Limit of a Geometric Sequence

The behavior of a geometric sequence $ a_n = a_1 \cdot q^{n-1} $ is entirely determined by its common ratio $ q $. In particular, its limit depends on how large $ q $ is in absolute value:

if $ |q| < 1 $, the terms get progressively smaller and approach zero, so the limit is $ 0 $;
if $ q = 1 $, every term is the same, so the sequence is constant and its limit is $ a_1 $;
if $ q > 1 $, the terms grow larger and larger without bound, diverging to $ +\infty $ or $ -\infty $ depending on the sign of $ a_1 $;
if $ q \leq -1 $, the terms keep switching sign and do not settle to a single value, so no limit exists.

In summary, a geometric sequence converges only when $ -1 < q \leq 1 $.

Examples

Let’s go through some examples to see how this works in practice.

Case |q| < 1

Take a geometric sequence with first term $ a_1 = 5 $ and common ratio $ q = \frac{1}{2} $.

$$ a_n = 5 \cdot \left(\frac{1}{2}\right)^{n-1} $$

The first few terms are:

$$ 5, 2.5, 1.25, 0.625, \dots $$

Each term is half the previous one, so the values shrink and get closer and closer to zero.

$$ \lim_{n \to \infty} 5 \cdot \left(\frac{1}{2}\right)^{n-1} = 0 $$

Case q = 1

Now consider a sequence with $ a_1 = 3 $ and $ q = 1 $.

$$ a_n = 3 \cdot 1^{n-1} = 3 $$

The terms are:

$$ 3, 3, 3, 3, \dots $$

Nothing changes from one step to the next, so the sequence stays fixed at 3.

$$ \lim_{n \to \infty} 3 \cdot 1^{n-1} = 3 $$

Case q > 1

Consider a sequence with $ a_1 = 2 $ and $ q = 2 $.

$$ a_n = 2 \cdot 2^{n-1} $$

The first few terms are:

$$ 2, 4, 8, 16, 32, \dots $$

Each term doubles the previous one, so the sequence grows rapidly and without bound.

$$ \lim_{n \to \infty} 2 \cdot 2^{n-1} = +\infty $$

Note. If the first term is negative, for example $ a_1 = -2 $, the terms become $$ -2, -4, -8, -16, \dots $$ and the sequence diverges to negative infinity $$ \lim_{n \to \infty} -2 \cdot 2^{n-1} = -\infty $$

Case q ≤ -1

Finally, consider a sequence with $ a_1 = 1 $ and $ q = -2 $.

$$ a_n = 1 \cdot (-2)^{n-1} $$

The first few terms are:

$$ 1, -2, 4, -8, 16, \dots $$

The sign alternates at each step while the magnitude increases, so the sequence never settles to a single value.

For this reason, the limit does not exist.

And so on.