Question

its very amazing question that
what is the sum of no. 1+2+3+4+5+6+7+8+9................so on to infinity

Answer 1

Let S = 1+2+3+4+5+6+7

Consider S1= 1-1+1-1+1-1+1-1…..

Now, this sum should be 0 or 1 based on number of natural numbers taken. If infinite numbers are even, S1=0, if odd S1=1. But, Riemann zeta function gives it a value of ½. Mathematical community too agrees that the sum is ½. How? Serious mathematical work went into the proof. If you are interested to know, please go through Ramanujan’s summation principles and zeta function.

S1=1-1+1-1+1-1+1…..

1-S1=1-(1-1+1-1+1-1+1…)

1-S1=1-1+1-1+1-1+1…..

1-S1=S1

So, S1=1/2

We need much powerful tools in mathematics like zeta functions to come to unique solution of ½. For now, we could agree S1=1/2.

Let S2=1-2+3-4+5-6+7…..

So, S2=1-2+3-4+5-6+7-8+9…..

S2= 1-2+3-4+5-6+7-8……. I have shifted RHS by a unit position

+ 2S2=1-1+1-1+1-1+1…..

Hence, 2S2=S1

Therefore, S2=1/4

Let’s come back to our sum of infinite numbers.

S=1+2+3+4+5+6+7+8+9…..

S2=1-2+3-4+5-6+7-8+9….

So, S-S2=4+8+12+16+20…..

Hence, S-S2=4(1+2+3+4+5+6+7+8….)

S-S2=4S

So, -S2=3S

And, S=-S2/3=-1/12

Hope it helps! :)