Playing around with geometric series I got confused. First, consider the following equation

$$\sum_{i=0}^n 1^i = n+1$$

Now, replacing $1$ by $\frac{a}{a}$ gives

$$\sum_{i=0}^n \left(\frac{a}{a}\right)^i = n+1$$

Also, replacing $\frac{a}{a}$, by $r$ we have

$$\sum_{i=0}^n r^i = \frac{r^{n+1}-1}{r-1}$$

and replacing $r$ by $\frac{a}{b}$, we get

$$\sum_{i=0}^n \left(\frac{a}{b}\right)^i = \frac{\frac{a}{b}^{n+1}-1}{\frac{a}{b}-1} = \frac{a \left(\frac{a}{b}\right)^n-b}{a-b}$$

For the special case, where $a=b$,

$$\sum_{i=0}^n \left(\frac{a}{b}\right)^i =\sum_{i=0}^n \left(\frac{a}{a}\right)^i = \sum_{i=0}^n 1^i$$

and therefore,

$$n+1 = \frac{a \left(\frac{a}{b}\right)^n-b}{a-b} = \frac{a \left(\frac{a}{a}\right)^n-a}{a-a} = \frac{0}{0}$$

Where is my mistake?

  • $\begingroup$ You could make your "argument" way shorter by just using the first and third equation and putting $r=1$ in the third. $\endgroup$ Apr 20, 2015 at 21:08
  • $\begingroup$ r=1 is where you went wrong. $\endgroup$
    – Rammus
    Apr 20, 2015 at 21:09
  • 2
    $\begingroup$ The formula you used for $\sum_0^n r^i$ is not valid when $r=1$. $\endgroup$ Apr 20, 2015 at 21:09
  • $\begingroup$ Oh yes, this is very obvious now! Thank you! $\endgroup$
    – Remi.b
    Apr 20, 2015 at 21:10

1 Answer 1


The error is in the third line. The partial sum of a geometric series is valid only if the common ratio is not $1$.


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