How Do I show a sequence is bounded? 
Given the sequence $(S_n)$, such that $S_0 = 1$ and $$S_{n+1} = \frac{S_n}{1 + S_n}$$ show that it is convergent?

We were able to show that it was monotone but we are not sure how to show that it is bounded.
 A: $$
{1 \over S_{n + 1}} = {1 \over S_{n}} + 1 = {1 \over S_{n - 1}} + 2
=
\cdots = {1 \over S_{0}} + n + 1 = n + 2
$$
$$
S_{n} = {1 \over n + 1}\quad\Longrightarrow\quad
\color{#0000ff}{\large\lim_{n \to \infty}S_{n} = 0}
$$
A: If you showed the sequence is monotone; decreasing, I assume, you can see that all the terms of the sequence are non-negative, since each term is a ratio of non-negative terms. Then 0 is a bound for your sequence, though not necessarily the limit of the sequence. Then your sequence is a monotone, bounded sequence of Real numbers, so that it converges, to the greatest lower bound of the sequence.  
A: $S_n$ is positive since it is recursively defined as a ration of positive quantities. Now, $S_{n+1} = 1 - \frac{1}{1 + S_n} < 1$, hence bounded.
A: Hint: Consider the function 

$$ f(x)=\frac{x}{1+x},\quad x\geq 1. $$

A: Your sequence is positive by using induction right? Thus, $0 < S_n < S_n + 1$. Dividing gives an upper bound of 1 for $S_{n+1}$.
A: $s_0 =1$ so $s_1 = 1/2$ and $s_2 = 1/(1/2 + 1) = \frac{1/2}{3/2} = 1/3$.  this looks like a pattern.  If you can show that $s_n = 1/(n+1)$ you definitely know the sequence is bounded.
This can be done by induction.  We already have established this is true for $s_1$.  If $s_{n-1} = 1/n$ then $s_n = \frac{1/n}{1 + 1/n} = \frac{1/n}{(n+1)/n} = (1/n)(n+1)/n = 1/(n+1).  
That should do it.
