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In the present article we explain the \ essence of the phenomenon in an elementary and general method, and generate \ linear fraction terms in order to accelerate the convergence speed for \ various infinite series by virtue of ", StyleBox["Mathematica", FontSlant->"Italic"], " commands. For example the following acceleration by linear fractions for \ \[Zeta](2) almost equals L.Euler's ingenious acceleration ", Cell[BoxData[ \(TraditionalForm\`\((log\ 2)\)\^2\)]], " + ", Cell[BoxData[ \(\[Sum]\+\(n = 1\)\%\[Infinity] 1\/\(\(2\^\(n - 1\)\) n\^2\)\)]], " in his great analysis on Basel problem." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ RowBox[{ RowBox[{\(f[n_]\), ":=", RowBox[{ StyleBox[\(\(1\/60\) \((1\/\(n - 2\) + 1\/\(n + 3\))\)\), FontColor->RGBColor[1, 0, 0]], StyleBox["-", FontColor->RGBColor[1, 0, 0]], StyleBox[\(\(2\/15\) \((1\/\(n - 1\) + 1\/\(n + 2\))\)\), FontColor->RGBColor[1, 0, 0]], StyleBox["+", FontColor->RGBColor[1, 0, 0]], StyleBox[\(\(37\/60\) \((1\/n + 1\/\(n + 1\))\)\), FontColor->RGBColor[1, 0, 0]], "+", \(\[Sum]\+\(k = 1\)\%n 1\/k\^2\)}]}], ";"}], "\[IndentingNewLine]", RowBox[{ RowBox[{"N", "[", RowBox[{ RowBox[{"{", RowBox[{\(f[20]\), ",", \(Zeta[2]\), ",", StyleBox[\(Log[2]\^2 + \[Sum]\+\(n = 1\)\%20 1\/\(2\^\(n - 1\)\ \ n\^2\)\), FontSize->14, FontColor->RGBColor[0, 0, 1]]}], "}"}], ",", "12"}], "]"}], "//", "TableForm"}]}], "Input", CellLabel->"In[14]:="], Cell[BoxData[ InterpretationBox[GridBox[{ {"1.6449340702982661778562801332`12.000000000000004"}, {"1.6449340668482264364724151666`12."}, {"1.6449340628651162540816900397`11.999999999999998"} }, RowSpacings->1, ColumnSpacings->3, RowAlignments->Baseline, ColumnAlignments->{Left}], TableForm[ {1.6449340702982661778562801332`12.000000000000004, 1.6449340668482264364724151666`12., 1.6449340628651162540816900397`11.999999999999998}]]], "Output", CellLabel->"Out[15]//TableForm="] }, Open ]], Cell["\<\ The following is another example of acceleration by linear fraction \ terms. A few fraction terms make the approximation accurate. \ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{\(g[n_]\), ":=", RowBox[{ StyleBox[\(\(1\/180\) \((1\/\(n - 1\) - 1\/\(n + 2\))\)\), FontColor->RGBColor[1, 0, 0]], StyleBox["+", FontColor->RGBColor[1, 0, 0]], StyleBox[\(\(11\/60\) \((1\/\(n + 1\) - 1\/n)\)\), FontColor->RGBColor[1, 0, 0]], "+", \(\[Sum]\+\(k = 1\)\%n 1\/k\), "-", \(\(Log[n] + Log[n + 1]\)\/2\)}]}], ";", \(N[{g[20], EulerGamma}, 12] // TableForm\)}]], "Input", CellLabel->"In[23]:="], Cell[BoxData[ InterpretationBox[GridBox[{ {"0.5772156659767604409329616551`12."}, {"0.5772156649015328737698695872`12."} }, RowSpacings->1, ColumnSpacings->3, RowAlignments->Baseline, ColumnAlignments->{Left}], TableForm[ {0.5772156659767604409329616551`12., 0.5772156649015328737698695872`12.}]]], "Output", CellLabel->"Out[23]//TableForm="] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Goldbach's method", "Section", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "In 1729 Goldbach had an inequality estimation for \[Zeta](2) = ", Cell[BoxData[ \(\[Sum]\+\(k = 1\)\%\[Infinity] 1\/k\^2\)], TextAlignment->Center, TextJustification->0], "," }], "Text"], Cell[BoxData[ \(\(\(\[Sum]\+\(k = 7\)\%\[Infinity] 1\/\(\(\ \)\(\((\ k - 7\/16)\)\ \((\ \ k + 9\/16)\)\)\) + \[Sum]\+\(k = 1\)\%6 1\/k\^2 < \[Sum]\+\(k = 1\)\%\ \[Infinity] 1\/k\^2 < \[Sum]\+\(k = 7\)\%\[Infinity] 1\/\(\(\ \)\(\((\ k - \ 1\/2)\)\ \((\ k + 1\/2)\)\)\) + \[Sum]\+\(k = 1\)\%6 1\/k\^2\)\(,\)\)\)], \ "NumberedEquation", TextAlignment->Center, TextJustification->0], Cell["that is,", "Text"], Cell[BoxData[ \(41423\/25200 < \[Sum]\+\(k = 1\)\%\[Infinity] 1\/k\^2 < \ \(\(76997\/46800\)\(.\)\)\)], "NumberedEquation", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "Applying his method to more general situation, we can improve the \ convergence of infinite series. That is , if the n-th partial sum of ", Cell[BoxData[ StyleBox[\(\[Sum]\+\(k = 1\)\%\[Infinity] a\_k\), Background->None]], TextAlignment->Center, TextJustification->0, FontSize->18, FontWeight->"Bold", Background->GrayLevel[1]], " is revised by the term ", Cell[BoxData[ StyleBox[\(\[Sum]\+\(k = n + 1\)\%\[Infinity] b\_k\), FontColor->RGBColor[1, 0, 0]]], TextAlignment->Center, TextJustification->0, FontSize->18, FontWeight->"Bold", Background->GrayLevel[1]], " , the convergence speed of a sequence {", Cell[BoxData[ RowBox[{ StyleBox[\(\[Sum]\+\(k = n + 1\)\%\[Infinity] b\_k\), FontColor->RGBColor[1, 0, 0]], "+", StyleBox[\(\[Sum]\+\(k = 1\)\%n a\_k\), FontColor->RGBColor[0, 0, 1]]}]], TextAlignment->Center, TextJustification->0, FontSize->18, FontWeight->"Bold", Background->GrayLevel[1]], "} is equal to that of {", Cell[BoxData[ StyleBox[\(\[Sum]\+\(k = 1\)\%n\((a\_k - b\_k)\)\), FontColor->RGBColor[0, 0, 1]]], TextAlignment->Center, TextJustification->0, FontSize->18, FontWeight->"Bold", Background->GrayLevel[1]], "}." }], "Text"], Cell[BoxData[ RowBox[{ RowBox[{ StyleBox[\(\[Sum]\+\(k = n + 1\)\%\[Infinity] b\_k\), FontColor->RGBColor[1, 0, 0]], "+", StyleBox[\(\[Sum]\+\(k = 1\)\%n a\_k\), FontColor->RGBColor[0, 0, 1]]}], "=", RowBox[{\(\[Sum]\+\(k = 1\)\%\[Infinity] b\_k\), "+", StyleBox[\(\[Sum]\+\(k = 1\)\%n\((a\_k - b\_k)\)\), FontColor->RGBColor[0, 0, 1]]}]}]], "NumberedEquation", TextAlignment->Center, TextJustification->0, FontSize->18, FontWeight->"Bold", Background->GrayLevel[0.900008]], Cell[TextData[{ "In the case of ", Cell[BoxData[ \(a\_k\)], TextAlignment->Center, TextJustification->0, FontSize->18, FontWeight->"Bold", Background->GrayLevel[1]], "= ", Cell[BoxData[ \(TraditionalForm\`1\/k\^2\)]], ", ", Cell[BoxData[ \(b\_k\)], TextAlignment->Center, TextJustification->0, FontSize->18, FontWeight->"Bold", Background->GrayLevel[1]], " = ", Cell[BoxData[ \(TraditionalForm\`1\/\(k\^2 - 1\/4\)\)]], " for the above, we observe that the linear fraction of n," }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Sum]\+\(k = n + 1\)\%\[Infinity] 1\/\(k\^2 - 1\/4\)\)], "Input", CellLabel->"In[11]:="], Cell[BoxData[ \(2\/\(1 + 2\ n\)\)], "Output", CellLabel->"Out[11]="] }, Open ]], Cell[TextData[{ "improves the convergence speed of ", Cell[BoxData[ \(TraditionalForm\`lim\+\(n \[Rule] \[Infinity]\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(k = 1\)\%n 1\/k\^2\)]], "." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"Table", "[", RowBox[{ RowBox[{"N", "[", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{ StyleBox[\(2\/\(1 + 2\ n\)\), FontColor->RGBColor[1, 0, 0]], "+", StyleBox[\(\[Sum]\+\(k = 1\)\%n 1\/k\^2\), FontColor->RGBColor[0, 0, 1]]}], ",", \(Zeta[2]\)}], "}"}], ",", "15"}], "]"}], ",", \({n, 30, 40}\)}], "]"}], "//", "TableForm"}]], "Input", CellLabel->"In[26]:="], Cell[BoxData[ TagBox[GridBox[{ {"1.644937002847499242714029184`15.", "1.644934066848226436472415167`15."}, {"1.644936732073956092377579518`15.", "1.644934066848226436472415167`15."}, {"1.644936493597155115576602717`15.", "1.644934066848226436472415167`15."}, {"1.6449362827421274355828440258`15.", "1.644934066848226436472415167`15."}, {"1.6449360956229617176103475493`15.", "1.644934066848226436472415167`15."}, {"1.6449359289917041931676249611`15.", "1.644934066848226436472415167`15."}, {"1.6449357801194413585921014667`15.", "1.644934066848226436472415167`15."}, {"1.6449356467020550718981509445`15.", "1.644934066848226436472415167`15."}, {"1.6449355267850376479555192457`15.", "1.644934066848226436472415167`15."}, {"1.6449354187031379399603876868`15.", "1.644934066848226436472415167`15."}, {"1.6449353210316267663395094245`15.", "1.644934066848226436472415167`15."} }, RowSpacings->1, ColumnSpacings->3, RowAlignments->Baseline, ColumnAlignments->{Left}], Function[ BoxForm`e$, TableForm[ BoxForm`e$]]]], "Output", CellLabel->"Out[26]//TableForm="] }, Open ]], Cell[TextData[{ "We give another example of acceleration of the convergence speed of ", Cell[BoxData[ \(TraditionalForm\`lim\+\(n \[Rule] \[Infinity]\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(k = 1\)\%n 1\/k\^2\)]], "." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Series[ 1\/n\^2 - \((p\/\(\((n - 1)\) \((n + 1)\)\) + q\/\(\((n - 2)\) \((n + 2)\)\) + r\/\(\((n - 3)\) \((n + 3)\)\))\), {n, \[Infinity], 8}]\)], "Input", CellLabel->"In[1]:="], Cell[BoxData[ InterpretationBox[ RowBox[{\(\((1 - p - q - r)\)\ \((1\/n)\)\^2\), "+", \(\((\(-p\) - 4\ q - 9\ r)\)\ \((1\/n)\)\^4\), "+", \(\((\(-p\) - 16\ q - 81\ r)\)\ \((1\/n)\)\^6\), "+", \(\((\(-p\) - 64\ q - 729\ r)\)\ \((1\/n)\)\^8\), "+", InterpretationBox[\(O[1\/n]\^9\), SeriesData[ n, DirectedInfinity[ 1], {}, 2, 9, 1], Editable->False]}], SeriesData[ n, DirectedInfinity[ 1], { Plus[ 1, Times[ -1, p], Times[ -1, q], Times[ -1, r]], 0, Plus[ Times[ -1, p], Times[ -4, q], Times[ -9, r]], 0, Plus[ Times[ -1, p], Times[ -16, q], Times[ -81, r]], 0, Plus[ Times[ -1, p], Times[ -64, q], Times[ -729, r]]}, 2, 9, 1], Editable->False]], "Output", CellLabel->"Out[1]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(values = Solve[{1 - p - q - r\ \[Equal] 0, \(-p\) - 4\ q - 9\ r\ \[Equal] 0, \(-p\) - 16\ q - 81\ r \[Equal] 0}, {p, q, r}]\)], "Input", CellLabel->"In[2]:="], Cell[BoxData[ \({{p \[Rule] 3\/2, q \[Rule] \(-\(3\/5\)\), r \[Rule] 1\/10}}\)], "Output", CellLabel->"Out[2]="] }, Open ]], Cell["\<\ The finite sum of linear fractions,\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Sum]\+\(k = n + 1\)\%\[Infinity]\((p\/\(\((k - 1)\) \((k + 1)\)\) + q\/\(\((k - 2)\) \((k + 2)\)\) + r\/\(\((k - 3)\) \((k + 3)\)\))\) /. values[\([1]\)] // Apart\)], "Input", CellLabel->"In[3]:="], Cell[BoxData[ \(1\/\(60\ \((\(-2\) + n)\)\) - 2\/\(15\ \((\(-1\) + n)\)\) + 37\/\(60\ n\) + 37\/\(60\ \((1 + n)\)\) - 2\/\(15\ \((2 + n)\)\) + 1\/\(60\ \((3 + n)\)\)\)], "Output", CellLabel->"Out[3]="] }, Open ]], Cell[TextData[{ "improves the convergence speed of ", Cell[BoxData[ \(TraditionalForm\`lim\+\(n \[Rule] \[Infinity]\)\)]], " ", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(k = 1\)\%n 1\/k\^2\)]], " much more." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"Table", "[", RowBox[{ RowBox[{"N", "[", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{ StyleBox[\(\(1\/60\) \((1\/\(n + 3\) + 1\/\(n - 2\))\)\), FontColor->RGBColor[1, 0, 0]], StyleBox["-", FontColor->RGBColor[1, 0, 0]], StyleBox[\(\(2\/15\) \((1\/\(n + 2\) + 1\/\(n - 1\))\)\), FontColor->RGBColor[1, 0, 0]], StyleBox["+", FontColor->RGBColor[1, 0, 0]], StyleBox[\(\(37\/60\) \((1\/n + 1\/\(n + 1\))\)\), FontColor->RGBColor[1, 0, 0]], "+", StyleBox[\(\[Sum]\+\(k = 1\)\%n 1\/k\^2\), FontColor->RGBColor[0, 0, 1]]}], ",", \(Zeta[2]\)}], "}"}], ",", "15"}], "]"}], ",", \({n, 30, 40}\)}], "]"}], "//", "TableForm"}]], "Input", CellLabel->"In[1]:="], Cell[BoxData[ TagBox[GridBox[{ {"1.6449340670596293750834073456`15.", "1.644934066848226436472415167`15."}, {"1.6449340670167982981917149271`15.", "1.644934066848226436472415167`15."}, {"1.6449340669836042136006533027`15.", "1.644934066848226436472415167`15."}, {"1.644934066957674827657467118`15.", "1.644934066848226436472415167`15."}, {"1.6449340669372693490535288646`15.", "1.644934066848226436472415167`15."}, {"1.6449340669210983908215356923`15.", "1.644934066848226436472415167`15."}, {"1.6449340669081984811208003981`15.", "1.644934066848226436472415167`15."}, {"1.6449340668978436887393993647`15.", "1.644934066848226436472415167`15."}, {"1.6449340668894827588191153468`15.", "1.644934066848226436472415167`15."}, {"1.6449340668826939695676026826`15.", "1.644934066848226436472415167`15."}, {"1.6449340668771524229925307532`15.", "1.644934066848226436472415167`15."} }, RowSpacings->1, ColumnSpacings->3, RowAlignments->Baseline, ColumnAlignments->{Left}], Function[ BoxForm`e$, TableForm[ BoxForm`e$]]]], "Output", CellLabel->"Out[1]//TableForm="] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["EulerGamma", "Section", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "For Euler's constant ", Cell[BoxData[ \(TraditionalForm\`lim\_\(n \[Rule] \[Infinity]\)\)]], "(", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(k = 1\)\%n 1\/k\)]], " - log n ) we utilize the following equality." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(FullSimplify[\[Sum]\+\(k = 1\)\%n 1\/k - \(Log[n] + Log[n + 1]\)\/2 == 1 - \(1\/2\) Log[2] + \[Sum]\+\(k = 2\)\%n\((1\/k + \(1\/2\) Log[\(1 - 1\/k\)\/\(1 + 1\/k\)])\), Assumptions \[Rule] {n > 0, n \[Element] Integers}]\)], "Input", CellLabel->"In[57]:=", Background->None], Cell[BoxData[ \(True\)], "Output", CellLabel->"Out[57]="] }, Open ]], Cell["\<\ We can find linear fraction terms for the acceleration of \ convergence by the same method.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Series[ 1\/k + \(1\/2\) Log[\(1 - 1\/k\)\/\(1 + 1\/k\)] - \((s\/\(\((k - 1)\) k \((k + 1)\)\ \) + t\/\(\((k - 2)\) k \((k + 2)\)\))\), {k, \[Infinity], 8}]\)], "Input", CellLabel->"In[58]:="], Cell[BoxData[ InterpretationBox[ RowBox[{\(\((\(-\(1\/3\)\) - s - t)\)\ \((1\/k)\)\^3\), "+", \(\((\(-\(1\/5\)\) - s - 4\ t)\)\ \((1\/k)\)\^5\), "+", \(\((\(-\(1\/7\)\) - s - 16\ t)\)\ \((1\/k)\)\^7\), "+", InterpretationBox[\(O[1\/k]\^9\), SeriesData[ k, DirectedInfinity[ 1], {}, 3, 9, 1], Editable->False]}], SeriesData[ k, DirectedInfinity[ 1], { Plus[ Rational[ -1, 3], Times[ -1, s], Times[ -1, t]], 0, Plus[ Rational[ -1, 5], Times[ -1, s], Times[ -4, t]], 0, Plus[ Rational[ -1, 7], Times[ -1, s], Times[ -16, t]]}, 3, 9, 1], Editable->False]], "Output", CellLabel->"Out[58]="] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(val = Solve[{\(-\(1\/3\)\) - 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[2] Jonathan Borwein and David Bailey, Mathematics by Experiment, A K Peters, \ 2004. [3] Nico M. 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