Question

What is the sum of all natural numbers to infinity?

Answer 1

Hint: In the number system, the positive integers i.e. all the integers greater than zero are called as natural numbers. In the above question, we have to find the sum of all natural numbers.
So we have to find the value of \[1 + 2 + 3 + 4 + 5 + ...\] to infinity. This is a series with infinite terms which is called an infinite series.

Complete step by step solution:
We have to find the value of \[1 + 2 + 3 + 4 + 5 + ... + \infty = ?\]
To approach our solution, we will consider 3 infinite series.
First, consider a series \[{S_1}\] with infinite terms as
\[ \Rightarrow {S_1} = 1 - 1 + 1 - 1 + 1 - 1 + ...\]
Since we do not know if infinite is an even or odd number, we will consider both one after another.
Now if \[{S_1}\] has even number of terms, then all terms will cancel out and we will get,
\[
   \Rightarrow {S_1} = 1 - 1 + 1 - 1 + 1 - 1 + ... \\
   \Rightarrow {S_1} = 0 \\
 \]
If \[{S_1}\] has odd number of terms, then only one term will remain as
\[
   \Rightarrow {S_1} = 1 - 1 + 1 - 1 + 1 - 1 + ... \\
   \Rightarrow {S_1} = 1 \\
 \]
Taking average of both values of \[{S_1}\] , we get
\[
   \Rightarrow {S_1} + {S_1} = 0 + 1 \\
   \Rightarrow 2{S_1} = 1 \\
 \]
Therefore,
\[ \Rightarrow {S_1} = \dfrac{1}{2}\]
Now consider another series \[{S_2}\] as,
\[ \Rightarrow {S_2} = 1 - 2 + 3 - 4 + 5 - 6 + ...\]
We can also write is as,
\[ \Rightarrow {S_2} = 0 + 1 - 2 + 3 - 4 + 5 - 6 + ...\]
Now adding both values of \[{S_2}\] , we get,
\[ \Rightarrow {S_2} + {S_2} = \left( {1 + 0} \right) + \left( { - 2 + 1} \right) + \left( {3 - 2} \right) + \left( { - 4 + 3} \right) + \left( {5 - 4} \right) + \left( { - 6 + 5} \right) + ...\]
That gives,
\[ \Rightarrow 2{S_2} = 1 - 1 + 1 - 1 + 1 - 1 + ...\]
Now since, \[{S_1} = 1 - 1 + 1 - 1 + 1 - 1 + ...\]
Therefore, we can write
\[ \Rightarrow 2{S_2} = {S_1}\]
Now since \[{S_1} = \dfrac{1}{2}\] ,
Therefore,
\[ \Rightarrow 2{S_2} = \dfrac{1}{2}\]
Hence,
\[ \Rightarrow {S_2} = \dfrac{1}{4}\]
Now consider \[{S_3}\] as
\[ \Rightarrow {S_3} = 1 + 2 + 3 + 4 + 5 + ...\]
Since \[{S_2} = 1 - 2 + 3 - 4 + 5 - 6 + ...\]
Then subtracting \[{S_2}\] from \[{S_3}\] gives
\[ \Rightarrow {S_3} - {S_2} = \left( {1 - 1} \right) + \left( {2 + 2} \right) + \left( {3 - 3} \right) + \left( {5 - 5} \right) + \left( {6 + 6} \right)...\]
That gives,
\[ \Rightarrow {S_3} - {S_2} = 4 + 8 + 12 + 16 + 20 + ....\]
Or
\[ \Rightarrow {S_3} - {S_2} = 4\left( {1 + 2 + 3 + 4 + 5 + ...} \right).\]
Since, \[{S_3} = 1 + 2 + 3 + 4 + 5 + ..\]
Therefore,
\[ \Rightarrow {S_3} - {S_2} = 4{S_3}\]
We can write is as,
\[
   \Rightarrow - {S_2} = 4{S_3} - {S_3} \\
   \Rightarrow - {S_2} = 3{S_3} \\
 \]
Putting the value of \[{S_2} = \dfrac{1}{4}\] in the above equation, we get
\[ \Rightarrow - \dfrac{1}{4} = 3{S_3}\]
Therefore, we get
\[ \Rightarrow {S_3} = - \dfrac{1}{{12}}\]
Or we can write,
\[ \Rightarrow 1 + 2 + 3 + 4 + 5 + ... = - \dfrac{1}{{12}}\]
That is the required sum of all the natural numbers.
Therefore, the sum of all natural numbers to infinity is \[ - \dfrac{1}{{12}}\] .

Note:
The above obtained sum \[1 + 2 + 3 + 4 + 5 + ... = - \dfrac{1}{{12}}\] is known as the Ramanujan’s sum of natural numbers, named after the great Indian mathematician Srinivasa Ramanujan.
A sum of numbers is called a series and when there are infinite terms in the series then it is called an infinite series. If the sum of the series is finite then it is called a convergent series, otherwise it is called a divergent series if the sum is infinity or not defined.