# Explanation for repeating numbers $\displaystyle{\sum_{n=1}^{\infty}(-1)^n(n^{1/n}-1)}$ added with smaller and smaller steps of negative powers of 10?

I noticed the following about the original MRB constant sum. Can you offer an explanation of why there are several repeating numbers in the following expansions?

$$\displaystyle{\sum_{n=1}^{\infty}(-1)^n(n^{1/n}-1)}$$ is a sum for the MRB constant.

I wondered, (what would happen to the sum if you added smaller and smaller steps?)

(using Mathematica 13.1)

 ReImPlot[ NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/n},  WorkingPrecision -> 30, Method ->"AlternatingSigns"], {n, 1,  Infinity}]




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.

I don't know if it can have anything about the repeating decimals, but here is an overview, where we have a jump discontinuity at n=1.

ListPlot[Table[Re[ NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/n},   WorkingPrecision -> 30, Method -> "AlternatingSigns"]], {n, .9, 1.1, .01}]]


Show[{ReImPlot[NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/n},
WorkingPrecision -> 30, Method -> "AlternatingSigns"], {n, .9,
1.1}], Plot[ NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/1},
WorkingPrecision -> 30, Method -> "AlternatingSigns"], {n, .9,
1.1}]}]


Some steps of multiples of powers of 10 also give solutions that have a lot of repeating 9's.

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 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[216]:=

N[Table[Im[
NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/10^n},
WorkingPrecision -> 180, Method -> "AlternatingSigns"]], {n, 20,
50}], 100]/Pi

Out[216]= \
{-1.014999999999999999889749999999999999985380680380635051103404220561\
131468438815891294117640691851961*10^-37, \
-1.0149999999999999999889749999999999999998538068038063505110089362727\
15953551936726556669955860834527*10^-39, \
-1.0149999999999999999988974999999999999999985380680380635051100642567\
94264174386914539123932192756267*10^-41, \
-1.0149999999999999999998897499999999999999999853806803806350511006174\
62009746382736692535387904947180*10^-43, \
-1.0149999999999999999999889749999999999999999998538068038063505110061\
49514164568466234472485086767530*10^-45, \
-1.0149999999999999999999988974999999999999999999985380680380635051100\
61470035712789301212271980781307*10^-47, \
-1.0149999999999999999999998897499999999999999999999853806803806350511\
00614675251194997650990266937597*10^-49, \
-1.0149999999999999999999999889749999999999999999999998538068038063505\
11006146727406017081148770216506*10^-51, \
-1.0149999999999999999999999988974999999999999999999999985380680380635\
05110061467248954237916126569712*10^-53, \
-1.0149999999999999999999999998897499999999999999999999999853806803806\
35051100614672464436446265904565*10^-55, \
-1.0149999999999999999999999999889749999999999999999999999998538068038\
06350511006146724619258529763685*10^-57, \
-1.0149999999999999999999999999988974999999999999999999999999985380680\
38063505110061467246167479364741*10^-59, \
-1.0149999999999999999999999999998897499999999999999999999999999853806\
80380635051100614672461649687715*10^-61, \
-1.0149999999999999999999999999999889749999999999999999999999999998538\
06803806350511006146724616471771*10^-63, \
-1.0149999999999999999999999999999988974999999999999999999999999999985\
38068038063505110061467246164693*10^-65, \
-1.0149999999999999999999999999999998897499999999999999999999999999999\
85380680380635051100614672461647*10^-67, \
-1.0149999999999999999999999999999999889749999999999999999999999999999\
99853806803806350511006146724616*10^-69, \
-1.0149999999999999999999999999999999988974999999999999999999999999999\
99998538068038063505110061467246*10^-71, \
-1.0149999999999999999999999999999999998897499999999999999999999999999\
99999985380680380635051100614672*10^-73, \
-1.0149999999999999999999999999999999999889749999999999999999999999999\
99999999853806803806350511006147*10^-75, \
-1.0149999999999999999999999999999999999988974999999999999999999999999\
99999999998538068038063505110061*10^-77, \
-1.0149999999999999999999999999999999999998897499999999999999999999999\
99999999999985380680380635051101*10^-79, \
-1.0149999999999999999999999999999999999999889749999999999999999999999\
99999999999999853806803806350511*10^-81, \
-1.0149999999999999999999999999999999999999988974999999999999999999999\
99999999999999998538068038063505*10^-83, \
-1.0149999999999999999999999999999999999999998897499999999999999999999\
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}
`
• Hint: MathJax can and should be used in titles. Nov 13, 2022 at 9:54
• I just tried something. How about that? Nov 13, 2022 at 9:59
• Horrendous. Is it a mathematica question ? If so: nearly off-topic. If it is a mathematics question use mathematics notation and content/context and so on and so forth. Strangely you used proper $\sum_{n=1}^{\infty}(-1)^n(n^{1/n}-1)$ in the body. Nov 13, 2022 at 10:03
• What is a mathematical notation for a sum over steps? Nov 13, 2022 at 10:07
• @TymaGaidash It is named after them, I suppose Nov 13, 2022 at 14:07