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Question Number 2138 by Filup last updated on 06/Nov/15

Is the following proof correct ?

Δ = \underset{i = 0}{\sum^{\infty}} {(- 1)}^{i} 2^{i} = 1 - 2 + 4 - 8 + 16 - 32 + \dots

Let :

Δ_{1} = 1 - 2 + 4 - 8 + 16 - 32 + \dots

Δ_{2} = 1 - 2 + 4 - 8 + 16 - 32 + \dots

Δ_{1} + Δ_{2} = 1 + (- 2 + 1) + (4 - 2) + (- 8 + 4) + \dots

∴ Δ_{1} + Δ_{2} = 1 - (1 - 2 + 4 - 8 + \dots)

∴ Δ_{1} + Δ_{2} = 1 - Δ_{1}

Δ_{1} = Δ_{2} = Δ

3 Δ = 1

∴ Δ = \frac{1}{3}

∴ Δ = \underset{i = 0}{\sum^{\infty}} {(- 1)}^{i} 2^{i} = \frac{1}{3}

Commented by prakash jain last updated on 04/Nov/15

You can NOT add 2 divergent series and expect

reesult to be double . Same for subtraction .

Commented by Filup last updated on 04/Nov/15

A proof of 1 + 2 + 3 + 4 + 5 + \dots = - \frac{1}{12}

involves this method, though .

Also, according to {wikipedia}^{'} s page for

this sequence, this answer is correct .

Commented by Filup last updated on 04/Nov/15

Z e r o s r e p r e s e n t s p a c e s

S_{1} = 1 + 2 + 3 + 4 + 0

S_{2} = 0 + 1 + 2 + 3 + 4

I am usure as to why the attempted

proof is incorrect dispite getting the

same result as even Euler

Commented by prakash jain last updated on 04/Nov/15

Ok . You are talking about a different sum .

I misunderstood .

Commented by 123456 last updated on 04/Nov/15

other sumations methods and analitic clntinuation

Commented by Filup last updated on 04/Nov/15

Because i am only an amature mathematian,

I {don}^{'} t really understand what it is

I have done . Please explain how this is

a different type of sum ?

Commented by prakash jain last updated on 04/Nov/15

My mistake . I misread your question . Also

ignore my previous comment about

divergent series .

Commented by Filup last updated on 04/Nov/15

Ok thank you very much

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