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# Infinity..........

$${ 1 }^{ 2 }-{ 2 }^{ 2 }+{ 3 }^{ 2 }-{ 4 }^{ 2 }+{ 5 }^{ 2 }-{ 6 }^{ 2 }.........$$

Note by Spandan Barhai
2 years, 10 months ago

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Read Grandi's Series for questions like these.

The sum you've written isn't defined, since its a divergent series. But, if you assume it has a value $$S$$, you can find its value.

$$S = 1^2 - 2^2 + 3^2 - 4^2 \dots$$

$$S = 0^2 + 1^2 - 2^2 + 3^2 \dots$$

$$2S = 1 - 3 + 5 - 7 + \dots$$.(by adding downwards)

Define $$T = 1 - 3 + 5 - 7 + \dots$$

Therefore $$2S = T$$, now

$$T = 1 - 3 + 5 - 7 + \dots$$

$$T = 0 + 1 - 3 + 5 + \dots$$

$$2T = 1 - 2 + 2 - 2 + 2 + \dots$$ (adding downwards)

$$2T = 1 - 2(1 - 1 + 1 - 1 + \dots)$$

$$2T = 1 - 2(V)$$

Where $$V$$ is Grandi's series, which has a value of $$\frac{1}{2}$$

$$\therefore 2T = 1 - 2(\frac{1}{2})$$

$$\implies 2T = 0$$

$$\implies T = 0$$

Now, $$2S = T$$

$$\implies 2S = 0$$

$$\implies S = 0$$

$$\implies1^2 - 2^2 + 3^2 - 4^2 \dots = 0$$

- 2 years, 10 months ago

Why is $$( 1 - 1 + 1 - 1 + \ldots ) = \frac {1 }{2}$$
It is really nonsense

- 2 years, 10 months ago

Read the link I gave in the original post. It deals with the same question. TL;DR of the link,

The series 1 − 1 + 1 − 1 + … has no sum.

...but its sum should be 1/2

Simple way to see this, Let the sum be $$A$$.

$$A = 1 - 1 + 1 - 1 + \dots$$

$$A = 1 - (1 - 1 + 1 - 1 + \dots$$

$$A = 1 - A$$

$$2A = 1$$

$$A = \frac{1}{2}$$

- 2 years, 10 months ago

Isn't that infinite sequence is a geometric sequence?

A= 1 + 1(-1)+1(-1)^2+1(-1)^3 .... and the formula of infinite geometric sequence is S = a/(1-r) where a = 1 and r = -1

therefore :

A = 1/(1 - (-1)) = 1/2

- 2 years ago

Kind of...

The formula of the sum of an infinite geometric sequence is only valid when $$|r| < 1$$. So you can not use it here.

But, since the method used here to find the value of A is similar to how the formula is derived, they both will give the same answer.

Thank you

- 2 years, 10 months ago

Thanks Everyone!!!

- 2 years, 10 months ago

bro which class r u in

- 2 years, 10 months ago

class 9. I got this question in a math quiz...

- 2 years, 10 months ago

We can write 1^2-2^2+3^2-4^2+5^2-6^2+.... as (1+2)(1-2)+(3+4)(3-4)+(5+6)(5-6)+... = -(1+2+3+4+5+6+....)

=-(n(n+1)/2)=-n(n+1)/2

- 2 years, 10 months ago

$$1^2-2^2+3^2-4^2+5^2-6^2...\\=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+...\\=-1(1+2+3+4+5+6+...)\\=-\infty$$

- 2 years, 10 months ago

You could also do this

$$1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 +7^2 \dots$$

$$= 1^2 +(-2^2 + 3^2) + (- 4^2 + 5^2) - (6^2 +7^2) \dots$$

$$= 1 + (3-2)(3+2) + (5-4)(5+4) + (7-6)(7+6) \dots$$

$$= 1 + 5 + 9 + 13 + \dots$$

$$= \infty$$

- 2 years, 10 months ago

It depends on the parity of number of terms. If ∞ is taken as even then mine is correct otherwise yours.

- 2 years, 10 months ago

when you find the differences of the pairs like 1^2-2^2 and 3^2 -4^2, you get 3 - 7 -11 - 15 -19... so on. So, u can solve it using it as an arithmatic progression.

- 2 years, 10 months ago

its easy ,just use $$a^2-b^2=[a-b][a+b]$$ and take -1 common from all terms and then u will get a AP series

- 2 years, 10 months ago