3-,1 .50,1 479,9 999}9}9
999999999999999999999999999999998897499999999999999999999\
99999999999999999985380680380635*10^-85, \
-1.0149999999999999999999999999999999999999999889749999999999999999999\
99999999999999999999853806803806*10^-87, \
-1.0149999999999999999999999999999999999999999988974999999999999999999\
99999999999999999999998538068038*10^-89, \
-1.0149999999999999999999999999999999999999999998897499999999999999999\
99999999999999999999999985380680*10^-91, \
-1.0149999999999999999999999999999999999999999999889749999999999999999\
99999999999999999999999999853807*10^-93, \
-1.0149999999999999999999999999999999999999999999988974999999999999999\
99999999999999999999999999998538*10^-95, \
-1.0149999999999999999999999999999999999999999999998897499999999999999\
99999999999999999999999999999985*10^-97}
Here is an answer to the previous question.
While mathematicians try to crack this nut, here's a physicist's point of view. I will focus on how to calculate this integral, keeping things as simple as possible, probably approximately.
This is an example of that how we can use diverging series to compute values. It is only important to guess where to truncate the series.
