How can I tell whether a geometric series converges?

The standard form of a geometric sequence is :

`u_n = u_1 * r^(n-1)`

And a geometric series can be written in several forms :

`sum_(n=1)^(+oo)u_n = sum_(n=1)^(+oo)u_1*r^(n-1) = u_1sum_(n=1)^(+oo)r^(n-1)`

`= u_1*lim_(n->+oo)(r^(1-1) + r^(2-1) + r^(3-1) + ... + r^(n-1))`

Let `r_n = r^(1-1) + r^(2-1) + r^(3-1) + ... + r^(n-1)`

Let's calculate `r_n - r*r_n` :

`r_n - r*r_n = r^(1-1) - r^(2-1) + r^(2-1) - r^(3-1) + r^(3-1) + ... - r^(n-1) + r^(n-1) - r^n = r^(1-1) - r^n`

`r_n(1-r) = r^(1-1) - r^n = 1 - r^n`

`r_n = (1 - r^n)/(1-r)`

Therefore, the geometric series can be written as :

`u_1sum_(n=1)^(+oo)r^(n-1) = u_1*lim_(n->+oo)((1 - r^n)/(1-r))`

Thus, the geometric series converges only if the series `sum_(n=1)^(+oo)r^(n-1)` converges; in other words, if `lim_(n->+oo)((1 - r^n)/(1-r))` exists.

• If |r| > 1 : `lim_(n->+oo)((1 - r^n)/(1-r)) = oo`

• If |r| < 1 : `lim_(n->+oo)((1 - r^n)/(1-r)) = 1/(1-r)`.

Therefore, the geometric series of geometric sequence `u_n` converges only if the absolute value of the common factor `r` of the sequence is strictly inferior to `1`.