Ramanujan Summation

An overated mistake ?

In 1918 ,In a paper submitted by Srinivasa Ramanujan there was some highly controversial summation which socks the world of mathematics at that point of time.

but,
we know that sum of “finite” consecutive natural numbers is n(n+1)/2 ,And when n →∞
it obviously should be infinity ,
but,
the answer as given by Ramanujan, and later verified by another notable Mathematician Reimann, is – 1/12.Yes, not only is the answer a real number, it is a "negative” rational number!

So, were Ramanujan and Reimann wrong!

Let us consider few series,

1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 …….. = 1/4

C = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 ………..

C = 0 + 1 - 2 + 3 - 4 + 5 - 6 + 7 …………

(Now add the two series in the certain way that 1 and 0, -2 and 1 , 3 and -2 , -4 and 3 and so on....)

2 C = 1 - 1 + 1 - 1 + 1 - 1 + 1 ………

-2 C = - 1 + 1 - 1 + 1 - 1 + 1 - 1 ……. (Multiplying -1 on both side )

1 - 2 C= 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 …… (Adding +1 on both side )

(Here the pattern re-appears)

1 - 2 C= 2C

1 = 4 C

C = 1/4

1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 …….. = 1/4

Using this result we obtain ,

1 + 2 + 3 + 4 + 5 + 6 +............ = -1/12

let, 

S= 1 + 2 + 3 + 4 + 5 + 6 + ..... _____________(1)

4S=     4       + 8    + 12 + ....._____________(2)

Now substracting (2) from (1) we get
(substract in a certain way that 2 and 4,4 and 8, 6 and 12 and so on.... )

S - 4S = 1 - 2 + 3 - 4 + 5 - 6 +.....

-3S = 1 /4    (from above series)

S= -1/12

Boom! 

Now, there are many steps may seem rather doubtful, but after multiple observations, we can find that they do make “some” sense.

Infinity is an extremely confusing Mathematical term, and when it is associated with a series, it does not generally give us the answer we expect. Such is the legacy of this term. Until someone actually comes up with a proper contradiction, we won’t take Ramanujan Summation as an overrated mistake, but rather as an astounding step in the ever-intriguing world of Mathematics!

-:THANKS:-

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Perfect Number

A number which is equal to the sum of all of its proper divisors is called Perfect number.

Proper divisor:- The positive divisor of a number excluding itself.

Let N be a number and d₁,d₂,d₃.....d_m be the proper divisors of N.

N is a perfect number if

N = d₁ + d₂ + d₃ + ......... + d_m

_{_________________________}

Example:  6, 28, 496, 8128, 33550336

The proper divisors of 6 are 1,2 and 3

6=1+2+3

Since 6 is the sum of all of its Proper divisors hence it’s a Perfect Number.

Question: Is 24 a Perfect Number?

The proper divisors of 24 are 1,2,3,4,6,8,12.

24 ≠ 1+2+3+4+6+8+12

Since the sum of proper divisors of 24 are nor equal to 24, It’s not a Perfect Number

_{_________________________}

Euclid proved that all even Perfect Number is of the form

N = 2^P(2^P-1)

Where P is a Prime for which (2^P-1) is a Mersenne Prime.

Mersenne Prime:- The Prime number which is of the form 2ⁿ-1; where n is an Integer.

Facts:

• 6 is the smallest Perfect number.
• There are 51 known perfect number.
• The latest Perfect Number is discovered in 2018 which has 49,724,095 digits.

                                                    

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How to check whether a number is Prime or not?

Let N be a number.

Do the following steps to check if N is a prime or not.

1. Find the square root of N. 

      Let the square root of N be m. 

           √N=m

2. If m is a natural number

   then N is a composite number since the root of N is a whole number and it's obvious that m is a factor of N.

else

3. If m is not a natural number (i,e. if m is a fraction)

→increase it to the next natural number.

    let it be m'.

→Then divide the given number (N) by all the prime numbers below m'.

→If N is divisible by any of these prime numbers then it is a composite number else it is a prime number.

Example 1:

Check if 289 is prime or not.

Find the square root of 289

√289=17

Since the root (17) is a natural number, it's also a divisor of 289.

Hence 289 is not a prime number.

Example 2:

Check whether 347 is prime or not.

Find the square root of 347

√347=18.627

The square root of 347 is a fractional number

So increase the root to the next natural number.

18.627≃19

Now divide 347 by all primes below 19.

The primes below 19 are 2,3,5,7,11,13,17

347/2 (Remainder 1)

347/3(Remainder 2)

347/5(Remainder 2)

347/7(Remainder 4)

347/11(Remainder 6)

347/13(Remainder 9)

347/17(Remainder 7)

So none of the primes below 19 divides 347.

Hence 347 is a Prime Number.

There is always 6174

Starting with any four digit number (that has at least two different digits) just follow the following steps:

1. Arrange the digits of the four digit number in descending/ascending order to make the largest and smallest numbers possible.
2. Subtract the smaller number from the larger one.
3. Take the answer and repeat the process.

Eventually you'll end up at 6174. 

it never takes more than seven stages to get there.

example:

Picking a number at random, 

let's try 4551,

Stage 1: 5541-1455 = 4086
Stage 2: 8640 - 0468 = 8172
Stage 3: 8721 - 1278 = 7443
Stage 4: 7443 - 3447 = 3996
Stage 5: 9963 - 3699 = 6264
Stage 6: 6642 - 2466 = 4176
Stage 7: 7641 - 1467 = 6174

Here the number "6174" is known as 'Kaprekar's Constant'

How to find the number of Positive Divisors of a Number?

Let N be the number.

Now express N as the product of Prime factors

N=P₁^A1 * P₂^A2 * P₃^A3 * .........       

(Where P_i are the primes 
              Aj are the powers of respective primes,

              i =1,2,3....   and   j= 1,2,3...)

Let  τ(N) denote the number of positive divisors of n, then

τ(N)= (A1 +1) (A2 +1) (A3 +1)......

Example 1:

Find the number of positive divisors of 108.

Here N=108 

→Now write the number as the product of prime factors.

So 108 can be written as

108=2²×3³

The number of positive divisors of 108 is 

τ(108)= (2+1)(3+1) = 3×4 = 12   

    

Example 2:

Find the number of positive divisors of 50000.

Let N=50000

 50000 =2⁴×5⁵

The number of positive divisors of 108 is

τ(50000) = (4+1)(5+1) = 5×6 =30


              
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Composite Number

A whole number greater than 1 which has more than 2 factors is called Composite number.

or

If a number has a factor other than 1 and itself then the number is said to be a Composite number.
Otherwise the number is called Primer Number.
Example:- 4,6,8,9,10,12,14,15......

FACTS:

• 4 is the smallest composite number.
• Every even number except 2 is a composite number.

                                     

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There Are Infinite Numbers Of Primes

Proof:

let us assume that there are finite numbers of primes,                                                         
       say it be  P₁,P₂,P₃,_{.............}P_K

       let  N=  P₁P₂P_{3.............}P_K + 1
       If N is a prime number and  N > P_i,
                    where i=1,2,3,4.............k
then it contradicts our assumption that we have finite number of primes.

So if N is a composite number, 
let P_j will divide N for some 1<j<k
=>P_j|N
=>N=P_j*t , where "t" is an integer
=>P₁,P₂,P₃,_{.............}P_K =P_j*t
=>P₁P₂P_3.....P_j......P_K + 1 = P_j*t

Devide both sides by P_j
=>(P₁P₂P_3.....P_j.......P_K + 1)/P_j  = t

=>P₁P₂P_3.....Pj-1.P_j+1.......P_K + 1/P_{j =   t}

since "t" is an integer so
P₁P₂P_3.....P_j-1.P_j+1.......P_K + /P_j will be an integer. 
which is not possible
So N must be a prime number
this is a contradiction to our assumption that N is composite.

=> N = P₁P₂P_{3.............}P_K + 1 is also a prime.
Hence there are infinite number of primes.


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