5.1 Sequences  (Page 4/25)

As defined above, if a sequence does not converge, it is said to be a divergent sequence. For example, the sequences $\{1+3n\}$ and $\left\{{\left(\mathrm{-1}\right)}^n\right\}$ shown in [link] diverge. However, different sequences can diverge in different ways. The sequence $\left\{{\left(\mathrm{-1}\right)}^n\right\}$ diverges because the terms alternate between $1$ and $\mathrm{-1},$ but do not approach one value as $n\to \infty .$ On the other hand, the sequence $\{1+3n\}$ diverges because the terms $1+3n\to \infty$ as $n\to \infty .$ We say the sequence $\{1+3n\}$ diverges to infinity and write $\underset{n\to \infty }{\text{lim}}(1+3n)=\infty .$ It is important to recognize that this notation does not imply the limit of the sequence $\{1+3n\}$ exists. The sequence is, in fact, divergent. Writing that the limit is infinity is intended only to provide more information about why the sequence is divergent. A sequence can also diverge to negative infinity. For example, the sequence $\{\text{−}5n+2\}$ diverges to negative infinity because $\mathrm{-5}n+2\to \text{−}\infty$ as $n\to \text{−}\infty .$ We write this as $\underset{n\to \infty }{\text{lim}}\left(\mathrm{-5}n+2\right)=\to \text{−}\infty .$

Because a sequence is a function whose domain is the set of positive integers, we can use properties of limits of functions to determine whether a sequence converges. For example, consider a sequence $\left\{a_n\right\}$ and a related function $f$ defined on all positive real numbers such that $f(n)=a_n$ for all integers $n\ge 1.$ Since the domain of the sequence is a subset of the domain of $f,$ if $\underset{x\to \infty }{\text{lim}}f(x)$ exists, then the sequence converges and has the same limit. For example, consider the sequence $\left\{\frac{1}{n}\right\}$ and the related function $f\left(x\right)=\frac{1}{x}.$ Since the function $f$ defined on all real numbers $x>0$ satisfies $f\left(x\right)=\frac{1}{x}\to 0$ as $x\to \infty ,$ the sequence $\left\{\frac{1}{n}\right\}$ must satisfy $\frac{1}{n}\to 0$ as $n\to \infty .$

We can use this theorem to evaluate $\underset{n\to \infty }{\text{lim}}{r}^n$ for $0\le r\le 1.$ For example, consider the sequence $\left\{{\left(1\text{/}2\right)}^n\right\}$ and the related exponential function $f(x)={\left(1\text{/}2\right)}^x.$ Since $\underset{x\to \infty }{\text{lim}}{\left(1\text{/}2\right)}^x=0,$ we conclude that the sequence $\left\{{\left(1\text{/}2\right)}^n\right\}$ converges and its limit is $0.$ Similarly, for any real number $r$ such that $0\le r<1,$ $\underset{x\to \infty }{\text{lim}}{r}^x=0,$ and therefore the sequence $\left\{r^n\right\}$ converges. On the other hand, if $r=1,$ then $\underset{x\to \infty }{\text{lim}}{r}^x=1,$ and therefore the limit of the sequence $\left\{1^n\right\}$ is $1.$ If $r>1,$ $\underset{x\to \infty }{\text{lim}}{r}^x=\infty ,$ and therefore we cannot apply this theorem. However, in this case, just as the function $r^x$ grows without bound as $n\to \infty ,$ the terms $r^n$ in the sequence become arbitrarily large as $n\to \infty ,$ and we conclude that the sequence $\left\{r^n\right\}$ diverges to infinity if $r>1.$

We summarize these results regarding the geometric sequence $\left\{r^n\right\}\text{:}$

Later in this section we consider the case when $r<0.$

We now consider slightly more complicated sequences. For example, consider the sequence $\{{(2\text{/}3)}^n+{(1\text{/}4)}^n\}.$ The terms in this sequence are more complicated than other sequences we have discussed, but luckily the limit of this sequence is determined by the limits of the two sequences $\left\{{(2\text{/}3)}^n\right\}$ and $\left\{{(1\text{/}4)}^n\right\}.$ As we describe in the following algebraic limit laws, since $\left\{{(2\text{/}3)}^n\right\}$ and ${\{1\text{/}4)}^n\}$ both converge to $0,$ the sequence $\{{(2\text{/}3)}^n+{(1\text{/}4)}^n\}$ converges to $0+0=0.$ Just as we were able to evaluate a limit involving an algebraic combination of functions $f$ and $g$ by looking at the limits of $f$ and $g$ (see Introduction to Limits ), we are able to evaluate the limit of a sequence whose terms are algebraic combinations of $a_n$ and $b_n$ by evaluating the limits of $\left\{a_n\right\}$ and $\left\{b_n\right\}.$