Question

prove that 1+2+3+4+infinity =. -1/12 it's proven by Ramanujan if u can't don't answer
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Answer 1

hi here's your 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⋯)

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⋯

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 = -1/12

❤️❤️ hope it helps you

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.