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The convergent Sum[n^(1/n)/n^x, {n, 1, Infinity}] and the like

Marvin Ray Burns A.G.S. (cum laude)

Marvin Ray Burns A.G.S. (cum laude), original investigator of the MRB constant

Posted 10 years ago

Consider 2 OEIS sequences: A250091 Decimal expansion of 1+2^(1/2)+3^(1/3)+4^(1/4)+5^(1/5)+6^(1/6)+7^(1/7). and A037077 Decimal expansion of upper limit of - 1^(1/1) + 2^(1/2) - 3^(1/3) + ... . (Also known as the MRB constant). What happens when you take Sum[n^(1/n)/n^x, {n, 1, Infinity}], for larger and larger x? And what is an interesting property of the results? Start with, ml = Table[N[Sum[n^(1/n)/n^x, {n, 1, Infinity}], 30], {x, 2, 20}] ({1.87957850179922452548780902030, \ 1.28433140829818311804978960294, 1.11649815901955280446360446952, \ 1.05230243601496483918417757240, 1.02455966710449268950894930761, \ 1.01181956555161661613504788502, 1.00577036801151528942775809119, \ 1.00284169266529443639880816265, 1.00140701129195096425768601368, \ 1.00069904553360492220816210358, 1.00034807151502562926666795815, \ 1.00017356043075113117799946483, 1.00008662379823500633540883697, \ 1.00004326025121731439904151647, 1.00002161302955915670254559274, \ 1.00001080084552597690154618049, 1.00000539854026299833844244956, \ 1.00000269864442366327406568283, 1.00000134911408655949085647188}) Then strip the one from the results so that you are left with the fractional part: Table[(ml[[x + 1]] - 1)/(ml[[x]] - 1), {x, 1, 18}] (* {0.32325870597856601335264674468, \ 0.40972666268855964950237000356, 0.4489550431967470748752560010, \ 0.4695702337356838690155488021, 0.4812591922084509040893131359, \ 0.488204747147076802354478580, 0.492462986697483171702127350, \ 0.495131408520977249601439190, 0.49683008061408269065187630, \ 0.49792395243647742451603934, 0.49863439913591189446856479, \ 0.49909877418555419122177077, 0.4994037677723545403535596, \ 0.4996048092875231735715817, 0.4997376927845338620069440, \ 0.499825708090576356557523, 0.499884096847404516964014, \ 0.499922877845513394172502}) It appears to go to 1/2. If you replace 1/n with x'n you get the following. nl = Table[N[Sum[n^(x/n)/n^x, {n, 1, Infinity}], 30], {x, 2, 50}]; Table[(nl[[x + 1]] - 1)/(nl[[x]] - 1), {x, 1, 48}] ( {0.46806917048762805165976009063, \ 0.58552410491421033048161264365, 0.63715313842141044277734720504, \ 0.66437715102933760259856891504, 0.6800914584690084265802970483, \ 0.6896517396623494068107583992, 0.6956645339242514154420027322, \ 0.6995308126570528621804163014, 0.7020551160673681928212474536, \ 0.7037211349305621509391174590, 0.704829272421506701581292293, \ 0.705570526527653090370011062, 0.706068434368063473135523002, \ 0.706403915885726363146867730, 0.706630475283771278322974287, \ 0.706783737809199814614774542, 0.706887548585582718794905601, \ 0.70695793059744011259680719, 0.70700568235807212426782039, \ 0.70703809747753197542542866, 0.70706011041940439375378651, \ 0.70707506372292522997028901, 0.70708522368715768677252204, \ 0.70709212797318195988585045, 0.7070968204128457956537715, \ 0.7071000098814048194928538, 0.7071021779224227875086932, \ 0.7071036517226710359058870, 0.7071046536265801869716379, \ 0.7071053347497099814427328, 0.707105797806408367696005, \ 0.707106112617028525990783, 0.707106326644497037070963, \ 0.707106472154654201526278, 0.707106571082786572444759, \ 0.707106638341462308732904, 0.707106684069050301189998, \ 0.70710671515823764830212, 0.70710673629513577042717, \ 0.70710675066569719374063, 0.70710676043596925704852, \ 0.70710676707859684098774, 0.70710677159479918693162, \ 0.7071067746652847275237, 0.7071067767528537886064, \ 0.7071067781721555549972, 0.7071067791371142100420, \ 0.7071067797931729427458}*) Which appears to go to 1/Sqrt[2]. You get the same final results if you change the series to alternating series, using Table[N[Sum[(-1)^n n^(1/n)/n^x, {n, 1, Infinity}], 30], {x, 2, 50}] and Table[N[Sum[(-1)^n n^(x/n)/n^x, {n, 1, Infinity}], 30], {x, 2, 50}] I hope you find this interesting, as well!

Consider 2 OEIS sequences:

A250091 Decimal expansion of 1+2^(1/2)+3^(1/3)+4^(1/4)+5^(1/5)+6^(1/6)+7^(1/7).

and

A037077 Decimal expansion of upper limit of - 1^(1/1) + 2^(1/2) - 3^(1/3) + ... . (Also known as the MRB constant).

What happens when you take Sum[n^(1/n)/n^x, {n, 1, Infinity}], for larger and larger x? And what is an interesting property of the results?

Start with,

  ml = 
 Table[N[Sum[n^(1/n)/n^x, {n, 1, Infinity}], 30], {x, 2, 20}]

(*{1.87957850179922452548780902030, \
1.28433140829818311804978960294, 1.11649815901955280446360446952, \
1.05230243601496483918417757240, 1.02455966710449268950894930761, \
1.01181956555161661613504788502, 1.00577036801151528942775809119, \
1.00284169266529443639880816265, 1.00140701129195096425768601368, \
1.00069904553360492220816210358, 1.00034807151502562926666795815, \
1.00017356043075113117799946483, 1.00008662379823500633540883697, \
1.00004326025121731439904151647, 1.00002161302955915670254559274, \
1.00001080084552597690154618049, 1.00000539854026299833844244956, \
1.00000269864442366327406568283, 1.00000134911408655949085647188}*)

Then strip the one from the results so that you are left with the fractional part:

 Table[(ml[[x + 1]] - 1)/(ml[[x]] - 1), {x, 1, 18}]

(* {0.32325870597856601335264674468, \
0.40972666268855964950237000356, 0.4489550431967470748752560010, \
0.4695702337356838690155488021, 0.4812591922084509040893131359, \
0.488204747147076802354478580, 0.492462986697483171702127350, \
0.495131408520977249601439190, 0.49683008061408269065187630, \
0.49792395243647742451603934, 0.49863439913591189446856479, \
0.49909877418555419122177077, 0.4994037677723545403535596, \
0.4996048092875231735715817, 0.4997376927845338620069440, \
0.499825708090576356557523, 0.499884096847404516964014, \
0.499922877845513394172502}*)

It appears to go to 1/2.

If you replace 1/n with x'n you get the following.

nl = Table[N[Sum[n^(x/n)/n^x, {n, 1, Infinity}], 30], {x, 2, 50}];

 Table[(nl[[x + 1]] - 1)/(nl[[x]] - 1), {x, 1, 48}]

(* {0.46806917048762805165976009063, \
0.58552410491421033048161264365, 0.63715313842141044277734720504, \
0.66437715102933760259856891504, 0.6800914584690084265802970483, \
0.6896517396623494068107583992, 0.6956645339242514154420027322, \
0.6995308126570528621804163014, 0.7020551160673681928212474536, \
0.7037211349305621509391174590, 0.704829272421506701581292293, \
0.705570526527653090370011062, 0.706068434368063473135523002, \
0.706403915885726363146867730, 0.706630475283771278322974287, \
0.706783737809199814614774542, 0.706887548585582718794905601, \
0.70695793059744011259680719, 0.70700568235807212426782039, \
0.70703809747753197542542866, 0.70706011041940439375378651, \
0.70707506372292522997028901, 0.70708522368715768677252204, \
0.70709212797318195988585045, 0.7070968204128457956537715, \
0.7071000098814048194928538, 0.7071021779224227875086932, \
0.7071036517226710359058870, 0.7071046536265801869716379, \
0.7071053347497099814427328, 0.707105797806408367696005, \
0.707106112617028525990783, 0.707106326644497037070963, \
0.707106472154654201526278, 0.707106571082786572444759, \
0.707106638341462308732904, 0.707106684069050301189998, \
0.70710671515823764830212, 0.70710673629513577042717, \
0.70710675066569719374063, 0.70710676043596925704852, \
0.70710676707859684098774, 0.70710677159479918693162, \
0.7071067746652847275237, 0.7071067767528537886064, \
0.7071067781721555549972, 0.7071067791371142100420, \
0.7071067797931729427458}*)

Which appears to go to 1/Sqrt[2].

You get the same final results if you change the series to alternating series, using

Table[N[Sum[(-1)^n n^(1/n)/n^x, {n, 1, Infinity}], 30], {x, 2, 50}]

and

Table[N[Sum[(-1)^n n^(x/n)/n^x, {n, 1, Infinity}], 30], {x, 2, 50}]

I hope you find this interesting, as well!

POSTED BY: Marvin Ray Burns A.G.S. (cum laude)

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Marvin Ray Burns A.G.S. (cum laude)

Marvin Ray Burns A.G.S. (cum laude), original investigator of the MRB constant

Posted 10 years ago

This is a little hard to put in words, but I found out concerning the partial sums of the MRB constant, for integer a>1, as x gets large the following seems to be true. In code that is (Sum[(-1)^n (n^(1/n) - 1), {n, 1, a^(x + 2)}] - Sum[(-1)^n (n^(1/n) - 1), {n, 1, a^(x + 1)}]) / (Sum[(-1)^n (n^(1/n) - 1), {n, 1, a^(x + 1)}] - Sum[(-1)^n (n^(1/n) - 1), {n, 1, a^x}]) -> ((a - 1) x + (a - 2))/(a ((a - 1) x - 1)) Here is a code that makes a table showing the relation to be true for many values: Clear[a, b, c]; TableForm[ Table[b[a] = Table[NSum[(-1)^n (n^(1/n) - 1), {n, 1, a^x}, Method -> "AlternatingSigns", WorkingPrecision -> 100, NSumTerms -> 300], {x, 1, 100}]; c[a] = Table[(b[a][[x + 1]]) - (b[a][[x]]), {x, 1, 98}]; Table[N[((a - 1) x + (a - 2))/(a ((a - 1) x - 1)) - N[c[a][[x + 1]]/c[a][[x]], 20], 10], {x, 2, 96}], {a, 2, 7}], TableHeadings -> {Table["a=" <> ToString[x], {x, 2, 7}], Table["x=" <> ToString[x], {x, 1, 100}]}]

This is a little hard to put in words, but I found out concerning the partial sums of the MRB constant, for integer a>1, as x gets large the following seems to be true.

enter image description here

In code that is

(Sum[(-1)^n (n^(1/n) - 1), {n, 1, a^(x + 2)}] - 
    Sum[(-1)^n (n^(1/n) - 1), {n, 1, 
      a^(x + 1)}])
/
(Sum[(-1)^n (n^(1/n) - 1), {n, 1, a^(x + 1)}] - 
    Sum[(-1)^n (n^(1/n) - 1), {n, 1, a^x}]) 
->
((a - 1) x + (a - 2))/(a ((a - 1) x - 1))

Here is a code that makes a table showing the relation to be true for many values:

Clear[a, b, c]; TableForm[
 Table[b[a] = 
   Table[NSum[(-1)^n (n^(1/n) - 1), {n, 1, a^x}, 
     Method -> "AlternatingSigns", WorkingPrecision -> 100, 
     NSumTerms -> 300], {x, 1, 100}]; 
  c[a] = Table[(b[a][[x + 1]]) - (b[a][[x]]), {x, 1, 98}]; 
  Table[N[((a - 1) x + (a - 2))/(a ((a - 1) x - 1)) - 
     N[c[a][[x + 1]]/c[a][[x]], 20], 10], {x, 2, 96}], {a, 2, 7}], 
 TableHeadings -> {Table["a=" <> ToString[x], {x, 2, 7}], 
   Table["x=" <> ToString[x], {x, 1, 100}]}]

POSTED BY: Marvin Ray Burns A.G.S. (cum laude)

Marvin Ray Burns A.G.S. (cum laude)

Marvin Ray Burns A.G.S. (cum laude), original investigator of the MRB constant

Posted 10 years ago

Going back to the convergent series, simplify the notation and define, Clear[m]; m = Table[N[Sum[n^(1/n)/n^x, {n, 1, Infinity}], 50], {x, 2, 100}]; n = Table[N[Sum[n^(x/n)/n^x, {n, 1, Infinity}], 50], {x, 2, 100}]; a = Table[((n[[x]] - 1)/(m[[x]] - 1)), {x, 1, 99}]; a1 = Table[a[[x + 1]]/a[[x]], {x, 1, 98}]; a1[[90 ;; 98]] ({1.4142135623730948387945, 1.414213562373094905421, \ 1.414213562373094950919, 1.414213562373094981986, \ 1.41421356237309500320, 1.41421356237309501768, \ 1.41421356237309502756, 1.41421356237309503431, 1.4142135623730950389}) We get Sqrt[2]. Reverse m and n and the reciprocal behaves half way like a linear function. in that f(a x) == a f(x): so that we get 1/Sqrt[2].. b = Table[(m[[x]] - 1)/(n[[x]] - 1), {x, 1, 99}]; b1 = Table[b[[x + 1]]/b[[x]], {x, 1, 98}]; b1[[90 ;; 98]] ({0.7071067811865476294044, 0.7071067811865475960914, \ 0.707106781186547573342, 0.707106781186547557809, \ 0.707106781186547547203, 0.70710678118654753996, \ 0.70710678118654753502, 0.70710678118654753165, \ 0.70710678118654752935})

Going back to the convergent series, simplify the notation and define,

 Clear[m]; m = 
 Table[N[Sum[n^(1/n)/n^x, {n, 1, Infinity}], 50], {x, 2, 100}];
n = Table[N[Sum[n^(x/n)/n^x, {n, 1, Infinity}], 50], {x, 2, 100}];

 a = Table[((n[[x]] - 1)/(m[[x]] - 1)), {x, 1, 99}];

a1 = Table[a[[x + 1]]/a[[x]], {x, 1, 98}]; a1[[90 ;; 98]]

(*{1.4142135623730948387945, 1.414213562373094905421, \
1.414213562373094950919, 1.414213562373094981986, \
1.41421356237309500320, 1.41421356237309501768, \
1.41421356237309502756, 1.41421356237309503431, 1.4142135623730950389}*)

We get Sqrt[2].

Reverse m and n and the reciprocal behaves half way like a linear function. in that f(a x) == a f(x): so that we get 1/Sqrt[2]..

 b = Table[(m[[x]] - 1)/(n[[x]] - 1), {x, 1, 99}];

b1 = Table[b[[x + 1]]/b[[x]], {x, 1, 98}]; b1[[90 ;; 98]]

(*{0.7071067811865476294044, 0.7071067811865475960914, \
0.707106781186547573342, 0.707106781186547557809, \
0.707106781186547547203, 0.70710678118654753996, \
0.70710678118654753502, 0.70710678118654753165, \
0.70710678118654752935}*)

POSTED BY: Marvin Ray Burns A.G.S. (cum laude)

Marvin Ray Burns A.G.S. (cum laude)

Marvin Ray Burns A.G.S. (cum laude), original investigator of the MRB constant

Posted 10 years ago

Concerning the Mathematica NSum error found in the previous post, It can be eliminated by NSumTerms. m = NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 200] (* 0.18785964246206712024851793405427323005590309490013878617200468408947\ 7231564660213703296654433107496903842345856258019061231370094759226630\ 438929348896184120837336626081613602738126379373435283224982*) slll = Table[ NSum[(-1)^n (n^(1/n) - w/n), {n, 1, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 100, NSumTerms -> 500] - (Log[2^w] - 1/2 + m), {w, 450}]; Print[ N[slll, 30]] Gives all near 0's ListPlot[slll] So by extending sum to divergent series in the regularization method used by Mathematica's NSum, Sum[(-1)^n (n^(1/n) - w/n),{n,1,Infinity}] == Log[2^w] - 1/2 + m. == w Log[2] - 1/2 +m Proof by adding two series: Sum[(-1)^n (n^(1/n)),{n,1,Infinity}] has a midpoint (the mean of the upper limit point and the lower middle point) at - 1/2 + m and treat that as the sum of a convergent series. Sum[(-1)^n (w/n), {n, 1, Infinity}] is the same as w Log[2]

Concerning the Mathematica NSum error found in the previous post, It can be eliminated by NSumTerms.

  m = NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}, 
  Method -> "AlternatingSigns", WorkingPrecision -> 200]

(* 0.18785964246206712024851793405427323005590309490013878617200468408947\
 7231564660213703296654433107496903842345856258019061231370094759226630\
 438929348896184120837336626081613602738126379373435283224982*)

slll = Table[
    NSum[(-1)^n (n^(1/n) - w/n), {n, 1, Infinity}, 
      Method -> "AlternatingSigns", WorkingPrecision -> 100, 
      NSumTerms -> 500] - (Log[2^w] - 1/2 + m), {w, 450}]; Print[
   N[slll, 30]]

Gives all near 0's

ListPlot[slll]

enter image description here

So by extending sum to divergent series in the regularization method used by Mathematica's NSum,

Sum[(-1)^n (n^(1/n) - w/n),{n,1,Infinity}] == Log[2^w] - 1/2 + m. == w Log[2] - 1/2 +m

Proof by adding two series:

Sum[(-1)^n (n^(1/n)),{n,1,Infinity}]

has a midpoint (the mean of the upper limit point and the lower middle point) at - 1/2 + m and treat that as the sum of a convergent series.

Sum[(-1)^n (w/n), {n, 1, Infinity}]

is the same as w Log[2]

POSTED BY: Marvin Ray Burns A.G.S. (cum laude)

Marvin Ray Burns A.G.S. (cum laude)

Marvin Ray Burns A.G.S. (cum laude), original investigator of the MRB constant

Posted 10 years ago

I checked for round off error in the following and found some evidence of an error on Mathematica's part.. To keep expanding our list of identities, Let m be the MRB constant: m = NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 200] (* 0.18785964246206712024851793405427323005590309490013878617200468408947\ 7231564660213703296654433107496903842345856258019061231370094759226630\ 438929348896184120837336626081613602738126379373435283224982) Then m- Sum[(-1)^n (n^(1/n) - x/n)]==1/2 - Log[2^x]) for x up to and including x=20, sl = Table[(m - NSum[(-1)^n (n^(1/n) - x/n), {n, 1, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 200]) - ( 1/2 - Log[2^x]), {x, 2, 350}];Print[N[sl[[1 ;; 19]], 30], N[sl[[340 ;; 349]], 30]] ( {1.242910^-194,1.242910^-194,1.242910^-194,1.242910^-194,1.242910^-194,1.242910^-194,1.242910^-194,1.242910^-194,1.242910^-194,1.242910^-194,1.242910^-194,1.242910^-194,1.242910^-194,1.242910^-194,1.242910^-194,1.242910^-194,1.242910^-194,1.242910^-194 ,0.0259183374259628836676682211919} {2.2607686481144959098698921022610^-14,-8.8831573509279809049302873562810^-15,3.4144069215196099053542098452110^-15,-1.2834252862273327031182856283410^-15,4.7162769057545867928937058722210^-16,-1.6938011200889555658680659044810^-16,5.9431365111675343691332666008610^-17,-2.0366168616615714966447435888710^-17,6.8137712133853436118957374310210^-18,-2.2247862762516659985547155134010^-18,7.0866805681602501076050324786610^-19} ) x=20 is the 18th element of sl. Notice that the 19th element is ,0.0259183374259628836676682211919. But for x>20 something beautiful happens: ListPlot[sl] The reason for the poor behavior is that Sum[(-1)^n (n^(1/n) - x/n)] for x/n not equal to 1 is divergent. and as x gets large x/n is further from 1 and NSum[(-1)^n (n^(1/n) - x/n), {n, 1, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 200] /. x -> 10^t for the most part goes to 10^(t - 1) Log[2^10] - 1/2 + m) as t gets large: Table[NSum[(-1)^n (n^(1/n) - 10^t/n), {n, 1, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 200] - (10^(t - 1) Log[2^10] - 1/2 + m), {t, 10}] (* {-1.390898310^-190, \ -0.0018047654004370285056973513785100572645865454457815534444526025649\ 6588183515222563305775295474928095349258443231532776865726011389856029\ 989033125798213015705226943409189152746974520337737698733944, \ -1.40209610^-190, -1.5038910^-190, -2.521810^-190, \ -1.270110^-189, -1.145010^-188, -1.132510^-187, -1.131210^-186, \ -1.131110^-185}) Notice that NSum[(-1)^n (n^(1/n) - s^t/n), {n, 1, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 200] for the most part goes to `(s^(t - 1) Log[2^s] - 1/2 + m)` Enter the following: Table[N[NSum[(-1)^n (n^(1/n) - s^t/n), {n, 1, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 200] - (s^(t - 1) Log[2^s] - 1/2 + m), 10], {t, 10}, {s, 10}] // TableForm In general we have, NSum[(-1)^n (n^(1/n) - w/n), {n, 1, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 200] going to ( Log[2^w] - 1/2 + m). See next: sll = Table[ NSum[(-1)^n (n^(1/n) - w/n), {n, 1, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 200] - (Log[2^w] - 1/2 + m), {w, 450}]; Print[ Print[N[sll[[1 ;; 20]], 30], N[sll[[340 ;; 349]], 30]] ({-1.390796510^-190,-1.390807910^-190,-1.390819210^-190,-1.390830510^-190,-1.390841810^-190,-1.390853110^-190,-1.390864410^-190,-1.390875710^-190,-1.390887010^-190,-1.390898310^-190,-1.390909710^-190,-1.390921010^-190,-1.390932310^-190,-1.390943610^-190,-1.390954910^-190,-1.390966210^-190,-1.390977510^-190,-1.390988810^-190,-1.391000110^-190, -0.0259183374259628836676682211919} {-1.39463110^-190,-1.39464210^-190,-1.39465310^-190,-1.39466510^-190,-1.39467610^-190,-1.39468710^-190,-1.39469910^-190,-1.39471010^-190,-1.39472110^-190,-1.39473310^-190}*) This is the same formula we started with, yet the results are a little different giving some evidence of an error on Mathematica's part. ListPlot[sll]

I checked for round off error in the following and found some evidence of an error on Mathematica's part..

To keep expanding our list of identities,

Let m be the MRB constant:

m = NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}, 
  Method -> "AlternatingSigns", WorkingPrecision -> 200]

    (* 0.18785964246206712024851793405427323005590309490013878617200468408947\
     7231564660213703296654433107496903842345856258019061231370094759226630\
     438929348896184120837336626081613602738126379373435283224982*)

Then m- Sum[(-1)^n (n^(1/n) - x/n)]==1/2 - Log[2^x]) for x up to and including x=20,

   sl = Table[(m - 
     NSum[(-1)^n (n^(1/n) - x/n), {n, 1, Infinity}, 
      Method -> "AlternatingSigns", WorkingPrecision -> 200]) - (
      1/2 - Log[2^x]), {x, 2, 350}];Print[N[sl[[1 ;; 19]], 30],
       N[sl[[340 ;; 349]], 30]]
    (* {1.2429*10^-194,1.2429*10^-194,1.2429*10^-194,1.2429*10^-194,1.2429*10^-194,1.2429*10^-194,1.2429*10^-194,1.2429*10^-194,1.2429*10^-194,1.2429*10^-194,1.2429*10^-194,1.2429*10^-194,1.2429*10^-194,1.2429*10^-194,1.2429*10^-194,1.2429*10^-194,1.2429*10^-194,1.2429*10^-194
,0.0259183374259628836676682211919}
{2.26076864811449590986989210226*10^-14,-8.88315735092798090493028735628*10^-15,3.41440692151960990535420984521*10^-15,-1.28342528622733270311828562834*10^-15,4.71627690575458679289370587222*10^-16,-1.69380112008895556586806590448*10^-16,5.94313651116753436913326660086*10^-17,-2.03661686166157149664474358887*10^-17,6.81377121338534361189573743102*10^-18,-2.22478627625166599855471551340*10^-18,7.08668056816025010760503247866*10^-19}
*)

x=20 is the 18th element of sl. Notice that the 19th element is ,0.0259183374259628836676682211919.

But for x>20 something beautiful happens:

ListPlot[sl]

enter image description here

The reason for the poor behavior is that Sum[(-1)^n (n^(1/n) - x/n)] for x/n not equal to 1 is divergent. and as x gets large x/n is further from 1 and

NSum[(-1)^n (n^(1/n) - x/n), {n, 1, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 200] /. x -> 10^t

for the most part goes to 10^(t - 1) Log[2^10] - 1/2 + m) as t gets large:

Table[NSum[(-1)^n (n^(1/n) - 10^t/n), {n, 1, Infinity}, 
   Method -> "AlternatingSigns", 
   WorkingPrecision -> 200] - (10^(t - 1) Log[2^10] - 1/2 + m), {t, 
  10}]
(* {-1.3908983*10^-190, \
-0.0018047654004370285056973513785100572645865454457815534444526025649\
6588183515222563305775295474928095349258443231532776865726011389856029\
989033125798213015705226943409189152746974520337737698733944, \
-1.402096*10^-190, -1.50389*10^-190, -2.5218*10^-190, \
-1.2701*10^-189, -1.1450*10^-188, -1.1325*10^-187, -1.1312*10^-186, \
-1.1311*10^-185}*)

Notice that

NSum[(-1)^n (n^(1/n) - s^t/n), {n, 1, Infinity}, 
  Method -> "AlternatingSigns", 
  WorkingPrecision -> 
   200]

for the most part goes to (s^(t - 1) Log[2^s] - 1/2 + m) Enter the following:

Table[N[NSum[(-1)^n (n^(1/n) - s^t/n), {n, 1, Infinity}, 
     Method -> "AlternatingSigns", 
     WorkingPrecision -> 200] - (s^(t - 1) Log[2^s] - 1/2 + m), 
   10], {t, 10}, {s, 10}] // TableForm

In general we have, NSum[(-1)^n (n^(1/n) - w/n), {n, 1, Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 200]

going to ( Log[2^w] - 1/2 + m). See next:

 sll = Table[
   NSum[(-1)^n (n^(1/n) - w/n), {n, 1, Infinity}, 
     Method -> "AlternatingSigns", 
     WorkingPrecision -> 200] - (Log[2^w] - 1/2 + m), {w, 450}]; Print[
  Print[N[sll[[1 ;; 20]], 30], N[sll[[340 ;; 349]], 30]]

  (*{-1.3907965*10^-190,-1.3908079*10^-190,-1.3908192*10^-190,-1.3908305*10^-190,-1.3908418*10^-190,-1.3908531*10^-190,-1.3908644*10^-190,-1.3908757*10^-190,-1.3908870*10^-190,-1.3908983*10^-190,-1.3909097*10^-190,-1.3909210*10^-190,-1.3909323*10^-190,-1.3909436*10^-190,-1.3909549*10^-190,-1.3909662*10^-190,-1.3909775*10^-190,-1.3909888*10^-190,-1.3910001*10^-190,
-0.0259183374259628836676682211919}
{-1.394631*10^-190,-1.394642*10^-190,-1.394653*10^-190,-1.394665*10^-190,-1.394676*10^-190,-1.394687*10^-190,-1.394699*10^-190,-1.394710*10^-190,-1.394721*10^-190,-1.394733*10^-190}*)

This is the same formula we started with, yet the results are a little different giving some evidence of an error on Mathematica's part.

ListPlot[sll]

enter image description here

POSTED BY: Marvin Ray Burns A.G.S. (cum laude)

Marvin Ray Burns A.G.S. (cum laude)

Marvin Ray Burns A.G.S. (cum laude), original investigator of the MRB constant

Posted 10 years ago

Like the upper limit of the divergent Sum[(-1)^n n^(1/n)] can be found in the convergent Sum[(-1)^n (n^(1/n) - 1)], mm1=Table[N[Sum[(-1)^n (n^(1/n) - 1)/n^x, {n, 1, Infinity}], 30], {x, 2, 50}]; Table[(mm1[[x + 1]])/(mm1[[x]]), {x, 1, 48}] Still goes to 1/2 And nn1 = Table[ N[Sum[(-1)^n (n^(x/n) - 1)/n^x, {n, 1, Infinity}], 30], {x, 2, 50}]; Table[(nn1[[x + 1]])/(nn1[[x]]), {x, 1, 48}] Still goes to 1/Sqrt[2] And the computation is a lot faster with the "-1" in the summation: Timing[ Table[N[Sum[(-1)^n (n^(1/n) - 1)/n^x, {n, 1, Infinity}], 30], {x, 2, 50}];] (* {0.780005, Null}) Timing[ Table[N[Sum[(-1)^n (n^(1/n))/n^x, {n, 1, Infinity}], 30], {x, 2, 50}];] ( {6.864044, Null}*)

Like the upper limit of the divergent Sum[(-1)^n n^(1/n)] can be found in the convergent Sum[(-1)^n (n^(1/n) - 1)],

mm1=Table[N[Sum[(-1)^n (n^(1/n) - 1)/n^x, {n, 1, Infinity}], 30], {x, 2, 
  50}]; 
Table[(mm1[[x + 1]])/(mm1[[x]]), {x, 1, 48}]

Still goes to 1/2

And

nn1 = Table[
  N[Sum[(-1)^n (n^(x/n) - 1)/n^x, {n, 1, Infinity}], 30], {x, 2, 50}];
Table[(nn1[[x + 1]])/(nn1[[x]]), {x, 1, 48}]

Still goes to 1/Sqrt[2]

And the computation is a lot faster with the "-1" in the summation:

 Timing[
 Table[N[Sum[(-1)^n (n^(1/n) - 1)/n^x, {n, 1, Infinity}], 30], {x, 2, 
    50}];]

(* {0.780005, Null}*)

 Timing[
 Table[N[Sum[(-1)^n (n^(1/n))/n^x, {n, 1, Infinity}], 30], {x, 2, 
    50}];]

(* {6.864044, Null}*)

POSTED BY: Marvin Ray Burns A.G.S. (cum laude)

Marvin Ray Burns A.G.S. (cum laude)

Marvin Ray Burns A.G.S. (cum laude), original investigator of the MRB constant

Posted 10 years ago

mn1 = Table[ N[Sum[(-1)^n (n^(x/n) - 1)/n, {n, 1, Infinity}], 30], {x, 2, 200}]; Table[(mn1[[x + 1]])/(mn1[[x]]), {x, 1, 198}] Very, very slowly goes to 1/Log[2]

POSTED BY: Marvin Ray Burns A.G.S. (cum laude)

Marvin Ray Burns A.G.S. (cum laude)

Marvin Ray Burns A.G.S. (cum laude), original investigator of the MRB constant

Posted 10 years ago

Thank you Daniel, I might have a chance to work on that tomorrow.

POSTED BY: Marvin Ray Burns A.G.S. (cum laude)

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 10 years ago

This should get you started. The limiting behavior of the first is explained by the fact that the first term hugely dominates, and approaches 1; the rest get small quickly as `x` increases. The second (subtract off 1, take successive ratios) comes from the fact that the first terms approach 1 fast. The part to show is that it is fast enough that the difference between them and 1 becomes negligeable faster than the second terms decrease. So now the second terms dominate and we are looking at ratios approximating 2^(1/2-(x+1))/2^(1/2-x) which of course is 1/2. I think I'll lleave the last for homework.

POSTED BY: Daniel Lichtblau

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