Almost all the high school maths students who tried this problem said there has to be some mathematical error associated with the proof of this statement because in no way the sum of positive numbers can turn out to be a negative fraction. What do you think ? Is it incorrect? And if not, can you prove it? Try yourself before seeing the explanation.

**Explanation** : Before we start, one can make certain direct assumptions just by looking at the series :

- The given series is an
**arithmetic progression**; a kind of sequence where each succeeding term differs from it’s preceding term by a**common difference**. Almost every high school maths student knows that sum of**‘n’**terms of arithmetic progression that is given below would be**n(n+1)/2**.

1 + 2 + 3 + 4 + .... + n =n(n+1)/2

- When
**‘n’**tends to infinity, the sum would definitely tend to infinity but that is**NOT**the answer. If not that then atleast the sum of positive numbers should produce a**POSITIVE**answer but that is not the case either.

**So how come is it true?**

**So how come is it true?**

The idea behind this is that the sum of positive numbers can be negative only at one condition.

This can be true only if we redefine the equals sign meaning from

Physics Central.to“equals to”.“is associated with”

**Let’s do it!**

**Let’s do it!**

We will solve the given series by solving for following two different series and then combining their values to obtain the required result.

Sum1= 1 - 1 + 1 - 1 + 1 - ...

Sum2= 1 - 2 + 3 - 4 + 5 - ...

**Value of Sum1**

**Value of Sum1**

Sum1= 1 - 1 + 1 - 1 + 1 - ...

- Multiplying by -1 both sides :

-Sum1= -1 + 1 - 1 + 1 - ...

- Adding
**1**to both sides :

1 -Sum1= 1 - 1 + 1 - 1 + 1 - ...

- The R.H.S is nothing but
**Sum1**itself, hence :

1 -Sum1=Sum1

i.e 2Sum1= 1

i.eSum1= 1/2

Hence, **Sum1 = 1 – 1 + 1 – 1 + … = 1/2**. Now we will solve for **Sum2**.

*Value of Sum2*

*Value of Sum2*

As we know,

Sum1= 1 - 1 + 1 - 1 + ...

and

Sum2= 1 - 2 + 3 - 4 + 5 - ...

- Subtracting
**Sum2**from**Sum1**would give :

Sum1-Sum2=(1 - 1 + 1 - 1 + 1 - ...)-(1 - 2 + 3 - 4 + 5 - ...)

i.eSum1-Sum2=(1 - 1 + 1 - 1 + 1 - ...)- 1 + 2 - 3 + 4 - 5 + ...

- Rearranging the terms a little bit gives us :

Sum1-Sum2=(1 - 1)+(-1 + 2)+(1 - 3)+(-1 + 4)+(1 - 5)+ ...

i.eSum1-Sum2= 0 + 1 - 2 + 3 - 4 + ...

- The R.H.S is nothing but
**Sum**2 itself, hence :

Sum1-Sum2=Sum2

i.e 2Sum2=Sum1

- Putting the value of Sum1, i.e
**1/2**in above equation :

2Sum2= 1/2

i.eSum2= 1/4

Hence, **Sum2 = 1 – 2 + 3 – 4 + … = 1/4**. Now we can show you how Ramanujan arrived at the final sum.

*Ramanujan’s Proof*

*Ramanujan’s Proof*

Let** ‘S’** be the required sum. That is :

S= 1 + 2 + 3 + 4 + 5 + ...(I)

- Multiplying above equation by
**4**and shifting the terms :

4S= 4 + 8 + ...(II)

- Subtracting
**II**from**I**:

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

- The R.H.S is nothing but
**Sum2**i.e 1/4, hence :

-3S= 1/4

i.eS= -1/12

Hence we have finally proved that :

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

For more information on Ramanujan’s Sum, must refer to WIKIPEDIA article **here**.

Do share with your high school mathematics friends to try this out.

It should be infinite