​
Let m be the MRB constant the sum of
n
(-1)
(
1/n
n
-1)
In[]:=
m=Sum[
n
(-1)
(
1/n
n
-1),{n,1,Infinity}]
Out[]=
∞
∑
n=1
n
(-1)
-1+
1
n
n

Notice the value
of
n
(-1)
(
1/n
n
-1)
/. n -> -E
In[]:=
N[
n
(-1)
1/n
n
]/.n->-E
Out[]=
-0.666996+0.185088
In[]:=
Abs[-0.666996+0.185088]
Out[]=
0.692201
It is close to {-2/3,m}
In[]:=
N[-2/3]
Out[]=
-0.666667
In[]:=
N[m]
Out[]=
0.18786
In[]:=
Abs[%+%%I]
Out[]=
0.69263
In[]:=
N[Log[2]]
Out[]=
0.693147
Use the series
In[]:=
Table[Normal[Series[(-1)^k
1/k
k
,{k,Infinity,n}]]/.k->-E,{n,5,6}]
Out[]=

-
(-1)
1-
1+π

+
2
(1+π)
2
2

-
3
(1+π)
6
3

+
4
(1+π)
24
4

-
5
(1+π)
120
5

,
-
(-1)
1-
1+π

+
2
(1+π)
2
2

-
3
(1+π)
6
3

+
4
(1+π)
24
4

-
5
(1+π)
120
5

+
6
(1+π)
720
6


In[]:=
t=Table[N[Normal[Series[(-1)^k
1/k
k
,{k,Infinity,n}]]]/.k->-E,{n,1,6}]
Out[]=
{-1.29476+0.242646,-0.585634+0.437914,-0.597366+0.140784,-0.682138+0.171501,-0.6688+0.188835,-0.666279+0.185203}
to show it is this close:
In[]:=
ListLinePlot[​​Join[(Tooltip[{Re[#1],Im[#1]}]&)/@t,{Style[{-2/3,N[m]},Red]}],​​PlotRange->All​​]
Out[]=
-1.3
-1.2
-1.1
-1.0
-0.9
-0.8
-0.7
-0.6
0.1
0.2
0.3
0.4
It lies somewhat between t[[5]] and t[[6]]>
In[]:=
ListPlot[​​Join[(Tooltip[{Re[#1],Im[#1]}]&)/@t[[5;;6]],{Style[{-2/3,N[m]},Red]}],PlotRange->{{-.5,-.8},{.18,.19}}​​]
Out[]=
-0.80
-0.75
-0.70
-0.65
-0.60
-0.55
-0.50
0.182
0.184
0.186
0.188
0.190
In[]:=
p=Normal[Series[(-1)^k
1/k
k
,{k,Infinity,5}]]/.k->-E
Out[]=
-
(-1)
1-
1+π

+
2
(1+π)
2
2

-
3
(1+π)
6
3

+
4
(1+π)
24
4

-
5
(1+π)
120
5

A guess as to how close t[[5]] is:
In[]:=
NIm[p]-m-
5
5126
+
7
6
9
10

Out[]=
-3.14193×
-14
10
In[]:=
mm=NSum[
n
(-1)
(
1/n
n
-1),{n,1,Infinity},Method->"AlternatingSigns",WorkingPrecision->30]
Out[]=
0.18785964246206712024857897184
In[]:=
NIm[p]-mm+
3
10^14
-
5
5126
+
7
610^9

Out[]=
-6.61779×
-16
10
Another guess as to how close t[[5]] is:
In[]:=
Im[N[Normal[Series[(-1)^k
1/k
k
,{k,Infinity,5}]]]/.k->-E]
Out[]=
0.188835
In[]:=
m+(m-Sin[m])//N
Out[]=
0.188963
In[]:=
Im[N[Normal[Series[(-1)^k
1/k
k
,{k,Infinity,5}]]]/.k->-E]-(m+(m-Sin[m])//N)
Out[]=
-0.0001276