# Divergence of the geometric Series at r=1

1. Apr 29, 2012

### sid9221

Now for the proof of convergence/divergence of the geometric series we first deduce the Nth partial sum which is given by:

$$\frac{r(1-r^n)}{1-r}$$

Now for 0<r<1 this become $$\frac{1}{1-r}$$ which clearly converges by AOL

At r>1 it's similarly obvious why it diverges.

But at r=1, I'm a bit confused. The text say's that it diverges but when I try to work it out I get:

$$1^n -> 1$$

$$S_n = \frac{C*0}{1-1}$$

which measn S_n = 0 which is confusing as it seems to indicate convergence ?

Am I suppose to interpret 0/0 as implying divergence ?

2. Apr 29, 2012

### sharks

For a geometric series, |r|<1 for convergence. If r has any other value, including 1 and -1, the series diverges.

If you find the limit at r=1, you will get a/(1-1)=a/0=∞, hence the series diverges.

3. Apr 29, 2012

### Ray Vickson

You are supposed to avoid 0/0 like the plague. The point is that in the LIMIT as r --> 1, the sum r/(1-r) --> infinity. Of course, you can also write down the sum directly when r = 1, to see that S = 1 + 1 + 1 + 1 + .... does not have a finite value.

RGV

4. Apr 29, 2012

### HallsofIvy

Staff Emeritus
What you should learn from this is "Don't just use formulas thoughtlessly- think!

If r= 1, then the sum is $\sum_{i=0}^\infty a(1)^n= a+ a+ a+ \cdot\cdot\cdot$. The nth partia series is na so that clearly does not converge.

If r= -1, then the sum is $\sum_{i= 0}^\infty a(-1)^n= a- a+ a- a+ \cdot\cdot\cdot$. The nth partial series is a if n is odd, 0 if n is even so, again, that does not converge.