Waste less time on Facebook — follow Brilliant.
×

Infinite summmations

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

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

and the proof is as given below:

Take \(S=a+a r+a r^2+a r^3+\text{...}\)

then \(r S=a r+a r^2+a r^3+a r^4+\text{...}\)

subtracting these two we get

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

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

My question is: why isn't this formula valid for all \(r\) excluding \(1\)?

Note by Danish Mohammed
3 years, 4 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

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

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

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

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

Just to say that you must be VERY careful when you sum something to infinity! Riccardo Zanotto · 3 years, 4 months ago

Log in to reply

@Riccardo Zanotto 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 \[S=1-1+1-1+....=\lim_{n\rightarrow \infty }\left | cos(\frac{n\pi }{2}) \right |\]

this limit doesn't equal \[\frac{1}{4}\] , Its obvious when you check the pure definition of the limit of a sequence. I think that limit here doesn't exist. Ahmed Taha · 3 years, 4 months ago

Log in to reply

@Ahmed Taha 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 · 3 years, 4 months ago

Log in to reply

@Riccardo Zanotto It doesn't have a value in the usual sense.But by Cesaro summing it you will arrive at \(\frac{1}{2}\). See here.Grandi's sum actually has no sum, but it should be \(\frac{1}{2}\) Bogdan Simeonov · 3 years, 4 months ago

Log in to reply

@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, \(0^0\) is impossibile, but taking the (right) limit of \(x^x\) one can say that it should be \(1\). Riccardo Zanotto · 3 years, 4 months ago

Log in to reply

@Riccardo Zanotto Exactly Bogdan Simeonov · 3 years, 4 months ago

Log in to reply

@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 \(\zeta (-1)\) (The riemann-zeta function) in mathematica; it returned \(\displaystyle \frac{-1}{12}\)

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

so \(\displaystyle \zeta (-1)=1+2+3+... = -\frac{1}{12}\) This seems so bizarre that my head hurts. :P Danish Mohammed · 3 years, 4 months ago

Log in to reply

@Danish Mohammed I have written a post about it check here Bogdan Simeonov · 3 years, 4 months ago

Log in to reply

@Riccardo Zanotto If you want more about infinit geometric sequences, read my summation notes (summation, summation part two and summation part two method two). Sharky Kesa · 3 years, 4 months ago

Log in to reply

@Riccardo Zanotto I was referring to this 1 and -1 fallacy ....... Nice post..... Eddie The Head · 3 years, 4 months ago

Log in to reply

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 \(\infty\).

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

With \(n\to\infty\) the above limit exists only if \(r<1\) as then \(r^{something \hspace{1mm} big}=0\) and the numerator becomes 1. So the result follows. Krishna Jha · 3 years, 4 months ago

Log in to reply

@Krishna Jha Exactly,..that is the interval of convergence... Eddie The Head · 3 years, 4 months ago

Log in to reply

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 · 3 years, 4 months ago

Log in to reply

@Eddie The Head 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 · 3 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...