We determine if a sequence converges or diverges.

Some examples of sequences are \begin{align*} i) \;\; & \{a_n\}_{1}^\infty = \{n^2\} = 1^2, 2^2, 3^2, 4^2, \dots = 1, 4, 9, 16, \dots \\ ii) \;\; & \{a_n\}_{1}^\infty = \{(-1)^n\} = (-1)^1, (-1)^2, (-1)^3, \dots = -1, 1, -1, 1, \dots \\ iii) \;\; & \{a_n\}_{1}^\infty = \left \{\frac{1}{2^n}\right \} = \frac{1}{2^1}, \frac{1}{2^2}, \frac{1}{2^3}, \frac{1}{2^4}, \dots = \frac 12, \frac 14, \frac 18, \frac{1}{16}, \dots \\ iv) \;\; & \{a_n\}_{1}^\infty = \left \{\frac{n}{n+1}\right \} = \frac{1}{1+1}, \frac{2}{2+1}, \frac{3}{3+1}, \frac{4}{4+1}, \dots = \frac 12, \frac 23, \frac 34, \frac 45, \dots \end{align*}

The question we will have about a given sequence is if it has a limit.

If a sequence has a (finite) limit, we say that the sequence converges. Otherwise, we say that it diverges.

We compute Since as , we can conclude that and so the sequence converges to . The diagram below shows the terms of the sequence approaching .

In general, a sequence of the form converges if and diverges if .

We compute Since oscillates between and we can conclude that and so the sequence diverges. The diagram below shows the oscillation of the terms of the sequence.

Since we conclude that the sequence diverges. The diagram below shows that the terms of the sequence grow without bound.

We compute \begin{align*} \lim _{n \to \infty } \frac{n}{n+1} &=\lim _{n \to \infty } \frac{n}{n\left (1+\frac{1}{n}\right )} \\ &=\lim _{n \to \infty } \frac{1}{1+\frac{1}{n}}\\ &=\frac{1}{1+0}\\ &= 1 \end{align*}

Hence, the sequence converges to . The diagram below shows the terms of the sequence approaching .

To compute the limit we use the fact that if a function, , has a limit, then the associated sequence has the same limit. Using L’Hopital’s rule twice, we have \begin{align*} \lim _{x \to \infty } \frac{x^2}{e^x} &= \frac{\infty }{\infty } = \lim _{x \to \infty } \frac{2x}{e^x} \;\text{(L'Hopital)}\\ &=\frac{\infty }{\infty } = \lim _{x \to \infty } \frac{2}{e^x} \;\;\text{(L'Hopital again)}\\ &=0 \end{align*}

Hence, as well, and the sequence converges to .

We will use the fact that if then . Since Taking reciprocals gives and hence the sequence is decreasing.

We will show that the associated function is an increasing function. To do this recall that if for , then is decreasing on the interval . The quotient rule gives which is negative if . Thus is decreasing on the interval and is a decreasing sequence.

### 1 Problems

You can type DNE or infinity into the answer box as needed.

The sequence convergesdiverges

The sequence convergesdiverges

The sequence convergesdiverges

The sequence convergesdiverges

The sequence convergesdiverges

The sequence convergesdiverges

The sequence convergesdiverges

The sequence convergesdiverges

The sequence convergesdiverges

The sequence convergesdiverges

a) increasing decreasing neither

b) increasing decreasing neither

c) increasing decreasing neither

d) increasing decreasing neither

e) increasing decreasing neither