Expressing a simple sequence in closed form I just started sequences, and I'm having trouble understanding how to finda pattern. My teacher didn't really explain the technique of finding the closed form. I there any formulas that I should memorize?
I have 3 problems that I couldn't solve. Any help will be appreciated.
BTW, how do I format my input?
1) $1, .1, .01, .001, .0001, .00001, \dots$
2) $\dfrac{1}{1}, \dfrac{2}{2},  \dfrac{4}{6},  \dfrac{8}{24},  \dfrac{16}{120},  \dfrac{32}{720}, \dots$
3) $-\dfrac{1}{\sqrt{3}}, \dfrac{2}{\sqrt{5}}, -\dfrac{3}{\sqrt{7}}, \dfrac{4}{\sqrt{9}}, -\dfrac{5}{\sqrt{11}}, \dots$
 A: Normally I’d give hints, but it sounds to me as if you could benefit from a more detailed discussion at this point.
There are two steps: the first is recognizing the pattern, and the second is expressing that pattern with a formula.


*

*The pattern here is obvious: these are successive decreasing powers of $10$. A generic power of $10$ is $10^n$; what values of $n$ do we have here? Clearly $n=0,-1,-2,\dots\,$. We’d like our sequence to be $a_0,a_1,a_2,\dots$, not $a_0,a_{-1},a_{-2},\dots$, but that’s easily managed: replace $n$ in the exponent by $-n$ to get $$a_n=10^{-n}\;.$$

*The numerators of the fractions are $1,2,4,8,16$, and $32$; clearly they’re doubling at each stage. In fact, they’re immediately recognizable as powers of $2$, specifically, $2^0,2^1,2^2,2^3,2^4$, and $2^5$. Thus, our $n$-th term $a_n$ should be a fraction with numerator $2^n$. The first denominator is wrong: it should be $1$. Once that error is corrected, we have $1,2,6,24,120$, and $720$. These are perhaps less familiar than powers of $2$, but they’re numbers that you should recognize as factorials: $1!,2!,3!,4!,5!$, and $6!$. Notice that the exponent $3$, say, in the numerator matches $4!$ in the denominator, and so on through each of the six terms that we’re given: if the numerator is $2^n$, the denominator is $(n+1)!$, not $n!$. Thus, $$a_n=\frac{2^n}{(n+1)!}\;.$$

*The first thing to notice is the alternating sign: that’s a sure sign that you’ll have a factor of $(-1)^n$ or $(-1)^{n+1}$. It’s actually a little nicer to index this sequence starting with $a_1$ instead of $a_0$: then the numerator of $a_n$ (ignoring the sign) is simply $n$, rather than $n+1$, and the alternating sign is correctly produced by $(-1)^n$ instead of $(-1)^{n+1}$. Okay: the numerator of $a_n$ (where $n$ starts at $1$) is going to be $(-1)^nn$. What about the denominator? For $n=1,2,3,4,5$ it’s $3,5,7,9,11$. The key point is that it’s following an arithmetic progression, so it can be described by a linear polynomial in $n$. Specifically, the $n$-th term of an arithmetic progression with constant difference $d$ always has the form $a+dn$ for a suitable starting value $a$. Here $d=2$, so we want to describe those denominators as the square root $\text{something}+2n$; clearly the $\text{something}$ must be $1$, and the denominator of $a_n$ is $\sqrt{2n+1}$. Thus, $$a_n=\frac{(-1)^nn}{\sqrt{2n+1}}\;.$$
A: You may format your input using Latex; there are many online resources on how to use it.
Regarding the question about finding patters, I think it is mostly experience. In many cases, usually a fixed number is added, or multiplied, etc.
In your case, for the first sequence, check out the first few elements, $1$, $0.1$, $0.01$, it appears that you form the next element by dividing the current element by $10$. Well, the rest of the series verifies this observation, and there you go...
For series with fractions, and especially if they are not written in their simplified forms (as in your second example), you may look at the numerators and the denominators separately. The same goes for the third sequence.
A: Not easy to help without solving it completely. I assume your sequences start at rank $n=0$.
For 1., these are powers of $1/10$. First term is $10^{-0}$. Second term is $10^{-1}$. Third term is $10^{-2}$. Etc...

$$ x_n=10^{-n} $$

For 2., you get powers of $2$ at the numerator and factorials in the denominator. Your first term should probably be $1$. Otherwise, just consider the following formula for $n\geq 1$.

$$ y_n=\frac{2^n}{(n+1)!} $$

For 3., there is an alternate sign, that is $(-1)^n$ or $(-1)^{n+1}$. Choose the right one. The numerator is transparent: $1,2,3,\ldots$. The denominator is a square root of an odd number.

$$ z_n=(-1)^{n+1}\frac{n+1}{\sqrt{2n+3}} $$

A: Hints:
$$10^{-n}=\frac{1}{10^n}$$
$$\frac{2^{n-1}}{n!}\ldots\text{but check the first one: it should be }\;1$$
$$\frac{(-1)^nn}{\sqrt{2n+1}}$$
