Question

Show that :-
\(\lim _{n \to\infty }\frac{2^{n-1}-1}{n}(\mod 13)=0\)
you can use anything you want to solve , have fun !!
Hint1 :-
\(\frac{2^{n-1}-1}{n}(\mod 13)=k \mod 13\)
\(\frac{2^{n-1}-1}{n}(\mod 13)=13 m+k\)
Hint2:-
\(\lim _{n \to\infty }\frac{2^{n-1}-1}{n}=?\)

Answer 1

question, what is mod 13 of a decimal for example
if n = 10 , plugging in
(2^(10-1) -1 ) /10 = 51.1

Answer 2

how do you calculate
51.1 mod 13 ?

Answer 3

hhmmm mod always means residues , dont give it much important
but give it a range of (0,1,.....,12)

Answer 4

ok since u know then act smart any try what u know ;
u asked

Answer 5

i'm not sure how to evaluate these decimals mod 13 , since that is what you will get

Answer 6

mod is not defined for decimals as far as i know.

Answer 7

ok
so
51.1=51+1/10=511/10 mod 13

Answer 8

ok :)

Answer 9

im asking because im not sure if the question makes sense . did you make up this problem ?

Answer 10

hey @perl if u dont know that does not mean its wrong :D
just feed ur knowledge a bit

Answer 11

i'll assume it is a well defined problem, since im no expert on number theory.
maybe ganeshie can give it a try

Answer 12

well its more analysis than being number theory

Answer 13

ok can you give a hint

Answer 14

i will make a table and plug in n = 1,2,3,...

Answer 15

:D
ok so what ever n is , u would have range of (0,1,....,12)
which is sounds like a line right ?

Answer 16

one way to do this is trying to map new function ( line ) and see

Answer 17

so what did you get for 51.1 mod 13 ? the value must be an integer
i will need this in order to proceed.

Answer 18

can we ignore the decimal .1

Answer 19

ok i added hint to the question

Answer 20

and btw residues is integer when u work with integers and have integers stuff this is how ppl used to it but it also can be in real number :D
and it might be infinite residue of some mod for irrational numbers

Answer 21

ok i have to go have fun ...

Answer 22

maybe @ganeshie8 can take a look
your hint didn't really help me

Answer 23

hmmm

Answer 24

by wolfram test the limit is DNE
http://www.wolframalpha.com/input/?i=lim+%282%5E%28n-1%29-1%29%2Fn+mod+13

Answer 25

the non modular limit does not exist

Answer 26

i have never seen a \(mod\) symbol in limit

Answer 27

hahaha! looks like this is a new dessert prepared by @Marki

Answer 28

yeah my new paper !horri

Answer 29

but woli gives me 0 take snap shot for ur result

Answer 30

here madam
http://gyazo.com/ed5bc4e71d8a1c68629c24b75c3bd3cc

Answer 31

aww yeah does not contradict to what im saying :P
ok so result from 0 to 13 try to find why it should be 0

Answer 32

"0 to 13" is a range of integers so the limit does not converge to any single number

Answer 33

or am i missing something hmm

Answer 34

hmm idk :|
this is what im thinking of anyway xD

Answer 35

idk what you're thinking hmm
why would an alternating sequence of 0->12 ever converge to 0 ?

Answer 36

some of these have defined modular output
n = 11
(2^(11-1) -1 ) / 11 = 93 = 2 mod 13

Answer 37

others are decimals , like n = 10
(2^(10-1) -1 ) / 10 = 51.1

Answer 38

ahh decimals are not a problem we can just work them as remainder :

51.1 mod 13 = 13*3 + 12.1mod 13 = 12.1

Answer 39

so the terms in the sequence will be between [0, 13)

Answer 40

you can* modify the mod formula to allow real numbers

Answer 41

http://www.wolframalpha.com/input/?i=51.1+mod+13

Answer 42

ok this is what im thinking off to be honest , so u wont think im fooling you
from hint 1 we can see that we can express mod operation as line in some space , which give me alternate function lets say \(\psi(n)\)

Answer 43

it sounded nice idea for first time xD

Answer 44

interesting, one sec im still trying to make sense of that..

Answer 45

what do these lines represent ?

Answer 46

\[\frac{2^{n-1}-1}{n}(\mod 13)=k \mod 13\]

Answer 47

Ok my system shuted down i'll resume some other time

Answer 48

Ok my system shuted down i'll resume some other time

Answer 49

Okay, the idea looks interesting but i don't fully get it yet..

Answer 50

*

Answer 51

i wonder since it is mod if we should only consider when (2^(n-1)-1)/n is an integer
so if it is an integer then ni=2^(n-1)-1 for some integer i
So we have
\[\lim_{n \rightarrow \infty}\frac{2^{n-1}-1}{n} ( \mod 13)=\lim_{n \rightarrow \infty}\frac{n i}{n} ( \mod 13)=\lim_{n \rightarrow \infty} i (\mod 13)= \\ i (\mod13) \text{ but this still isn't 0 : ( }\]

Answer 52

unless the only value i can be is 0
but that is something else i have to show (if it is even true)

Answer 53

or a multiple of 13

Answer 54

wait can we use Fermat's little theorem...
\[\text{ consider } n \text{ as odd } \\ 2^{n-1} \equiv 1 (\mod n) \\ \lim_{n \rightarrow \infty}\frac{2^{n-1}-1}{n}=\lim_{n \rightarrow \infty}\frac{1-1}{n}=\lim_{n \rightarrow \infty}\frac{0}{n}=0\]
this is probably completely wrong

Answer 55

i wanted to study number theory
just never really got much experience :(

Answer 56

or maybe I should have just said 2^(n-1) is congruent to 1 mod 13

Answer 57

@ganeshie8 does this seem right to you?

Answer 58

or is it like wolfram says

Answer 59

0 to 13

Answer 60

dne whatever

Answer 61

fermat's little theorem is applicable only when \(n\) is prime
\[2^{p-1} \equiv 1 \pmod p\]
is true for all odd primes \(p\)

in other words, \(2^{p-1}-1\) is divisible by \(p\) for all odd primes

Answer 62

for our present problem we can only say
\(2^{n-1}-1\) is divisble by \(13\) only when \(n-1\) is a multiple of \(12\)

Answer 63

for example

\(2^{13-1}-1\) is divisbible by \(13\)

Answer 64

\(2^{2(13-1)}-1\) is divisbible by \(13\)

Answer 65

\(2^{3(13-1)}-1\) is divisbible by \(13\)

Answer 66

etc...

Answer 67

but the remainders for other powers can be anything..

Answer 68

I see

Answer 69

fermat tells us only this much :

\(2^{k(13-1)}-1\) is divisible by \(13\) , for all \(k \in \mathbb{N}\)

Answer 70

so wolfram is most likely right in this case

Answer 71

since that one thingy will probably not be congruent to 1 for other integer n (well you know besides the odd integer n)

Answer 72

yes the remainder could be any real number between 0 and 13

Answer 73

But if we were given that n was odd
and n approaches large odd numbers then we could say the limit is 0