Let us take a geometric progression \(T_n= 1,2,4,8,16....\infty\) and let \(S\) be it's sum

I.e \(S = 1+2+4+8+16+....\infty ----(1)\)

\(2S=2+4+8+16+....\infty\)

adding 1 on both the sides we have \(1+2S = 1+2+4+8+16+....\infty ----(2)\)

Now \((2)-(1)\) will give\(1+2S-S= 0\)

and thus \(S= “-1" \)
So where is the fallacy \(“comment"\)

## Comments

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TopNewestI guess that you cannot let \(S=1+2+4+8+16+\ldots \infty\) because the sum of the geometric progression does not converge (approaches infinity). Also, you cannot subtract equation \((2)\) with \((1)\) because both \(S\) and \(2S\) do not exist. – Svatejas Shivakumar · 1 year, 1 month ago

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– Atul Shivam · 1 year, 1 month ago

I think you mean to say ,we can't find sum of geometric progression which converges to infinity!!! Is it so ???Log in to reply

– Svatejas Shivakumar · 1 year, 1 month ago

Yes sorry for not being clear.Log in to reply

Got a reason. If S is infinity. Then 2*infinity is also infinity. This means there will be nothing like 2S for infinity.Also you can't apply operations like addition or subtraction etc to infinity. If you will say that assuming n terms where n tends to infinity and apply operations. Then the result will be something which is likely to be. – Sachin Vishwakarma · 1 year, 1 month ago

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Before writing the statement "S is equal to.." you must prove the convergence of the series. If the series converges, then only it has a limiting value, which you may call S. If you assume that S exists before proving its existence, then anomalous results as the above may pop up. Here the series does not converge (rather it diverges) simply because its the sequence of its partial sums is not bounded. – Kuldeep Guha Mazumder · 1 year ago

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It looks like to be on the same pattern as we derive formula for infinite sum of G.P. But there was |r| < 1. We can't use this pattern in such case where |x| > 1. Because the terms are increasing at a large rate and uncertainty of one last infinitely large term may result in false value. – Sachin Vishwakarma · 1 year, 1 month ago

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– Atul Shivam · 1 year, 1 month ago

I don't think so as we can add or subtract anything to infinity and it will not affect anythingLog in to reply

– Sachin Vishwakarma · 1 year, 1 month ago

For |x|> 1 G.P. diverges. So, sum can't be found.Log in to reply

– Atul Shivam · 1 year, 1 month ago

What do you want to say, Can you please elaborateLog in to reply

– Sachin Vishwakarma · 1 year, 1 month ago

Well, i just want to say that infinity doesn't follow elementary operations. I need some time to give a proper explanation. I am very sorry that i couldn't explain it.Log in to reply

– Atul Shivam · 1 year, 1 month ago

Oh!!! Is it soLog in to reply

Uncertainty of one term may result in this false value. Because if you get to same approach by using sigma instead of value. You will get the desired result. Explain me if you know the exact reason. – Sachin Vishwakarma · 1 year, 1 month ago

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Because infinite does not follow general Algebraic rule of operations(+,-,x,÷). – Krishna Sharma · 1 year, 1 month ago

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– Atul Shivam · 1 year, 1 month ago

Hey are you still in ISM dhanbad ???Log in to reply

I.e \(S= (1+\frac {1}{2}+\frac {1}{4}+\frac {1}{8}+\frac {1}{16}......)----(1)\) \((2S =2+1+\frac {1}{2}+\frac {1}{4}+\frac {1}{8}+\frac {1}{16}......)----(2) \) Now \(2S-2= (1+\frac {1}{2}+\frac {1}{4}+\frac {1}{8}+\frac {1}{16}......)\)

\(2S=4\) and hence \(S= 2\) which is absolutely correct as \(S\) has actual value of \(+2\). Why it seems to be correct and not the above one

sorry for my poor English I am not very much fluent \(:-) \) – Atul Shivam · 1 year, 1 month ago

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– Krishna Sharma · 1 year, 1 month ago

That's what I said earlier your procedure is correct when S is finite but same thing is not applicable when S is infinite because the value of \(\displaystyle \infty - \infty \) cannot be determined. I am in 1st year of college(my age is 18).Log in to reply

– Atul Shivam · 1 year, 1 month ago

I was just asking because I am in the same city:)Log in to reply

– Atul Shivam · 1 year, 1 month ago

Congratulations for ur first step towards successLog in to reply