Prove the series converges, which sequence is defined inductively Define $t_n$ inductively by $t_1=1$ and $\displaystyle t_{n+1}= \frac {t_{n}}{1+t_{n}^ \beta}$, where $\beta$ is fixed, $0 \le \beta \lt1$. Prove that $\sum_{k=1}^ \infty t_n$ converges.
[Hint : Find a constant $C$ such that $t_n \le Cn^{-1/\beta}$.]

I am not sure this is right, so if you give me advice, that will be very appreciated.
since $\displaystyle t_{n+1}= \frac {t_{n}}{1+t_{n}^ \beta}$
$t_{n+1}({1+t_{n}^ \beta})= t_{n}$
So, $\sum_{k=1}^ \infty t_n$ =  $\sum_{k=1}^ \infty t_{n+1}({1+t_{n}^ \beta}) $
Let $g(n)=t_{n+1}, f(n)= ({1+t_{n}^ \beta})$  
By Abel's test, if $g(n)$ is decreasing sequence and uniformly bounded, and $\sum_{k=1}^\infty f(n)$ converges uniformly, then $\sum_{k=1}^\infty f(n)g(n)$ converges uniformly.
So, since $\displaystyle g(n)=t_{n+1} = \frac {t_{n}}{1+t_{n}^ \beta} $ and $t_n$  is non-negative,
$g(n)$ is a decreasing sequence and is bounded by $t_1=1$.
$\sum_{k=1}^\infty f(n)$ converges by Ratio test,
$\lim_{n\to \infty}|(1+t_{n+1}^{\beta})/(1+t_n^{\beta})| < 1$, since $t_n$ is a decreasing sequence.
So by Abel's test,
$\sum_{k=1}^ \infty t_n$ =  $\sum_{k=1}^ \infty t_{n+1}({1+t_{n}^ \beta}) = \sum_{k=1}^ \infty f(x)g(x)$ converges.
 A: If you can prove the hint, it is easy to finish (without using Abel's test).
For example, if you know that $t_{n}\leq Cn^{-1/\beta}$ and $0<\beta<1$ this is the same as saying $t_{n}\leq \frac{C}{n^{\alpha}}$ where $\alpha=\frac{1}{\beta}>1$.  Then, you can use direct comparison to say that:
$$\sum_{n=1}^{\infty}t_{n} \leq \sum_{n=1}^{\infty}\frac{C}{n^{\alpha}}$$
which latter sum converges since it is a $p$-series with $p>1$.
Trying to show the hint I've gotten stuck (which is why I made this post a community wiki, so that anyone else who sees how to finish can easily edit this and put the finishing touches on).  Here's what I have:
Base Case
$t_{1}=1 \leq \frac{C}{1^{\alpha}}$ for any $C\geq 1$.
Inductive Step
Suppose $t_{i}\leq \frac{C}{i^{\alpha}}$ for all $1\leq i\leq n$ (for some $C\geq 1$, yet to be determined).
Then
\begin{align*}
t_{n+1} &= \frac{t_{n}}{1+t_{n}^{\beta}} \\
        &\leq \frac{\frac{C}{n^{\alpha}}}{1+t_{n}^{1/\alpha}} \\
        &= \frac{C}{n^{\alpha}(1+t_{n}^{1/\alpha}) } \\
        &\vdots \:\:\:\:(???) \\
        &\leq \frac{C}{(n+1)^{\alpha}}
\end{align*}
The final result will hold if you can show that $(n+1)^{\alpha}\leq n^{\alpha}(1+t_{n}^{1/\alpha})$.  If it helps, $n^{\alpha}(1+t_{n})\leq n^{\alpha}(1+t_{n}^{1/\alpha})$, so you are also done if you can show that $(n+1)^{\alpha}\leq n^{\alpha}(1+t_{n})$.
