Convergence of recursive sequence convergence iff 
Let $\{b_n\}_n \subseteq \mathbb{R}^+$. If the sequence $\{a_n\}_n$ is defined as $$a_n = \begin{cases}a_{n-1}+\frac{b_{n-1}}{a_{n-1}}&\text{if n>1}\\k&\text{if n=0}\end{cases}$$
  where $k$ is a positive number. Then $$\{a_n\}_n \text{ is convergent  }\Leftrightarrow \sum b_n < \infty$$

I prove the $(\Leftarrow)$. For the other implication I says
$$\sum b_n = \sum a_n(a_{n+1}-a_n)$$
I know that $a_n$ is bounded, but how can I control $(a_{n+1}-a_n)$?
 A: Note that $a_n-a_{n-1}=\frac{b_{n-1}}{a_{n-1}}$. Since $a_0\gt0$, $a_n$ is increasing.
Suppose $a_n\to A$, then
$$
\begin{align}
\sum_{n=1}^\infty\frac{b_{n-1}}{A}
&\le\sum_{n=1}^\infty\frac{b_{n-1}}{a_{n-1}}\\
&=\sum_{n=1}^\infty(a_n-a_{n-1})\\
&=A-k
\end{align}
$$
Thus,
$$
\sum_{n=0}^\infty b_n\le A(A-k)
$$
Suppose that 
$$
\sum_{n=0}^\infty b_n=B\lt\infty
$$
Then
$$
\begin{align}
\lim_{n\to\infty}a_n
&=k+\sum_{n=1}^\infty(a_n-a_{n-1})\\
&=k+\sum_{n=1}^\infty\frac{b_{n-1}}{a_{n-1}}\\
&\le k+\sum_{n=1}^\infty\frac{b_{n-1}}{k}\\
&=k+\frac{B}{k}
\end{align}
$$
A: Assume $\sum a_n$ converges. Then $a_n \to 0$ as $n \to \infty$. So $\{a_n\}$ is Cauchy. Therefore for all $\epsilon > 0$ there exists $N = N(\epsilon) > 0$ such that $|a_{n+1} - a_n| < \epsilon$ for all $n \geq N$. So
$$
\left|\sum_{n=N(1)}a_n(a_{n+1} - a_n)\right| \leq \epsilon \sum_{n=N(1)}^\infty |a_n| = \epsilon \sum_{n=N(1)}^\infty a_n.
$$
A: We find $b_n=a_n(a_{n+1}-a_n)$ by rearranging the recurrence equation.
By induction, $a_n>a_{n-1}>0$, i.e. $\{a_n\}_n$ is strictly increasing.
Thus convergence of $\{a_n\}_n$ is equivalent to boundedness.
If $a_n\to a$, then $0<b_n<a(a_{n+1}-a_n)$ and so by telescoping $\sum b_n<a(a-a_0)<\infty$.
On the other hand, if we assume that  $\{a_n\}_n$ is unbounded. Then $\sum_{n=m}^\infty(a_{n+1}-a_n)=+\infty$ for any $m$ and (with the convention $a_{-1}=0$)
$$\begin{align}\sum_{n=0}^\infty b_n&=\sum_{n=0}^\infty a_n(a_{n+1}-a_n)\\ &= \sum_{n=0}^\infty\sum_{m=0}^n(a_m-a_{m-1})(a_{n+1}-a_n)\\
&\stackrel{(1)}= \sum_{m=0}^\infty\sum_{n=m}^\infty(a_m-a_{m-1})(a_{n+1}-a_n)\\
&= \sum_{m=0}^\infty\left(\underbrace{(a_m-a_{m-1})}_{>0}\cdot\underbrace{\sum_{n=m}^{\infty}(a_{n+1}-a_n)}_{+\infty}\right)\\
&=+\infty\\
\end{align}$$
Note that the rearrangement at ($1)$ is allowed because "everything" is positive.
