Sequence that is neither increasing, nor decreasing, yet converges to 1 Give an example of a sequence which is neither increasing after a while, nor
decreasing after a while, yet which converges to 1. 
My solution: $1.01,\ .99,\ 1.001,\ .999,\ 1.0001,\ .9999,\ \text{etc}\dots$
Does that satisfy all the conditions? Also, judging by the instructions, do you think I would have to define that sequence? In which case, I could do $\{x_n\} = 1 + .01^n$ for odd $n$ and $1 - .01^n$ for even $n$ (which would change the sequence, but just increases the rate at which it approaches $1$). 
The definition of an increasing sequence used is the next term being bigger than OR equal to the preceding term. And dually for decreasing.
 A: Yes, your sequence satisfies all required conditions.
Another example sequence:
$$a_n = \begin{cases}
1+\frac 1n & \text{if } \log_{10}n \in \mathbb N \cup \{0\} \\
1 & \text{otherwise}
\end{cases}$$
has terms:
$$\begin{align}
a_1 & = 2 \\
a_{10} & = 1.1 \\
a_{100} & = 1.01 \\
a_{1000} & = 1.001 \\
a_{10000} & = 1.0001 \\
\ldots
\end{align}$$
and all remaining terms equal $1$.
A: I think defining it the way you did is fine, but you could also use
$$s_n = 1+\left(-\frac{1}{10}\right)^n$$
which is a handy little trick to express alternating sequences nicely. 
A: Your solution works.
I like $1+\dfrac{\sin n}{n}$.  But it's possible I'm weird.
A: EDIT: As the comments note, the original formulation of the question was not specific about whether the inequalities in the definitions were strict.  With the additional specification now that "increasing" and "decreasing" allow for equality, this answer of course doesn't fit anymore.
You could just take the sequence 1, 1, 1, 1, 1, ... .
A: It's so easy to find such a series, for example:
$$a_n=q^n+1\ ,\ -1<q<0$$
A: Continued fraction representations of irrational numbers can be represented as an infinite sequence of rational partial convergents which oscillate above and below the number being represented.
Thus if we take $a_n$ to be the to be the nth partial convergent of any irrational number divided by that number, it will converge to 1 in an oscillatory manner.
$$
a_n = \frac{[\zeta]_n}{\zeta}, \zeta \in \mathbb{I}
$$
For $\zeta = \sqrt{2}$, we have
$$
a_0 = \frac{1}{\sqrt{2}} \approx 0.707\\
a_1 = \frac{3}{2\sqrt{2}} \approx 1.061\\
a_2 = \frac{7}{5\sqrt{2}} \approx 0.990\\
a_3 = \frac{17}{12\sqrt{2}} \approx 1.002\\
...
$$
A: I think it is possible to use a well known Fibonacci sequence property:
$$\lim_{n\to\infty}\frac{F_{n}}{F_{n-1}} = \varphi$$
where $\varphi$  is the golden ratio, and also it is well known that each term of the sequence, i.e. let us call it $a_n=\frac{F_{n}}{F_{n-1}}$, oscillates between a value over and under $\varphi$. For that reason the sequence $$\left\{n \gt 2:a_n=\frac{F_{n}}{F_{n-1}\cdot \varphi}\right\}$$ will be neither increasing after a while, nor decreasing after a while, and will converge to $1$.
