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The formula for an infinite geometric series with $r \leq 1$ is $\frac{a_1}{1-r}$. My question is where does this come from and why must $r \leq 1$ for this formula to work? Why does anything above $1$ for $r$ make this infinite sum diverge instead of converge?

afedder
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King Squirrel
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    refer to this post: http://math.stackexchange.com/questions/370662/infinite-geometric-series-formula-derivation @King Squirrel – afedder May 15 '14 at 02:05

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Hint: Let $S=1+x+x^2+x^3+\ldots+x^n$. Now multiply S with $1-x$, and see what you get. Then ask yourself what is the limit of $x^{n+1}$ when n tends to infinity if $(a)~x<1$, and $(b)~x>1$.

Lucian
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So $S=a_0+a_0x+a_0x^2...$ Now multiply both sides by $x$. $xS=a_0x+a_0x^2+a_0x^3...=S-a_0.$ If $xS=S-a_0, S=\frac{a_0}{1-x}.$ If $x\geq1$, then the sum won't converge because the actual element $a_0x^n$ will keep getting bigger and bigger. Otherwise, it becomes near $0$.

Sidd Singal
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  • Oh you know I think I see. This is a formula for this but we have to look at the sum closer before saying what is computes. If r is >=1, then it diverges, and if it is <1 then it essentially goes away and converges to a number which is given by this. Is this correct? – King Squirrel May 15 '14 at 02:33
  • I can't completely understand what you wrote, but yes, that is correct. Remember that just because a sequence converges to 0, it does not mean that the sum will converge. For example $\sum \frac 1x$ does not converge, even though $\lim_{x\rightarrow\infty}\frac 1x = 0$ – Sidd Singal May 15 '14 at 03:00
  • Pretty much I'm asking if this only works if the sum converges in some way and is inapplicable to divergent sums. – King Squirrel May 15 '14 at 05:01
  • Do you mean "sum" or "sequence" – Sidd Singal May 15 '14 at 05:08
  • Well hold on, you say otherwise it becomes zero. I thought it could converge to some value. In class we got answers such as 3/2. – King Squirrel May 15 '14 at 20:07
  • I am talking about a sequence converging to 0. If we have the SEQUENCE $1,1/2,1/4,1/8,...1/2^n...$, then $1/2^n$ converges to 0. However, if we take the SUM $1+1/2+1/4+1/8+...+1/2^n$, then the sum converges to 2 – Sidd Singal May 15 '14 at 20:39
  • Ok I got it, yeah I was talking about sums being convergent or divergent. – King Squirrel May 16 '14 at 00:14
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By cancellation

$(1-r)\cdot(1+r+r^2+\ldots +r^{n-1}) = 1-r^n$

Taking limits when $0\leq r<1$

$$(1-r) \cdot \sum_{i=0}^{\infty}r^i = 1$$

so

$$ \sum_{i=0}^{\infty}r^i = (1-r)^{-1} $$

The argument fails for $r\geq 1$, since the sum diverges. The sum diverges because

$$ \sum_{i=0}^{n-1}r^i > (n-1)\cdot r \geq n-1 $$

which diverges.

Callus - Reinstate Monica
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