# Mathematics - Mathematical Analysis Special Sequences and Divergence

Hello its a me again. Today we continue with **Mathematical Analysis** getting into some **Special Sequences **and also** **more things about the **Divergence of Sequences**. You should check out my previous post about Subsequences and Convergence here before getting into this post! So, without further do, let's get started!

### Bernoulli's inequality:

Suppose x that is a real number and > -1 and also n that is a natural number then:

**(1 + x)^n >= 1 + n*x**

Using this "equation" we can find the limit of an geometric sequence.

## Geometric Sequence:

A sequence (an) with **generic term a(n) = r^n**, where n a natural number and r a real number is called a geometric sequence.

**If |r| < 1** then this sequence is also a null sequence and so:

**lim n-> +∞ [r^n] = 0.**

**If r equals 0 then the limit equals 1 **and **if |r| is greater than 0 then the limit equals +-∞** .

Try proving it using Bernoulli's Inequality.

Just suppose a real number x and set |r| = 1 / (1 + x) and try forming the inequality so that you end up with a limit that gives 0.

## D'Alembert Convergence/Divergence Theorem:

Suppose a sequence (an) with positive terms and the limit:

* lim n->∞ [a(n+1)/a(n)] = l*, a real number.

- If
**0 < a(n+1)/a(n) < l < 1**then the sequence (an) is a**null sequence**and so lim n->∞ (an) = 0 - If
**a(n+1)/a(n) > l > 1**then the**sequence**(an)**diverges**and so lim n->∞ (an) = ∞

**Example:**

Calculate the limit:

Suppose the limit is an sequence (an) then:

lim n-> ∞ (1/10)^n equals 0, cause 0 < 1/10 < 1 (from D'Alembert) and so:

### Root Sequence:

Suppose a sequence (an) with **generic term a(n) = n-root(a)** for every natural n and a positive real number 'a' then the limit:

* lim n-> ∞ [n-root(a)] = 1*.

- If a = 1 then we can directly see that it equals 2
- If a>1 then this can be proven easily using Bernoulli's Inequality and the Squeeze Theorem.

So, we can generalize saying that:

* lim n-> ∞ [n-root(n)] = 1* for every natural number n > 2.

### Napier's (Euler's) constant e:

We are all familiar with the number * e = 2.718*... But did you know that this value is actually the result of a sequence's limit?

So, if you take the **sequence **(an) with **generic term a(n) = (1 + 1/n)^n**, a sequence that is strictly increasing, bounded and converging and find the limit of this sequence you will end up with:

* lim n->∞ (1 + 1/n)^n = e*, where e is the napier constant e = 2.71828....

This can again be proven by using Bernoulli's Inequality.

We can also calculate it using a Series, something that we will start talking about from the next post on. So, for you to get an idea the series "1+ 1/2! + 1/3! + ... + 1/n!" also gives us 'e', but much faster and so we can use a smaller 'n' value (cause infinity is not feasible) and get a value closer to the 'e' constant.

The sequence (an) with generic term * a(n) = (1 + a/n)^n* has a limit equal to e^a and so:

**lim n->∞ (1 + a/n)^n = e^a**

### Sum up in form of a table:

The table also contains some other useful limits that may come handy.

## Divergence:

We already said that a sequence diverges when the limit equals to +-∞. But, we don't covered why they are useful and which other properties they have.

**Properties of Divergent Sequences:**

- A increasing and non-bounded sequence diverges to +∞
- A decreasing and non-bounded sequence diverges to -∞
- A sequence that is not upper-bounded is build up of at least one subsequence that diverges to +∞
- A sequence that is not lower-bounded is build up of at least one subsequence that diverges to -∞
- If (an), (bn) are sequences with a(n) <= b(n) for every natural number n and lim n->∞ (an) = +-∞ (an diverges) then lim n->∞ (bn) = +-∞ and so (bn) also diverges.
- If (an), (bn), (cn) are sequences with b(n) <= a(n) <= c(n) and the (bn), (cn) diverge which means that lim n->∞ (bn) = lim n->∞ (cn) = +-∞ then (an) also diverges and so lim n->∞ (an) = +-∞

Also, remember the indeterminate forms of limits ∞/∞ and ∞-∞, cause if you end up with something like that you will have to use the tips and tricks we already talked about to get an solvable form.

**Example:**

Suppose we want to calculate the limit:

We can clearly see that we have a indeterminate form ∞-∞.

This means that we will use conjugate multiplication and so:

The limits of 1/n and 3/n to infinity of course give us 0 and that's why we end up with simpler calculations. I think that I don't have to explain more than that.

And this is actually it for today and I hope you learned something!

Next time we will get started with Series and their Convergence!

Bye!

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