# 5.6. Calculation Rules for Convergent Sequences

Convergence splits the set of all sequences completely in two parts, namely the convergent sequences and the divergent ones. On the other hand the set of all sequences is equipped with an algebraic structure: Sequences can be added, multiplied etc. It is a natural and rewarding question to ask whether calculating with convergent sequences gives convergent results or not. We have a positive answer to that, noted as limit theorems. With many 'difficult' sequences they allow a simple decision as to convergence and, as a real advantage, the limit theorems calculate the limit itself, so we don't need to know its value first.

Theorem (Limit Theorems):  If $\left({a}_{n}\right)$ and $\left({b}_{n}\right)$ are convergent sequences so are $\left({a}_{n}\right)+\left({b}_{n}\right)\text{,}$  $\left({a}_{n}\right)-\left({b}_{n}\right)$ and $\left({a}_{n}\right)\cdot \left({b}_{n}\right)\text{.}$ The convergence of  $\frac{\left({a}_{n}\right)}{\left({b}_{n}\right)}$ depends on the denominator limit.

More precisely the following holds:

 ${a}_{n}\to a\wedge {b}_{n}\to b⇒{a}_{n}+{b}_{n}\to a+b$ [5.6.1] ${a}_{n}\to a\wedge {b}_{n}\to b⇒{a}_{n}-{b}_{n}\to a-b$ [5.6.2] ${a}_{n}\to a\wedge {b}_{n}\to b⇒{a}_{n}\cdot {b}_{n}\to a\cdot b$ [5.6.3] ${a}_{n}\to a\wedge {b}_{n}\to b⇒\frac{{a}_{n}}{{b}_{n}}\to \frac{a}{b}$ ,  if  $b\ne 0$ [5.6.4]

We prove the limit theorems separately.

Consider:

• The condition $b\ne 0$ in [5.6.4] also guarantees that the quotient is sequence (at least of the generalized type ${\left({a}_{n}\right)}_{n\ge k}$ ). Use [5.5.4] to find a  $k\in {ℕ}^{\ast }$ such that .

• The fourth limit theorem [5.6.4] cannot be used with $b=0$. The examples $\frac{\left(\frac{1}{{n}^{2}}\right)}{\left(\frac{1}{n}\right)}=\left(\frac{1}{n}\right)$  and  $\frac{\left(\frac{1}{n}\right)}{\left(\frac{1}{{n}^{2}}\right)}=\left(n\right)$ show both, convergent and divergent quotients.

• Constant sequences are always convergent. Thus we have the following special case of [5.6.3]:

 ${a}_{n}\to a⇒c\cdot {a}_{n}\to c\cdot a$ [5.6.5]

The third limit theorem will also add to our inventory of convergent and divergent sequences.

Proposition:  For all $k\in {ℕ}^{\ast }$ and  $c\in ℝ$ we have:

 $\frac{c}{{n}^{k}}=c\cdot {n}^{-k}\to 0$ [5.6.6] $\left(c\cdot {n}^{k}\right)$ is divergent for  $c\ne 0$ [5.6.7]

Proof:
 1. ► As $\frac{1}{n}\to 0$ (cf. [5.4.6]) the third limit theorem [5.6.3] proves the following convergence:  $\frac{1}{{n}^{k}}=\underset{k\text{-mal}}{\underbrace{\frac{1}{n}\cdot \cdots \cdot \frac{1}{n}}}\to 0\cdot \dots \cdot 0=0$ From [5.6.5] we see that this is our assertion. 2. ► Indirectly: If $\left(c\cdot {n}^{k}\right)$ is convergent, so is  $\left(n\right)=\left(c\cdot {n}^{k}\right)\cdot \left(\frac{1}{c}\right)\cdot \left(\frac{1}{{n}^{k-1}}\right)\text{.}$ Contradiction!

With these new examples we can capture all sequences of a certain type, like e.g. $\left(\frac{6{n}^{5}-2}{3{n}^{5}+{n}^{2}}\right)$. At first glance the limit theorems are unlikely to be employed successfully in this case. Numerator and denominator are unbounded and thus divergent sequences! With a little 'trick' however we will succeed: Cancelling down all sequence members by ${n}^{5}$ will give us another, more advantageous representation of the same sequence:

$\left(\frac{6{n}^{5}-2}{3{n}^{5}+{n}^{2}}\right)=\left(\frac{6-\frac{2}{{n}^{5}}}{3+\frac{1}{{n}^{3}}}\right)$.

Now the numerator is the difference and the denominator the sum of two convergent sequences. And according to the second resp. first limit theorem numerator and denominator are convergent now with $6-0=6$ being the limit of the first and $3+0=3$ the limit of the latter. Especially the denominator limit is non-zero and we can use the fourth limit theorem:

$\frac{6{n}^{5}-2}{3{n}^{5}+{n}^{2}}=\frac{6-\frac{2}{{n}^{5}}}{3+\frac{1}{{n}^{3}}}\to \frac{6}{3}=2$.

Let us practise this method a bit more.

 Example:   $\frac{4{n}^{3}+2{n}^{2}-1}{3{n}^{3}+5n}=\frac{\stackrel{\stackrel{4}{↑}}{4}+\stackrel{\stackrel{0}{↑}}{\frac{2}{n}}-\stackrel{\stackrel{0}{↑}}{\frac{1}{{n}^{3}}}}{\underset{\underset{3}{↓}}{3}+\underset{\underset{0}{↓}}{\frac{5}{{n}^{2}}}}\underset{3\ne 0}{\to }\frac{4+0-0}{3+0}=\frac{4}{3}$  $\frac{3{n}^{2}-n+1}{{n}^{4}+7{n}^{2}-2}=\frac{\stackrel{\stackrel{0}{↑}}{\frac{3}{{n}^{2}}}-\stackrel{\stackrel{0}{↑}}{\frac{1}{{n}^{3}}}+\stackrel{\stackrel{0}{↑}}{\frac{1}{{n}^{4}}}}{\underset{\underset{1}{↓}}{1}+\underset{\underset{0}{↓}}{\frac{7}{{n}^{2}}}-\underset{\underset{0}{↓}}{\frac{2}{{n}^{4}}}}\underset{1\ne 0}{\to }\frac{0-0+0}{1+0-0}=0$   $\left(\frac{{n}^{4}}{{n}^{3}+1}\right)$ is divergent.  Proof:  $\left(\frac{{n}^{3}+1}{{n}^{3}}\right)=\left(1\right)+\left(\frac{1}{{n}^{3}}\right)$ is obviously convergent. Thus if  $\left(\frac{{n}^{4}}{{n}^{3}+1}\right)$ is convergent, the sequence $\left(\frac{{n}^{4}}{{n}^{3}+1}\right)\cdot \left(\frac{{n}^{3}+1}{{n}^{3}}\right)=\left(n\right)$ is convergent as well due to the third limit theorem.   Contradiction!

Such sequences are completely predictable in their behaviour! The determining factor is the position of the highest n-power:

A sequence with its highest n-power solely located in the numerator is divergent.

A sequence with its highest n-power solely located in the denominator is a zero sequence.

A sequence with its highest n-power simultaneously located in the numerator and denominator as well converges to the quotient of the two relevant coefficients.

 Exercise:   $\phantom{.}\frac{2{n}^{4}+3n-1}{4{n}^{6}+{n}^{2}}{?}=\frac{\frac{2}{{n}^{2}}+\frac{3}{{n}^{5}}-\frac{1}{{n}^{6}}}{4+\frac{1}{{n}^{4}}}{?}=\frac{\frac{2}{{n}^{2}}+\frac{3}{{n}^{5}}-\frac{1}{{n}^{6}}}{4+\frac{1}{{n}^{4}}}\underset{4\ne 0}{\to }\frac{0+0-0}{4+0}=0$ $\phantom{.}\frac{3{n}^{2}-\sqrt{2}}{\sqrt{3}{n}^{2}+4n+3}{?}=\frac{3-\frac{\sqrt{2}}{{n}^{2}}}{\sqrt{3}+\frac{4}{n}+\frac{3}{{n}^{2}}}{?}=\frac{3-\frac{\sqrt{2}}{{n}^{2}}}{\sqrt{3}+\frac{4}{n}+\frac{3}{{n}^{2}}}\underset{\sqrt{3}\ne 0}{\to }\frac{3-0}{\sqrt{3}+0+0}=\frac{3}{\sqrt{3}}=\sqrt{3}$

 5.5. 5.7.