Limits of Sequences

Limits of Sequences

How to find the limits of sequences: meanings of convergence, divergence, how to solve the problems, examples, and their solutions.

Convergent Sequences

an = 1/n is a convergent sequence.

See how the terms of an = 1/n change
as n increases.

As n goes to infinity (∞),
an goes to a constant value: 0.

Then you can say that
an is a convergent sequence.

Then you can write the limit of an like this.

It is read as
'the limit of an as n goes to infinity equals 0'.

an = 1/n

So the limit of 1/n as n goes to infinity equals 0.

an = n/(n + 1) is a convergent sequence.

Here's another converging sequence: an = n/(n + 1).

As n goes to infinity,
an goes to a constant value: 1.

So an is also a convergent sequence.

So the limit of an as n goes to infinity equals 1.

an = n/(n + 1)

So the limit of n/(n + 1) as n goes to infinity equals 0.

Divergent Sequences: to ∞

an = n - 2 is a divergent sequence.

A divergent sequence is a sequence
that doesn't go to a constant value
as n goes to infinity.

There are three types of divergent sequences.

The first type is the ones
whose limit equal infinity (∞).

an = n - 2 is the first type.

As n goes to infinity,
an goes to infinity:
it doesn't go to a constant value.

Then you can say that
an is a divergent sequence.

So the limit of an as n goes to infinity equals ∞.

∞ is not a limit 'value'.
It symbols that an 'goes to infinity'.

an = n/(n + 1)

So the limit of n/(n + 1) as n goes to infinity equals ∞.

Divergent Sequences: to -∞

an = -2^n is a divergent sequence.

The second type is the ones
whose limit equal negative infinity (-∞).

an = -2n is an example of the second type.

As n goes to infinity,
an goes to negative infinity (-∞):
it doesn't go to a constant value.

So an is also a divergent sequence.

So the limit of an as n goes to infinity equals -∞.

an = -2n

So the limit of an = -2n as n goes to infinity equals -∞.

Divergent Sequences: Oscillation

an = (-1)^n oscillates between -1 and 1. So an is a divergent sequence.

The third type is the ones
whose terms oscillate.

The terms of an = (-1)n changes like -1, 1, -1, 1, ... .

As n goes to infinity,
an oscillates between -1 and 1:
it doesn't go to a constant value.

So an is also a divergent sequence.

So the limit of an as n goes to infinity is oscillation.

How to Solve

1/(infinity) = 0

To find the limit values of the given expresssions,
keep this principle in mind:
1/∞ = 0.

Example 1

Find the limit of the given expression if it exists. The limit of (5 + 4/n^2) as n goes to infinity

Recall that 1/∞ = 0.

As n → ∞,
n2 → ∞.

So 4/n2 → 4/∞ = 0.

So (given) = 5 + 0
= 5.

Example 2

Find the limit of the given expression if it exists. The limit of (7 + 1/2^n)(4 - 3/(ln n)) as n goes to infinity

As n → ∞,
2n → ∞ and ln n → ∞.

Graphing exponential functions

Logarithmic functions

So 1/2n → 1/∞ = 0
and 3/(ln x) → 3/∞ = 0.

So (given) = (7 + 0)(4 - 0).

Example 3

Find the limit of the given expression if it exists. The limit of sqrt(n - 2) as n goes to infinity

As n → ∞,
n - 2 → ∞.

Graphing square root functions

So the limit is equal to ∞.

Some teachers might want you to write
'Limit doesn't exist.'
to see if you know that ∞ is not a limit value.

So if your teacher wants you to write this,
just write this below the answer.

Example 4

Find the limit of the given expression if it exists. The limit of (cos n*pi) as n goes to infinity

To find the first few terms of an,
sketch the cosine graph.

cos 1π = -1
cos 2π = 1
cos 3π = -1
cos 4π = 1
...

As n → ∞,
cos oscillates.

So (given): oscillation.

Don't forget to write this
if your teacher wants it.