Evaluate this limit:
limit {(2+√2)n}
n→∞
where, {x} = x - floor(x)
As n gets larger the value of this fractional part becomes 1 - ε where ε approaches zero. For example,When n = 1000, the full (2+sqrt(2))
n is
19529142296286760679268323119916571338495126477834424524628601889682092840798095
39016344151687443348340370309980573782394761389389389899383440438171101824955415
45873951120855737792902146444289225094086638682907477039483129623514289156872892
57744634061380472425602852865710637623745075582380595132793944728714251387112023
80656299085683810928133792208678901022721108099955662163433403594258944062874262
75348093420792428239641529980982303364080744270642133557637353757989473214815170
926301375358866671270738127533577148860973029601050623.9999999999999999999999999
99999999999999999999999999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999451328383535818558201173088465510
46068957849436583055625877948072491030077798465862342981453232969742983959129227
85512770994478152681775786147665301654695950736397939638723230590619788074237548
39800929236435189992954389526169755731470328271782588105459939439714564176087836
94722612818887965467690818639358989339786590671879010960941443890620035744029336
95471388537718334746150840109956460682439329242846843329226064438090403848130392
11103652184612466036621178367879861719832289473001576641554622196475808915388538
84386566497383177169302623906353729016702809934902649573412783883598765448741533
49420601345854223730764314647251884366765333841152267892639267390425074721871338
57899018435196865476846307896079242525597797571546939479118272310203047595068932
40547168422522840667683721844094783001066485384301874395225235853944309323915369
81347084543706677222499194405448142055700115674257557250398714198823301785649890
65030614682419601676113806401364078748959937185915487455392322144874156992860655
22406693470961020255813694024576015645833664564834774223770938303462745400064018
88767654145682866559857904941734701442051890814831093330051632392281174679905741
39159967296465692215018467570497336280991262379062917247517106605957713756769652
80938920870059575082635457007138398456906212499734904600035807274021014943960729
44803491353505082545281782818839977007289073319924870108210742508573783678121127
36433899850885606378273792382884472975105527747135333042746062490796347551601473
97136080797816698307609145681062261794770266422850798217502599284656469241553807
99579226277116001938966150787795787835244067687513450139995541146146253518824388
15123354614314899888268417230584391897914603920306049415923661818...
(3000-digit precision was used and the last 535 digits were discarded to account for 534 digits precision lost to the portion before the decimal point.)
So the limit is the same as for.9
.99
.999
.9999
.99999
...
but is just sloppier with all those digits after the 9's.
This is a seeming paradox, as we're presumably looking for a fraction less than 1, but the limit is in fact 1, as it does not actually have to be achieved for a finite value of n.
Edited on November 17, 2022, 10:00 am
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Posted by Charlie
on 2022-11-17 09:46:11 |
solution
