# Playing around with Geometric series. Where is my algebraic mistake?

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?

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

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