# 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

See how the terms of *a*_{n} = 1/*n* change

as *n* increases.

As *n* goes to infinity (∞),*a*_{n} goes to a constant value: 0.

Then you can say that*a*_{n} is a convergent sequence.

Then you can write the limit of *a*_{n} like this.

It is read as

'the limit of *a*_{n} as *n* goes to infinity equals 0'.

*a*_{n} = 1/*n*

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

Here's another converging sequence: *a*_{n} = *n*/(*n* + 1).

As *n* goes to infinity,*a*_{n} goes to a constant value: 1.

So *a*_{n} is also a convergent sequence.

So the limit of *a*_{n} as *n* goes to infinity equals 1.

*a*_{n} = *n*/(*n* + 1)

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

## Divergent Sequences: to ∞

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 (∞).*a*_{n} = *n* - 2 is the first type.

As *n* goes to infinity,*a*_{n} goes to infinity:

it doesn't go to a constant value.

Then you can say that*a*_{n} is a divergent sequence.

So the limit of *a*_{n} as *n* goes to infinity equals ∞.

∞ is not a limit 'value'.

It symbols that *a*_{n} 'goes to infinity'.

*a*_{n} = *n*/(*n* + 1)

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

## Divergent Sequences: to -∞

The second type is the ones

whose limit equal negative infinity (-∞).*a*_{n} = -2^{n} is an example of the second type.

As *n* goes to infinity,*a*_{n} goes to negative infinity (-∞):

it doesn't go to a constant value.

So *a*_{n} is also a divergent sequence.

So the limit of *a*_{n} as *n* goes to infinity equals -∞.

*a*_{n} = -2^{n}

So the limit of *a*_{n} = -2^{n} as *n* goes to infinity equals -∞.

## Divergent Sequences: Oscillation

The third type is the ones

whose terms oscillate.

The terms of *a*_{n} = (-1)^{n} changes like -1, 1, -1, 1, ... .

As *n* goes to infinity,*a*_{n} oscillates between -1 and 1:

it doesn't go to a constant value.

So *a*_{n} is also a divergent sequence.

So the limit of *a*_{n} as *n* goes to infinity is oscillation.

## How to Solve

To find the limit values of the given expresssions,

keep this principle in mind:

1/∞ = 0.

## Example 1

Recall that 1/∞ = 0.

As *n* → ∞,*n*^{2} → ∞.

So 4/*n*^{2} → 4/∞ = 0.

So (given) = 5 + 0

= 5.

## Example 2

As *n* → ∞,

2^{n} → ∞ and ln *n* → ∞.

Graphing exponential functions

Logarithmic functions

So 1/2^{n} → 1/∞ = 0

and 3/(ln *x*) → 3/∞ = 0.

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

## Example 3

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

To find the first few terms of *a*_{n},

sketch the cosine graph.

cos 1*π* = -1

cos 2*π* = 1

cos 3*π* = -1

cos 4*π* = 1

...

As *n* → ∞,

cos *nπ* oscillates.

So (given): oscillation.

Don't forget to write this

if your teacher wants it.