A sequence is a function defined on the set of positive integers. This means that if you give me a positive integer $5$, I should be able to give you some object $a_{5}$, labeled with the number you gave me, which is $5$ in this case.
For example, we can think about the "identity sequence" given by $a_{n} = n$. That is, this sequence gives you back the number your give it as an input. Another example can be $a_{n} = n^{2}$. Sometimes, you may not be able to write down the formula for $a_{n}$ explicitly. For instance, define $a_{n}$ to be the $n$-th digit of $\pi = 3.141592 \dots$ after the dot. We have $a_{1} = 1, a_{2} = 4, a_{3} = 1, a_{4} = 5, \dots.$ However, because $\pi$ is an irrational number, so there is no way you can write down a general formula for $a_{n},$ which we call the $n$-th term of the sequence $(a_{i})_{i \geq 1}.$
Remark. Note that most of the time, it is perfectly fine to extend the domain of the sequence from positive integersA sequence is a function defined on the set of positive integers. This means that if you give me a positive integer $5$, I should be able to give you some object $a_{5}$, labeled with the number you gave me, which is $5$ in this case.
For example, we can think about the "identity sequence" given by $a_{n} = n$. That is, this sequence gives you back the number your give it as an input. Another example can be $a_{n} = n^{2}$. Sometimes, you may not be able to write down the formula for $a_{n}$ explicitly. For instance, define $a_{n}$ to be the $n$-th digit of $\pi = 3.141592 \dots$ after the dot. We have $a_{1} = 1, a_{2} = 4, a_{3} = 1, a_{4} = 5, \dots.$ However, because $\pi$ is an irrational number, so there is no way you can write down a general formula for $a_{n},$ which we call the $n$-th term of the sequence $(a_{i})_{i \geq 1}.$
Remark. Note that most of the time, it is perfectly fine to extend the domain of the sequence from positive integers to non-negative integers. For example, the sequence given by the formula $a_{n} = n$ can be thought over all the non-negative integers or even negative integers. It is of course okay to extend this formula for all real numbers, but we usually don't call it a sequence if it is over all real numbers. There is a reason for this, but discussing this digresses too much from our main discussion, so we won't discuss this.
We can take the limit of sequence, when ever the limit exists. For example, we have
\[\lim_{n \rightarrow \infty}\frac{1}{n} = 0.\]
One important property of real numbers is the following. The name of this property is "completeness", although your book (p.477, Theorem 9.1) does not mention this terminology.
Completeness proeprty. Any increasing sequence bounded above is convergent.
Exercise. Determine whether the sequence given by the following formula is increasing:
\[a_{n} = \left(1 + \frac{1}{n}\right)^{n}.\]
For the problem above, you might find the following hint useful:
Consider $f(x) = (1 + 1/x)^{x},$ for $x > 0.$ Let $g(x) := \ln(f(x)) = x \ln(1 + 1/x)$ so that $$g'(x) = \ln\left(1 + \frac{1}{x}\right) + \frac{x}{1 + 1/x}\left(-\frac{1}{x^{2}}\right) = \ln\left(1 + \frac{1}{x}\right) - \frac{1}{x + 1}.$$
Hence $g'(x) > 0$ for $x \geq 1.$ This shows that $g(x) = \ln(f(x))$ is increasing for $x \geq 1,$ so $f(x)$ is increasing for $x \geq 1.$ This implies that $f(n) = a_{n}$ is increasing for $n = 1, 2, 3, \dots.$
Remark. How do we know that $g'(x) > 0$ for $x \geq 1$? First, note that $g'(1) = \ln(2) - 1/2 > 0$ because $2 > e^{1/2}$, because $4 > e.$ Now, take the derivative of $g'(x)$ to have $$(g'(x))' = g''(x) = \frac{1}{1 + 1/x}\left(- \frac{1}{x^{2}}\right) + \frac{1}{(x + 1)^{2}} = -\frac{1}{x(x + 1)} + \frac{1}{(x+1)^{2}} > 0,$$ for $x \geq 1.$ Thus, we must have $g'(x) \geq \ln(2) - 1/2 > 0$ for all $x \geq 1.$
We have checked that the sequence $a_{n} = (1 + 1/n)^{n}$ is increasing. The sequence is also bounded above by $3$. Checking this is same as checking $1 + 1/n \leq 3^{1/n},$ which can be checked by checking $1 + x \leq 3^{x}$ for $0 \leq x \leq 1,$ which can be checked by differentiation trick we have introduced above. Therefore, by the completeness property, the limit of $a_{n}$ exists, and we call this limit $e.$ That is, we define $$e := \lim_{n \rightarrow \infty}\left(1 + \frac{1}{n}\right)^{n}.$$
A geometric sequence is a sequence such that that each pair of the consecutive terms have the same ratio. If such a common ratio $r$ is non-zero, then we can consider this sequence to be given by $a_{n+1}/a_{n} = r$ for all $n \geq 1.$ More explicitly, you may deduce that
\[a_{n} = a_{1}r^{n-1}.\]
A series is a sum of the terms of a sequence. You can have finite series or infinite series. There is a version of $p$-test in the series world.
$p$-series. Given any real number $p,$ the series
\[\sum_{n=1}^{\infty}\frac{1}{n^{p}}\]
converges if and only if $p > 1.$ We refer to this as "$p$-series", although this is not a standard math terminology outside Math 116.
Question. Is there a more general phenomenon?
Answer. Yes! This is what we call the integral test. Please read the link, and at least get comfortable with the statement. Alternatively, you may read p.492 of your book.
Question. Is there a more general phenomenon?
Answer. Yes! This is what we call the integral test. Please read the link, and at least get comfortable with the statement. Alternatively, you may read p.492 of your book.
Exercise. Solve #4 of Exam 3, Fall 2016.
Optional reading. You may consider $p$-series as a function in $p$, and it turned out that this is one of the most important functions in mathematics, called "zeta function". I like thinking about it, since it often breaks data about integers into atomic pieces about prime numbers.
A geometric series is a sum of a geometric sequence. Say $a_{1}, a_{2}, a_{3}, \dots$ is the geometric sequence of ratio $r$. Then we may consider their partial sums $$s_{n} := a_{1} + \cdots + a_{n}.$$
Exercise. Show that $s_{n} - s_{n-1} = a_{n}$ for $n \geq 1.$ Note that this holds for any sequence, not just for geometric sequences. This proves that if $s_{n}$ converges, then $a_{n}$ converges to $0$, which is called the general term test. (You might have to call it "$n$-th term test" for your exam, but this terminology is misleading.) A more useful way of thinking about this test is the following.
Term test for divergence. If $a_{n}$ does not go to $0,$ then $\sum_{n=1}^{\infty}a_{n}$ diverges.
Term test for divergence. If $a_{n}$ does not go to $0,$ then $\sum_{n=1}^{\infty}a_{n}$ diverges.
Exercise. Using the fact that $a_{n} = a_{1}r^{n-1}$, show that if $r \neq 1,$ then
\[s_{n} = \frac{a_{1}(r^{n} - 1)}{r - 1} = \frac{a_{1}(1 - r^{n})}{1 - r}.\]
In particular, if $-1 < r < 1,$ then
$$a_{1} + a_{2} + a_{3} + \cdots = \lim_{n \rightarrow \infty}s_{n} = \frac{a_{1}}{r - 1},$$ and you must remember this fact about geometric series!
Exercise. Show that $0.99999 \dots = 1.$
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