Limit of a Sequence

To see if a_n is convergent,
find the limit of a_n
as n → ∞.

a_n = 5 + 4/n²

So the limit of a_n is
the limit of 5 + 4/n².

As n → ∞,
the constant 5 is still 5.

As n → ∞,
4/n² goes to
4/∞² = 4/∞ = 0.

If the denominator get bigger,
then the whole fraction gets smaller.
So 4/∞ = 0.

So the limit is 5 + 0.

5 + 0 = 5

The limit of a_n is 5.

This 5 is a constant number.

So a_n is convergent.

So a_n is convergent.

Find the limit of a_n
as n → ∞.

a_n = √n - 2

So the limit of a_n is
the limit of √n - 2.

As n → ∞,
√n - 2 goes to
√∞ - 2 = ∞.

Think of the graph of y = √n - 2.
As n → ∞,
the graph goes to ∞.

Square Root Function: Graph

The limit of a_n is ∞.

∞ is not a constant number.

So a_n is divergent.

So a_n is divergent.

Find the limit of a_n
as n → ∞.

a_n = -n² + 1

So the limit of a_n is
the limit of -n² + 1.

As n → ∞,
-n² goes to
-∞² = -∞.

As n → ∞,
the constant +1 is still +1.

So the limit is -∞ + 1.

-∞ + 1 = -∞

The limit of a_n is -∞.

-∞ is not a constant number.

So a_n is divergent.

So a_n is divergent.

Find the first few terms of a_n.

a₁ = -1
a₂ = 1
a₃ = -1
a₄ = 1

The terms of a_n are -1, 1, -1, 1, ... .

a_n does not go to one direction.

So a_n shows oscillation.

a_n is oscillating.

So a_n is divergent.

So a_n is divergent.