The following question is in relation to alternating series . Please show the wo
ID: 3124218 • Letter: T
Question
The following question is in relation to alternating series. Please show the working out, clearly. Much appreciated:
As we have seen, it is usually quite tricky to figure out how many terms of a convergent infinite series need to be added to be able to guarantee that this partial sum coincides with the true sum in the first k digits of its decimal expansion. However, for certain series this is a very easy task. Here is an example. We know that 1 1 3 5 9 As you can easily check this series has the following two properties: 1. The terms alternate between being positive and negative numbers; 2. the absolute values of the terms are decreasing; 3. the terms have the limit 0 3This series is of course much prettier than Ramanujan's. However, it converges VERY slowly and is absolutely useless when it comes to really approximating w.Explanation / Answer
The first 2000 digits of 1/ are:
0.318309886183789986450799312456667680370035994643
55914711875563296632077006466821840248895081249479
12560548543265194753809537933816633014208589648181
79240423070178347044029966705371947416879710093186
40235152047890057031577366743598527601543123554220
33950361445658778353194236518125075746925561174739
38592736411972378031983731990455477701942970278328
45397536071900322598823314007544570546317168348617
89896985258934764288258690431430998251555306661596
23213601605729442025449241172958026253758671530587
As given in the seemingly crazy formula,
when n = 0
1/ = [ 163096908 ] / [ 6403203/2 ] = 0.318309886183796652607706183787889598854733200434
99577265757664322059320118513690824594658037452742
74357549537695363633119805972526807289099551241308
77252483979893869649447762513739406139535567591951
55686631766657723015464351811052470555686647966896
28669035019215076922821272234291218839006997918801
62284088985031496456663306077929783955642489376304
17464192869417114193389937870149268498636032313600
91587952191173118660358005470092559889002595984985
52714534109109731875613123229702419387117209318194
So we find disparity in the 14th decimal place
when n=1
The second term
= [ (-1) * 12 * 6! (13591409 + 545140134) ] / [ 3! 1! 6403209/2 ]
= [ -8640 * 558731543 ] / [ 6 * 6403209/2 ]
= [ -804573421920] / [ 18095625621654356959022098935941777779064832000000
0001/2]
= 0.000000000000005981069938657074052378697026217828
50742959152122730842427987207054294832553747219318
90717928578674270606434349458713621488446223624574
40528818275371648675220825130478323870558659714087
48138138847477724341845004098931530923445389393452
22951674555808769756242456630031993466817523119264
61782827631873497944280668264444772652215585757540
55926359193248611112269394397009176702208611352896
86587373202491960257050854568009267865941170477128
97341428024404598776729865126294373037317662396656
Observe that the first fourteen decimal places are zeros.
The third term
= [ 12 * 12! (13591409 + 545140134*2) ] / [ 6! (2!)3 64032015/2 ]
= [ 12! * 13246460124] / [ 5760 * 64032015/2 ]
= [ 479001600 * 13246460124 ] / [ 5760 * 64032015/2 ]
= [ 1101575623911840 ] / [ 12472571560196196319592288504749153247964447949210
68388944443347008749568000000000000000 ]1/2
= 0.000000000000000000000000000031191503911212483199
57373094131992369267016532488344124638393247533229
67736367708837018105977911533451938568707235591951
37229844717808411098953461874324576088370744056940
78477263796212997460403696263765074046011728081417
79123591924923242358332126664878434359701766185551
24982082638270307736006621706873208853198025712591
75733639493606946127687455271082381067587997423867
15653940443639694034521344243095203685223322146761
69562360110634851932232557279843133390594117513478
This number has 28 leading zeros.
So each iteration corrects values to 14 decimal places.
[1000/14] = 71
So for n = 72, we can get a value for 1/ which is insanely accurate to 1000 decimal places.
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