# Convergent sequence theorem

In a convergent sequence theorem we can also notice many other important properties. We can find out what happens with the limit of two sequences and how exactly does sequence operations affect their limits.

For easier understanding, we need to remind ourselves about sequence algebra.

## Sequence algebra

Since we can define a set of all infinite sequences of real numbers, we can also define addition and multiplication.

Also, if every member of the sequence is different from , then we can define quotient of the sequences as

.

Example. We have two sequences and

Find and .

## Convergent sequence theorem

If we have two convergent sequences and . Also let’s presume that

and .

1. Sequence is convergent and:

.

2. Sequence is convergent and:

.

One special case is when sequence is constant sequence or . Then:

.

Now, for a convergent sequences (constant sequence) and we have:

.

3. Sequence If for every and if then:

From here we can notice that if then

4. Sequence is convergent and:

Now that you know the rules you need to know when calculating limits we can start with tasks.

Example 1. Find

On this example try to figure out yourself what would happen to this sequence. The numerator of members of this sequences always remains the same. When grows, the members of the sequence become smaller and smaller. When goes to infinity, meaning it becomes infinitely large, that member will become very small, but still larger than zero since it can’t be zero and can’t be negative.

Limit of this sequence will be .

Generally, limit of every sequence that has a form like this will be zero.

Examples:

,

For constants it is valid that:

Now, you know how to calculate limits that have only in denominator, what happens when you find it in numerator also? Sequences will appear with many powers, so when you see a sequence that has more than one ’s and its powers, you divide the numerator and denominator with the biggest power.

Example 2. Find .

The biggest power of is (\frac{2n + 3}{n + 1}) = lim_{n \rightarrow \infty} (\frac{\frac{2n}{n} + \frac{3}{n}}{\frac{n}{n} + \frac{1}{n}})\Rightarrow lim_{n \rightarrow \infty} (\frac{2 + \frac{3}{n}}{1 + \frac{1}{n}}) lim_{n \rightarrow \infty} (\frac{2 + \frac{3}{n}}{1 + \frac{1}{n}}) = \frac{lim_{n \rightarrow \infty} 2 + lim_{n \rightarrow \infty} \frac{3}{n} }{lim_{n \rightarrow \infty} 1 + lim_{n \rightarrow \infty} \frac{1}{n}} = \frac{2 + 0}{1 + 0} = 2 lim_{n \rightarrow \infty} n = \infty lim_{n \rightarrow \infty}(\frac{2n^4 + n^3 + n + 3}{n + 1}) n^4 lim_{n \rightarrow \infty}(\frac{2n^4 + n^3 + n + 3}{n + 1}) = lim_{n \rightarrow \infty}(\frac{2 + \frac{1}{n} + \frac{1}{n^2} + \frac{3}{n^4}}{\frac{1}{n^3} + \frac{1}{n^4}}) lim_{n \rightarrow \infty} (\sqrt{n^2 + 2n} – n) n \rightarrow \infty\sqrt{n^2 + 2n}n\infty and – \infty lim_{n \rightarrow \infty} (\sqrt{n^2 + 2n} – n) \cdot \frac{\sqrt{n^2 + 2n} + n}{\sqrt{n^2 + 2n} + n} = lim_{n \rightarrow \infty}\frac{n^2 + 2n – n^2}{\sqrt{n^2 + 2n} + n} = lim_{n \rightarrow \infty}\frac{2n}{\sqrt{n^2 + 2n} + n} = lim_{n \rightarrow \infty}\frac{2}{\sqrt{1 + \frac{2}{n}} + 1} = \frac{2}{2} = 1\$

Shares