how prove $\sum_{n=1}^\infty\frac{a_n}{b_n+a_n} $is convergent? Let$a_n,b_n\in\mathbb R$ and $(a_n+b_n)b_n\neq 0\quad \forall n\in \mathbb{N}$. The series $\sum_{n=1}^\infty\frac{a_n}{b_n} $ and $\sum_{n=1}^\infty(\frac{a_n}{b_n})^2 $ are convergent.
How to prove that$$\sum_{n=1}^\infty\frac{a_n}{b_n+a_n} $$ is convergent.
Thanks in advance 
 A: Note that
$$
\frac{a_n}{b_n}-\frac{a_n}{a_n+b_n}=\frac{a_n^2}{a_nb_n+b_n^2}\tag{1}
$$
Since $\sum\frac{a_n}{b_n}$ converges, there exists an $N$ so that for $n>N$, we have
$$
\left|\frac{a_nb_n}{b_n^2}\right|=\left|\frac{a_n}{b_n}\right|<\frac12\tag{2}
$$
Then for $n>N$ we also have that
$$
0\le\frac{a_n^2}{a_nb_n+b_n^2}\le2\frac{a_n^2}{b_n^2}\tag{3}
$$
So by comparison to $\sum2\frac{a_n^2}{b_n^2}$, we have that $\sum\frac{a_n^2}{a_nb_n+b_n^2}$ converges. Then using $(1)$ we have that
$$
\sum\frac{a_n}{a_n+b_n}=\sum\frac{a_n}{b_n}-\frac{a_n^2}{a_nb_n+b_n^2}\tag{4}
$$
also converges.
A: Hint: Following Thomas, let us denote $c_n=a_n/b_n$. Since $\sum c_n$ converges, $c_n$ tends to $0$ and 
$$
\frac{c_n}{1+c_n}=c_n(1-c_n+O(c_n^2))=c_n-c_n^2+O(c_n^3)=c_n+O(c_n^2).
$$
A: As suggested by Thomas let $c_n=a_n/b_n$. Then $\sum c_n$ and $\sum c_n^2$ converge. Since $\sum c_n$ converges, we have $\lim_{n\to\infty}c_n=0$. We may assume without loss of generality that $|c_n|\le1/2$. It is easy to check that
$$
x-2\,x^2\le\frac{x}{1+x}\le x,\quad |x|\le1/2.
$$
Then
$$
c_n-2\,c_n^2\le \frac{c_n}{1+c_n}\le c_n.
$$
The convergence of $\sum c_n/(1+c_n)$ follows from the (generalized) comparison theorem.
A: Here's a thought: (I'll let $c_n = a_n / b_n$ following Thomas Andrew's suggestion)
Reductions: assume that $|\sum_n c_n|, \sum_n c_n^2 < 1$, as well as $|c_n| < 1$ for all $n$. Then apply the geometric series formula to the summand, termwise in $n$:
$$
\sum_n \frac{c_n}{1+c_n} = \sum_n \sum_{m=0}^{\infty} (-1)^m c_n^{m+1}
=\sum_n c_n + \sum_n \sum_{m=1}^{\infty} (-1)^m c_n^{m+1}
$$
If we admit the second inequality for the time being, the first term on the RHS is a finite number, and using Tonelli we can show the second term is an absolutely convergent sequence by comparing $\sum_n c_n^{m+1}$ with products of $\sum_n c_n^2$.
I'm skeptical that this is perfectly rigorous, especially because we're dealing with a potentially highly non-absolutely-convergent situation (e.g. $c_n = (-1)^n/n)$.
