Divergence of the geometric Series at r=1

[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 -&gt; 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 ?

 

[DryRun]

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.

 

[Ray Vickson]

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 -&gt; 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 ?

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

 

[HallsofIvy]

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.