Explore Fraction Concepts with Examples and Questions

sum ${\displaystyle \underset{n=2}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{(-<!--\; -\; -->1{)}^n}{n^2+n-<!--\; -\; -->2}}$
I am not able to do the telescoping process in the above series. I converted it into the following partial fraction:
$$\underset{n=2}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{(-<!--\; -\; -->1{)}^n}{(n+2)(n-<!--\; -\; -->1)}$$
But nothing seems to cancel (as usually happens) using the telescoping method. How can I solve the above series? Is there any other method to do the above problem?

If $x,y,z$ are positive numbers, then prove that $\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}\le <!--\; \le \; -->2$
Though I have solved a lot of problems on AM-GM inequality, I am unable to solve this one. I am also not showing my working because I do not think that they will be of any help.

$\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\ge <!--\; \ge \; -->3\frac{a^2+b^2+c^2}{a+b+c}$
Proposition
For any positive numbers a, b, and c,
I am requesting an elementary, algebraic explanation to this inequality. (I suppose the condition for equality is that a=b=c.) I am not familiar with symmetric inequalities in three variables. I would appreciate any references.

How many $\frac{1}{8}$ cups are in $\frac{3}{4}$ cup?

Convert the decimal 0.8 to a fraction and reduce it to the lowest term

Simplify $\frac{(0.81{)}^2\times (0.6\times 0.21{)}^2}{{35}^5\times 0.252}$ into $2^n\times 3^m\times 5^p\times 7^k$ where n,m,p,k are integers.

Why not $1$/irrational?
An older math teacher told me that I shouldn't leave a fraction with an irrational in the denominator. But lately, I keep hearing this from every math teacher that I have ever had.
Thus if I have this fraction
$\frac{1}{2^{1/2}}$
I should always convert it to one that has no irrationals in the denominator
$\frac{2^{1/2}}{2}$
But why? My thought are that, just like the fact that we can't write out irrationals, we cant write fractions with irrational as denominators as decimals, but we can approximate them, so I don't see the problem. If we approximate $1/\sqrt{2}$, it would be around 0.70710678118, so I dont understand why I shouldn't include irrationals in denominators, at least if I don't want some points deducted from a test.

Inequality with a+b+c=1
Let $a,b,c$ be positive reals such that $a+b+c=1$. Prove that
${\displaystyle \frac{bc}{a+5}}+{\displaystyle \frac{ca}{b+5}}+{\displaystyle \frac{ab}{c+5}}\le <!--\; \le \; -->{\displaystyle \frac{1}{4}}.$
Progress: This is equivalent to show that
${\displaystyle \frac{1}{a^2+5a}}+{\displaystyle \frac{1}{b^2+5b}}+{\displaystyle \frac{1}{c^2+5c}}\le <!--\; \le \; -->{\displaystyle \frac{1}{4abc}}.$
I'm not sure how to proceed further. Also equality doesn't occur when $a=b=c$, which makes me more confused.
Edit: The original inequality is as follows: Let $a_1,a_2,\cdots <!--\; \cdots \; -->,a_n$ be $n>2$ positive reals such that $a_1+a_2+\cdots <!--\; \cdots \; -->+a_n=1$. Prove that,
$\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}{\displaystyle \frac{\underset{j\ne <!--\; \ne \; -->k}{\prod <!--\; \prod \; -->}{a}_j}{a_k+n+2}}\le <!--\; \le \; -->{\displaystyle \frac{1}{(n-<!--\; -\; -->1{)}^2}}.$

limit of frac function
I'm trying to find
$\underset{x\to <!--\; \to \; -->0}{lim}((e^x-<!--\; -\; -->1)\cdot <!--\; \cdot \; -->\mathrm{frac}<!--\; \; -->\left(\frac{1}{x}\right))$
I thought maybe use squeeze theorem but then I have
$(e^x-<!--\; -\; -->1)\cdot <!--\; \cdot \; -->(\left(\frac{1}{x}\right)-<!--\; -\; -->1)\le <!--\; \le \; -->(e^x-<!--\; -\; -->1)\cdot <!--\; \cdot \; -->\mathrm{frac}<!--\; \; -->\left(\frac{1}{x}\right)\le <!--\; \le \; -->(e^x-<!--\; -\; -->1)\cdot <!--\; \cdot \; -->\frac{1}{x}$
where $\mathrm{frac}<!--\; \; -->(x)=x-<!--\; -\; -->\lfloor <!--\; \lfloor \; -->x\rfloor <!--\; \rfloor \; -->$
p.s - I can't use L'Hopipal rule.

How to isolate algebraic fraction in this case
i have a question about of how to isolate an equation, when it has a multiplication in the denominator.
a/b/3 is same than a*(3/b)
but, when is:
a/(b/3 * 4/a)
If i want to isolate the 4/a that is multiplying the b/3 and in turn dividing to a, so how isolate only the 4/a ?

if C equals three fractions, what is the value of C?
$$c=\frac{30}{7}\cdot <!--\; \cdot \; -->\frac{999}{300}\cdot <!--\; \cdot \; -->\frac{699}{100}$$
What is the value of c ? Is it smaller than 10? between 100 and 1000 ? between 10 and 100? or smaller than 1000?
How can i find that ?

Fraction solving related
If
$$a+b+c=0$$
, then evaluate
$$\frac{a^4+b^4+c^4}{a^2{b}^2+c^2(a^2+b^2)}$$
I tried solving but could make to mentionworthy progress.

Finding the smallest fraction
Please see these fractions:
(A) $\frac{33}{128}$ (B) $\frac{45}{138}$ (C) $\frac{53}{216}$ (D) $\frac{83}{324}$ (E) $\frac{15}{59}$
I need to find out quickly the smallest of these fractions. I am not allowed to use a calculator, though a little rough calculation is allowed.
As I see the problem, without a calculator, converting these to decimal values is not an option. Moreover, since the numbers involved are fairly large, finding the least common denominator and changing each fraction to make their denominators the same as the least common denominator is again next to impossible without a calculator. For the same reason, we can not try the method of making the numerators identical.
If we try the approximate method of changing the numerators and denominators to easy numbers, we can get something like this:
(A) $\frac{30}{120}$ (B) $\frac{45}{135}$ (C) $\frac{50}{200}$ (D) $\frac{80}{320}$ (E) $\frac{15}{60}$
Unfortunately, this makes all of the fractions approximately equivalent to $\frac{1}{4}$ except B, which becomes $\frac{1}{3}$
This is the point where I can not proceed any further.

Three computers can print out a task in 14 min, 21 min, and 42 min, respectively. How long would it take to complete the task with all three computers working?

Brian had $$\frac{2}{5}$$ of a spool of yarn. He used $$\frac{4}{5}$$ of his yarn for a project. What fraction of the spool was used for the project?

Proof in Triangles
The product of the Arithmetic mean of the length of the side of a triangle and harmonic mean of the lengths of the altitude of the triangle is equal to 2*( Area of Triangle)

Solving without squaring
If $\sec <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->\csc <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}={\displaystyle \frac{4}{3}}$, then prove that $\mathrm{\theta <!--\; \theta \; -->}={\displaystyle \frac{1}{2}}\arcsin <!--\; \; -->{\displaystyle \frac{3}{4}}$
I have tried really hard to solve this without squaring both sides of the equation but it seems next to impossible.
The closest thing I reached is $\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}={\displaystyle \frac{3(1-<!--\; -\; -->2{\sin }^2<!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}/2))}{8{\sin }^2<!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}/2)-<!--\; -\; -->1}}$
I also obtained $\sin <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}={\displaystyle \frac{-<!--\; -\; -->3}{2}}{\displaystyle \frac{(1-<!--\; -\; -->\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}/2{)}^2}{1+{\tan }^2<!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}/2)}}$

$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+...\to <!--\; \to \; -->?$
I tried like below
$$\begin{gathered} \frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+\cdots <!--\; \cdots \; -->= \\ \underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{1^2+2^2+3^2+\cdots <!--\; \cdots \; -->+n^2}= \\ \underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{\frac{n(n+1)(2n+1)}{6}}= \\ \underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{6}{n(n+1)(2n+1)}= \end{gathered}$$
Then I can use the fraction ,but
$$\frac{1}{n(n+1)(2n+1)}=\frac{1}{n}+\frac{1}{n+1}+\frac{-<!--\; -\; -->4}{2n+1}$$
This is ugly to turn into telescopic series . Can you help me to find :series converge to ?
Thanks in advance .

Given that ${\log }_6<!--\; \; -->2=.387$ ,find ${\log }_6<!--\; \; -->9$
I don't know proper method to solve this problem . Please help me.

Unknown process in integration
For the integrand $$\frac{1}{y(1-<!--\; -\; -->\frac{y}{L})}$$, cancel common terms in the numerator and denominator:
$$=\int <!--\; \int \; -->-<!--\; -\; -->\frac{L}{y(y-<!--\; -\; -->L)}dy$$
This is part of WolframAlpha’s calculations before creating partial fractions. What exactly is this method/theorem? And how does it work?