Question

i gave u challenge prove that 1+2+3+4+5+6+.......(infinity) = -1/12​

Answer 1

Answer:

S=1+2+…+(n−1)+n.

Write it backwards:

S=n+(n−1)+…+2+1.

Add the two equations, term by term; each term is n+1, so

2S=(n+1)+(n+1)+…+(n+1)=n(n+1).

Divide by 2:

S=

n(n+1)

2

challenge excepted

now mark me as brainlest and give thanks to all my answers plszzzz

Answer 2

$\star\:\:\:\sf\large\underline\blue{Question:-}$

• Prove the Ramanujan's infinite series problem.

$\star\:\:\:\sf\large\underline\blue{Answer:-}$

We have to prove,

$\boxed{ \sf1 + 2 + 3 + ... = \frac{ - 1}{12} }$

This is first proven by our Indian Mathematician, Srinivasa Ramanujan.

So, here the proof comes.

Consider these given series,

$\sf S_{1} = 1 - 1 + 1 - 1 + 1 - 1...$

$\sf S_{2} = 1 - 2 + 3 - 4 + 5 - 6...$

$\sf S_{3} = 1 + 2 + 3 + 4...$

Using these series, we will prove this infinite series problem.

Consider the first Series, $\sf S_{1}$, given that,

$\sf S_{1} = 1 - 1 + 1 - 1 + 1 - 1...$

We will find the sum of the given series.

Since this is an infinite series, we may imagine that total number of terms is odd, so,

$\sf S_{1} = 1 \cancel{ - 1 + 1} + \cancel{ - 1 + 1} +....$

So,

$\sf S_{1} = 1 ....(i)$

Also, if we consider the total number of terms is even, then,

$\sf S_{1} = \cancel{1 - 1} + \cancel{1 - 1} + ...$

So,

$\sf S_{1} = 0 ....(ii)$

As there are two solutions, we will take the average of them,

$\boxed{ \sf S_{1} = \frac{1}{2} }$

So, the value of Series is 1/2.

Now, we will find the value of the second series.

$\sf S_{2} = 1 - 2 + 3 - 4 + 5 - 6...$

$\sf S_{2} = \: \: \: \: \: \: \: \: 1 - 2 + 3 - 4 + 5 - 6...$

Adding both the series, we get,

$\sf 2S_{2} = 1 - 1 + 1 - 1 + ..$

If we take a closer look at the right hand side, we get

$\sf 2S_{2} = S _{1}$

$\sf \implies2S_{2} = \frac{1}{2}$

$\sf \implies S_{2} = \frac{1}{4}$

So,

$\boxed{ \sf S_{2} = \frac{1}{4} }$

Now,

$\sf S_{3} = 1 + 2 + 3 + 4 + 5 + ....$

$\sf S_{2} = 1 - 2 + 3 - 4 + 5 - 6...$

Subtracting both the series, we get,

$\sf S_{3} - S_{2} = 4 + 8 + 12 + 16 + ...$

$\sf \implies S_{3} - S_{2} = 4(1 + 2 + 3 + 4 + ...)$

$\sf \implies S_{3} - S_{2} = 4S _{3}$

$\sf \implies - S_{2} = 3S _{3}$

$\sf \implies S_{3} = \frac{ - 1}{4} \times \frac{1}{3}$

$\sf \implies S_{3} = \frac{ - 1}{12}$

So,

$\boxed{ \sf S_{3} = \frac{ - 1}{12} }$

So,

$\sf1 + 2 + 3 + 4 + 5 + ... = \frac{ - 1}{12}$

Hence Proved.

This was first proven by Ramanujan but this prove is somewhat crazy.