Infinite summmations

A formula that I learnt in school to sum an infinite geometric series with a common ratio between 1-1 and 11 is,

n=0arn=a1r if r<1\displaystyle \sum _{n=0}^{\infty } a r^n=\frac{a}{1-r}\text{ if }|r| < 1

and the proof is as given below:

Take S=a+ar+ar2+ar3+...S=a+a r+a r^2+a r^3+\text{...}

then rS=ar+ar2+ar3+ar4+...r S=a r+a r^2+a r^3+a r^4+\text{...}

subtracting these two we get

(1r)S=a(1-r) S=a

Therefore, S=a1rS=\displaystyle \frac{a}{1-r}

My question is: why isn't this formula valid for all rr excluding 11?

Note by Danish Mohammed
5 years, 5 months ago

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1 vote

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Well, an infinite summation doesn't make sense in itself. It must have a definition, and usually we take i=0f(i)=limni=0nf(i)\sum_{i=0}^\infty f (i)=\lim_{n\to\infty}\sum_{i=0}^n f (i) which leads us to the abig0a^{big}\to0 argumentation.

Regarding your demonstration, again, we have to take the infinite summation as a finite one: then if we set S(n)=a+ar++arnS (n)=a+ar+\dots+ar^n we get rS(n)=ar+ar2++arn+1rS (n)=ar+ar^2+\dots+ar^{n+1} and so S(n)(1r)=arn+1S(n)(1-r)=a-r^{n+1}; if we take this to the limit, if r<1|r|<1 then again rnr^n converges to 0, otherwise it diverges to infinity.

For example: let S=1+2+4+8+S=1+2+4+8+\dots; then 2S=2+4+8+2S=2+4+8+\dots and subtracting we'd get S=1S=-1, which obviously is absurd.

Another example: how do you evaluate S=11+11+S=1-1+1-1+\dots? you could say that S=(11)+(11)+=0S=(1-1)+(1-1)+\dots=0, or S=1+(1+1)+(1+1)+=1S=1+(-1+1)+(-1+1)+\dots=1, or even S=1SS=1-S, which is S=1/2S=1/2.

Just to say that you must be VERY careful when you sum something to infinity!

Riccardo Zanotto - 5 years, 5 months ago

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I am not sure if the last sum you evaluated is correct , because if we write the infinite series as the limit of a sequence S=11+11+....=limncos(nπ2)S=1-1+1-1+....=\lim_{n\rightarrow \infty }\left | cos(\frac{n\pi }{2}) \right |

this limit doesn't equal 14\frac{1}{4} , Its obvious when you check the pure definition of the limit of a sequence. I think that limit here doesn't exist.

Ahmed Taha - 5 years, 5 months ago

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Yes, of course it doesn't exist... in fact the three "values" are there to show that summing to infinity without care is really dangerous.

Riccardo Zanotto - 5 years, 5 months ago

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@Riccardo Zanotto It doesn't have a value in the usual sense.But by Cesaro summing it you will arrive at 12\frac{1}{2}. See here.Grandi's sum actually has no sum, but it should be 12\frac{1}{2}

Bogdan Simeonov - 5 years, 5 months ago

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@Bogdan Simeonov Yes, you're right... many things don't have a real value, but it makes sense to give them a value derived from the limit... For example, 000^0 is impossibile, but taking the (right) limit of xxx^x one can say that it should be 11.

Riccardo Zanotto - 5 years, 5 months ago

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@Riccardo Zanotto Exactly

Bogdan Simeonov - 5 years, 5 months ago

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@Bogdan Simeonov So basically this isn't an infinite sum in the usual sense but a unique value that we attach to this sum so that we can work with it?

Also, I tried ζ(1)\zeta (-1) (The riemann-zeta function) in mathematica; it returned 112\displaystyle \frac{-1}{12}

ζ(s)=11s+12s+13s+...\displaystyle \zeta (s)=\frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + ...

so ζ(1)=1+2+3+...=112\displaystyle \zeta (-1)=1+2+3+... = -\frac{1}{12} This seems so bizarre that my head hurts. :P

Danish Mohammed - 5 years, 5 months ago

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@Danish Mohammed I have written a post about it check here

Bogdan Simeonov - 5 years, 5 months ago

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I was referring to this 1 and -1 fallacy ....... Nice post.....

Eddie The Head - 5 years, 5 months ago

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If you want more about infinit geometric sequences, read my summation notes (summation, summation part two and summation part two method two).

Sharky Kesa - 5 years, 5 months ago

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The proof may also be given as the following in which your answer becomes clear:

Sum of the infinite geometric series can be taken to be the sum as number of a geometric series whose tend to \infty.

Sn=limna(1rn)1r\displaystyle \large S_n=\displaystyle\lim_{n\to\infty} \frac{a(1-r^{n})}{1-r}

With nn\to\infty the above limit exists only if r<1r<1 as then rsomethingbig=0r^{something \hspace{1mm} big}=0 and the numerator becomes 1. So the result follows.

Krishna Jha - 5 years, 5 months ago

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Exactly,..that is the interval of convergence...

Eddie The Head - 5 years, 5 months ago

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If r is greater than 1, then clearly the sum becomes diverging,....yes in that case also it seems that we can apply this formula but clearly that finite answer has no meaning because as r tens to infinity the terms of the sequence becomes incresingly large and the whole expression that is the sum of these terms become infinity.....you can notice another interesting fallacy when r =1...in that case writing the sum in 2 different gives gives two different answers.....

Eddie The Head - 5 years, 5 months ago

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Also the expression infinity -infinity has no meaning....this crops up when r is greater than 1....but on the contrary in the convergence interval as the sequence is decresing we can say that after infinite steps all the terms will become zero and hence the sum becomes finite and hence we can subtract these two numbers...I would love to hear what others say about this,.,,

Eddie The Head - 5 years, 5 months ago

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