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 $$1+2+3+4+5+...+\mathrm{\infty }=?$$
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,
$$\begin{gathered} \Rightarrow S_1=1-1+1-1+1-1+... \\ \Rightarrow S_1=0 \end{gathered}$$
If $$S_1$$ has odd number of terms, then only one term will remain as
$$\begin{gathered} \Rightarrow S_1=1-1+1-1+1-1+... \\ \Rightarrow S_1=1 \end{gathered}$$
Taking average of both values of $$S_1$$ , we get
$$\begin{gathered} \Rightarrow S_1+S_1=0+1 \\ \Rightarrow 2S_1=1 \end{gathered}$$
Therefore,
$$\Rightarrow S_1={\displaystyle \frac{1}{2}}$$
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,
$$\Rightarrow 2S_2=1-1+1-1+1-1+...$$
Now since, $$S_1=1-1+1-1+1-1+...$$
Therefore, we can write
$$\Rightarrow 2S_2=S_1$$
Now since $$S_1={\displaystyle \frac{1}{2}}$$ ,
Therefore,
$$\Rightarrow 2S_2={\displaystyle \frac{1}{2}}$$
Hence,
$$\Rightarrow S_2={\displaystyle \frac{1}{4}}$$
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,
$$\Rightarrow S_3-S_2=4S_3$$
We can write is as,
$$\begin{gathered} \Rightarrow -S_2=4S_3-S_3 \\ \Rightarrow -S_2=3S_3 \end{gathered}$$
Putting the value of $$S_2={\displaystyle \frac{1}{4}}$$ in the above equation, we get
$$\Rightarrow -{\displaystyle \frac{1}{4}}=3S_3$$
Therefore, we get
$$\Rightarrow S_3=-{\displaystyle \frac{1}{12}}$$
Or we can write,
$$\Rightarrow 1+2+3+4+5+...=-{\displaystyle \frac{1}{12}}$$
That is the required sum of all the natural numbers.
Therefore, the sum of all natural numbers to infinity is $$-{\displaystyle \frac{1}{12}}$$ .

Note:
The above obtained sum $$1+2+3+4+5+...=-{\displaystyle \frac{1}{12}}$$ 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.