# Calculus: Sequences

## Sequence Limits

Just as functions can have limits as the independent variable approaches either a particular value or infinity, so, too, can sequences have limits. However, because sequences consist of discrete numbers, the only kind of limit they have are limits at infinity. When a sequence $\left\{a_n\right\}$ has a limit L, it is denoted as

$$\lim\limits_{n \rightarrow \infty} a_n = L$$

A sequence $\left\{a_n\right\}$ has a limit $L$ when $n$ can be made sufficiently large such that $a_n$ is arbitrarily close to $L$. More precisely, a sequence $\{a_n\}$ has a limit $L$ if for every $\epsilon > 0$ there is a corresponding integer $N$ such that if $n > N$, then $\left|a_n - L\right| < \epsilon$. Note that every term after the $N$th term must be within $\epsilon$ of $L$; it is insufficient for only the $N$th term to be within $\epsilon$ of $L$ and for subsequent terms to be more than $\epsilon$ away.

There is a theorem that let's us use our existing knowledge of finding limits to make finding limits of sequences much easier: If there is a function $f(x)$ such that $f(n) = a_n$, then $\lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} f(n)$.

A sequence that has a limit is said to converge or be convergent. A sequence that does not converge is said to diverge, or be divergent.

## Problems

1. Determine whether following sequences converge, and if so, find the limit:

1. $a_n = 2n^2-2n$
2. $a_n = \sin(3n)$
3. $a_n = e^{-n}$
4. $a_n = \dfrac{2}{n+1}$

Determine whether following sequences converge, and if so, find the limit:

1. $\lim\limits_{n \rightarrow \infty} 2n^2-n = \lim\limits_{n \rightarrow \infty} 2n(n-1) \\ \lim\limits_{n \rightarrow \infty} 2n^2-n = 2\infty(\infty-1) \\ \lim\limits_{n \rightarrow \infty} 2n^2-n = \infty \\$
The limit does not exist, so the sequence does not converge.
2. Periodic functions don't converge, so the sequence does not converge.
3. $\lim\limits_{n \rightarrow \infty} e^{-n} = e^{-\infty} \\ \lim\limits_{n \rightarrow \infty}a_n = 0 \\$
The limit exists, therefore the sequence converges.
4. $\lim\limits_{n \rightarrow \infty} \dfrac{2}{n+1} = \dfrac{2}{\infty+1} \\ \lim\limits_{n \rightarrow \infty} \dfrac{2}{n+1} = 0 \\$
The limit exists, therefore the sequence converges.
2. Determine whether the following sequence converges: $a_n = \cos\left(e^{-x}\right)$

$\lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} \cos\left(e^{-x}\right) \\ \lim\limits_{n \rightarrow \infty} a_n = \cos\left(e^{-\infty}\right) \\ \lim\limits_{n \rightarrow \infty} a_n = \cos\left(0\right) \\ \lim\limits_{n \rightarrow \infty} a_n = 1 \\$

The sequence converges to 1.
3. Determine whether the following sequence converges: $a_n = \sin(\ln(x))$

$\lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} \sin(\ln(x)) \\ \lim\limits_{n \rightarrow \infty} a_n = \sin(\ln(\infty)) \\ \lim\limits_{n \rightarrow \infty} a_n = \sin(\infty) \\$

Sine is a periodif function, so its limit as its argument approaches infinity is undefined. Therefore, the sequence does not converge.
4. Determine whether the following sequence converges and if so, find the limit: $a_n = \tan^{-1}(\ln(n))$

$a_n = \tan^{-1}(\ln(n)) \\ \lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} \tan^{-1}(\ln(n)) \\ \lim\limits_{n \rightarrow \infty} a_n = \tan^{-1}\left(\lim\limits_{n \rightarrow \infty} \ln(n)\right) \\ \lim\limits_{n \rightarrow \infty} a_n = \tan^{-1}(\infty) \\ \lim\limits_{n \rightarrow \infty} a_n = \dfrac{\pi}{2} \\$
5. Find the limit of the following sequence: $a_n = n^{10} e^{-n}$

$a_n = n^{10} e^{-n} \\ \lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} n^{10} e^{-n} \\ \lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} \dfrac{n^{10}}{e^n} \\ \lim\limits_{n \rightarrow \infty} a_n = \dfrac{\lim\limits_{n \rightarrow \infty} n^{10}}{\lim\limits_{n \rightarrow \infty} e^n} \\$

By the limit laws, we know that exponential functions always outpace polynomials. Since the denominator grows to infinity faster than the numerator, the limit is 0. Alternatively, applying L'Hopital's rule 10 times in a row yields the same result.

$\lim\limits_{n \rightarrow \infty} a_n = 0 \\$

6. Find the limit of the sequence: $a_n = \dfrac{\arctan(n)}{2n}$

$a_n = \dfrac{\arctan(n)}{2n} \\ \lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} \dfrac{\arctan(n)}{2n} \\ \lim\limits_{n \rightarrow \infty} a_n = \dfrac{\lim\limits_{n \rightarrow \infty} \arctan(n)}{\lim\limits_{n \rightarrow \infty} 2n} \\ \lim\limits_{n \rightarrow \infty} a_n = \dfrac{\frac{\pi}{2}}{\infty} \\ \lim\limits_{n \rightarrow \infty} a_n = 0 \\$

7. See this sequence?

$$a_n = \dfrac{\sin(n)}{n}$$

Your mission, should you choose to accept it, is to find its limit as $n$ approaches infinity.

$a_n = \dfrac{\sin(n)}{n} \\ \lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} \dfrac{\sin(n)}{n} \\ \lim\limits_{n \rightarrow \infty} a_n = \dfrac{\lim\limits_{n \rightarrow \infty} \sin(n)}{\lim\limits_{n \rightarrow \infty} n} \\$

While the sine term in the numerator does not converge, the linear term in the denominator diverges to infinity. We can use the Squeeze Theorem to solve this limit. Since sine is at most 1 and at least -1, we can squeeze the original function between $\dfrac{1}{n}$ and $\dfrac{-1}{n}$. Since both converge to 0 in the limit, we can conlude that

$\lim\limits_{n \rightarrow \infty} a_n = 0$

8. Find the limit of the following sequence. Or don't, it's up to you: $b_n = \dfrac{5^n}{6^n}$

$b_n = \dfrac{5^n}{6^n} \\ \lim\limits_{n \rightarrow \infty} b_n = \lim\limits_{n \rightarrow \infty} \dfrac{5^n}{6^n} \\ \lim\limits_{n \rightarrow \infty} b_n = \lim\limits_{n \rightarrow \infty} \left(\dfrac{5}{6}\right)^n \\$

Since the fraction is less than 1 and greater than -1, the limit converges to zero:

$\lim\limits_{n \rightarrow \infty} b_n = 0 \\$

9. Find the limit of the following sequence: $a_n = 12$

This sequence is extremely boring. It's just the number 12 repeated over and over again. Unless you really love the number 12, then I guess it's pretty great. Anywho, either way we have

$\lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} 12 \\ \lim\limits_{n \rightarrow \infty} a_n = 12$

10. Find the limit of the sequence: $a_n = 1 + 0.9^n$

$a_n = 1 + 0.9^n \\ \lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} \left(1 + 0.9^n \right) \\ \lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} 1 + \lim\limits_{n \rightarrow \infty} 0.9^n \\ \lim\limits_{n \rightarrow \infty} a_n = 1 + 0.9^{\infty} \\ \lim\limits_{n \rightarrow \infty} a_n = 1 + 0 \\ \lim\limits_{n \rightarrow \infty} a_n = 1\\$