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Convergent sequence theorem

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.

(a_n) + (b_n) = (a_n + b_n)

(a_n) - (b_n) = (a_n - b_n)

(a_n) \cdot (b_n) = (a_n \cdot b_n)

Also, if every member of the sequence (b_n) is different from 0, then we can define quotient of the sequences as

\frac{(a_n)}{(b_n)} = (\frac{a_n}{b_n}).

Example. We have two sequences (a_n) = \frac{1}{n} and (b_n)  = - \frac{1}{n + 1}

Find (a_n + b_n) and (a_n \cdot b_n).

(a_n + b_n) = (a_n) + (b_n) = \frac{1}{n} - \frac{1}{n + 1} = \frac{n+1-n}{n (n + 1)}=\frac{1}{n (n + 1)}

a_n \cdot b_n = \frac{1}{n} \cdot \frac{-1}{n + 1} = \frac{-1}{n (n + 1)}

Convergent sequence theorem

If we have two convergent sequences (a_n) and (b_n). Also let’s presume that

lim_{n \rightarrow \infty} a_n = a and lim_{n \rightarrow \infty} b_n = b.

1. Sequence (a_n) \pm (b_n) = (a_n \pm b_n) is convergent and:

lim_{n \rightarrow \infty}(a_n \pm b_n) = lim_{n \rightarrow \infty}(a_n) \pm lim_{n \rightarrow \infty}(b_n) = a \pm b.

2. Sequence (a_n) \cdot (b_n) = (a_n \cdot b_n) is convergent and:

lim_{n \rightarrow \infty}(a_n \cdot b_n) = lim_{n \rightarrow \infty}(a_n) \cdot lim_{n \rightarrow \infty}(b_n) = a \cdot b.

One special case is when sequence a_n is constant sequence or a_n = a. Then:

lim_{n \rightarrow \infty} (a_n) = a.

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

(a_n \cdot c_n) = a \cdot lim_{n \rightarrow \infty} c_n = a \cdot c.

3. Sequence If b \not= 0 for every n \in N and if lim_{n \rightarrow \infty}(b_n) \not= 0 then:

lim_{n \rightarrow \infty} (\frac{a_n}{b_n}) = \frac{lim_{n \rightarrow \infty} (a_n)}{lim_{n \rightarrow \infty} (b_n)} = \frac{a}{b}

From here we can notice that if lim_{n \rightarrow \infty} (a_n) = \infty then lim_{n \rightarrow \infty} (\frac{1}{a_n}) = 0

4. Sequence a_n^{b_n} is convergent and:

lim_{n \rightarrow \infty} (a_n^{b^n}) = (lim_{n \rightarrow \infty} a_n)^{lim_{n \rightarrow \infty} b_n} = a^b

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

Example 1. Find lim_{n \rightarrow \infty} (\frac{2}{n})

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 n grows, the members of the sequence become smaller and smaller. When n 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 0.

lim_{n \rightarrow \infty} (\frac{2}{n}) = 0

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


lim_{n \rightarrow \infty} (\frac{1}{5n}) = 0,       lim_{n \rightarrow \infty} (\frac{20}{n^4}) = 0...

For constants c, d, k, e \in \mathbb{R} it is valid that:

lim_{n \rightarrow \infty} (\frac{c}{d \cdot n^k}) = 0

Now, you know how to calculate limits that have n 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 n’s and its powers, you divide the numerator and denominator with the biggest power.


Example 2. Find lim_{n \rightarrow \infty} (\frac{2n + 3}{n + 1}).

The biggest power of n is 1 (\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}})Now we can use convergent sequence theorem and separate this limit into four different limits. 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   <strong>Example 3</strong>. Limits can also be infinity. lim_{n \rightarrow \infty} n = \infty    <strong>Example 4</strong>. Find lim_{n \rightarrow \infty}(\frac{2n^4 + n^3 + n + 3}{n + 1}).  Now, the biggest power is four which means that we'll divide numerator and denominator with 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}})   <strong>Example 5</strong>. Find lim_{n \rightarrow \infty} (\sqrt{n^2 + 2n} – n).  For calculating this limit first you should transform this sequence because when n \rightarrow \inftyso does\sqrt{n^2 + 2n}andn, which brings us to expansion\infty and – \inftywhich is undefined.  For this we'll use a simple trick and multiply this with a fraction whose value is equal to one but one with which we'll set the difference of squares. 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$