Bounded Sequences I came across the following problems during the course of my self-study of real analysis:

Show that the sequence $(x_n)$ defined by $x_n = 1+ \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}$ is unbounded. 

I know a sequence $(x_n)$ is bounded if there exists a positive number $K$ such that $|x_n| \leq K$ for all $n$. So suppose for contradiction that it is bounded. Maybe we can define sequences $a_n = x_n-1$, $b_n = a_n-\frac{1}{2}$, $c_n = b_n- \frac{1}{3} \dots$ and try to come up with a contradiction?

Show that the sequence $(x_n)$ defined by $x_1 = x$,  $x_{n+1} = x_{n}+ 1/x_n$ is unbounded. 

Suppose for contradiction that $(x_n)$ is bounded by $K$ for all $n$. Then show that there is some $K' < K$ which is also an upper bound? 

Show that the sequence $(x_n)$ defined by $x_n = 1+ \frac{1}{2!}+ \frac{1}{3!} + \dots + \frac{1}{n!}$ is bounded above by $2$.

So there is some relationship between $n!$ and $2^{n-1}$. I think $n! \geq 2^{n-1}$ and we can prove this by induction on $n$? So $x_n \leq 1+1+ \frac{1}{8} + \dots + \frac{1}{2^{n-1}}$? 
 A: *

*The simplest way to show that a sequence is unbounded is to show that for any $K\gt 0$ you can find $n$ (which may depend on $K$) such that $x_n\geq K$.
The simplest proof I know for this particular sequence is due to one of the Bernoulli brothers Oresme. I'll get you started with the relevant observations and you can try to take it from there:
Notice that $\frac{1}{3}$ and $\frac{1}{4}$ are both greater than or equal to $\frac{1}{4}$, so
$$\frac{1}{3}+\frac{1}{4}\geq \frac{1}{4}+\frac{1}{4} = \frac{1}{2}.$$
Likewise, each of $\frac{1}{5}$, $\frac{1}{6}$, $\frac{1}{7}$, and $\frac{1}{8}$ is greater than or equal to $\frac{1}{8}$, so
$$\frac{1}{5}+\frac{1}{6}+\frac{1}{7} + \frac{1}{8} \geq \frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8} = \frac{1}{2}.$$
Now look at the fractions $\frac{1}{n}$ with $n=9,\ldots,16$; compare them to $\frac{1}{16}$; then compare the fractions $\frac{1}{n}$ with $n=17,\ldots,32$ to $\frac{1}{32}$. And so on.
See what this tells you about $x_1$, $x_2$, $x_4$, $x_8$, $x_{16}$, $x_{32}$, etc.

*Your proposal does not work as stated. For example, the sequence $x_n = 1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n-1}}$ is bounded by $K=10$; but it's also bounded by $K=5$. Just because you can find a better bound to some proposed upper bound doesn't tell you the proposal is contradictory. It might, if you specify that you want to take $K$ to be the least upper bound of the sequence. Even so, it's hard to establish that a sequence is unbounded that way. (Note also that you haven't really defined the sequence very well: it is undefined for $x=0$, though that is the only problem.)
To get you started: Show that if you start the sequence with $-x$ instead of $x$, then you just get the same sequence multiplied by $-1$. That is, if you fix $x\neq 0$, and you let $y_1=-x$, $y_{n+1}= y_n + (1/y_n)$, then $y_k = -x_k$; so the sequence $(x_n)$ is bounded if and only if the sequence $y_k$ is bounded, and so you may assume $x\gt 0$.
Then show that if $0\lt x\lt 1$, and you let $y_1 = \frac{1}{x}$, $y_{n+1} = y_n+(1/y_n)$, then $y_k=x_k$ for $k\geq 2$; so you may assume that $x\geq 1$. 
Now you have that the sequence is increasing. If it were bounded, it would converge, say to $L\gt 0$. Then
$$L = \lim_{n\to\infty}x_n = \lim_{n\to\infty}x_{n+1} = \lim_{n\to\infty}\left(x_n + \frac{1}{x_n}\right) = \lim_{n\to\infty}x_n + \frac{1}{\lim_{n\to\infty}x_n} = L+\frac{1}{L}.$$
I think that's a very big problem for $L$...

*Yes, if you can prove that $n!\gt 2^{n-1}$ for all $n\geq 1$, then you can bound your sequence by a sequence of powers of $\frac{1}{2}$; if you can show that sequence is bounded, you'll be done. And, yes, you can prove the inequality in question by induction on $n$. It's very simple to do. But you messed up your computations later (that second $1$ should be a $\frac{1}{2}$). If $n!\geq 2^{n-1}$, then
$$\begin{align*}
x_n &= 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots + \frac{1}{n!} \\
&\leq 1 + \frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots + \frac{1}{2^{n-1}}\\
 &= \frac{1 - \frac{1}{2^n}}{1 - \frac{1}{2}}\\
 &= \frac{2^n-1}{2^{n-1}}\\
& = 2 - \frac{1}{2^{n-1}}.\end{align*}$$
Almost done! 
A: The first part was answered in Grotaur's answer. The third part you did it yourself.
For the sequence $x_{n+1}=x_n+\frac{1}{x_n}$, first note that if the first term is positive so are all the others; if the first term is negative, so are all the others. Suppose that the first term is positive, and therefore the sequence is increasing ($x_{n+1}-x_n=\frac{1}{x_n}>0$.)
Suppose now that the sequence is bounded. A bounded increasing sequence is convergent, and denote by $L$ it's limit. Take $n \to \infty $ in the recurrence relation, and see what values could $L$ have.
A: *

*Consider terms $s_k = x_{2^k-1} - x_{2^{k-1}}$. Prove that $s_k\geq \frac{1}{2}$ and note that $x_{2^n}\geq s_1+s_2+...s_n\geq \frac n 2$.

