Question

i gave u challenge prove that 1+2+3+4+5+6+.......(infinity) = -1/12. it's proven by Ramanujan​

Answer 1

Answer:

Now for the icing on the cake, the one you’ve been waiting for, the big cheese. Once again we start by letting the series C = 1+2+3+4+5+6⋯, and you may have been able to guess it, we are going to subtract C from B.

B-C = (1–2+3–4+5–6⋯)-(1+2+3+4+5+6⋯)

Because math is still awesome, we are going to rearrange the order of some of the numbers in here so we get something that looks familiar, but probably wont be what you are suspecting.

B-C = (1-2+3-4+5-6⋯)-1-2-3-4-5-6⋯

B-C = (1-1) + (-2-2) + (3-3) + (-4-4) + (5-5) + (-6-6) ⋯

B-C = 0-4+0-8+0-12⋯

B-C = -4-8-12⋯

. If you notice, all the terms on the right side are multiples of -4, so we can pull out that constant factor, and lo n’ behold, we get what we started with.

B-C = -4(1+2+3)⋯

B-C = -4C

B = -3C

And since we have a value for B=1/4, we simply put that value in and we get our magical result:

1/4 = -3C

1/-12 = C or C = -1/14

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.