TITLE: If $\sum_{n = 1}^\infty {{a_n}}$ converges, then is $\sum_{n = 1}^\infty (1+a_n)^{-1}$ a convergent series?
QUESTION [7 upvotes]: If $\sum\limits_{n = 1}^\infty  {{a_n}}$ is convergent (with ${a_n} > 0$, $\forall n\in\mathbb{Z}$), then is $\sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{1 + {a_n}}}} \right)}$ is a convergent series?
I tried with the quotient criterion and comparison theorem no avail.
But I figured something was this:
$${\frac{1}{{1 + {a_n}}}}<1$$
but still I find it useful.
Any suggestions please.

REPLY [2 votes]: A more interesting situation is to consider $$\tag 1 \sum_{i\geqslant 1}\frac{a_i}{1+a_i}$$
If $a_i>0$, and $$\tag 2 \sum_{i\geqslant 1} a_i$$ exists, we have that $\dfrac{1}{1+a_i}\to 1$. Since this is the quotient of the $i$-th term of $(1)$ with that of $(2)$, this means that $(1)$ converges if $(2)$ does.  But if $(1)$ converges, from $\dfrac{a_i}{1+a_i}\to 0$ and $1+a_i>0$ we get that $a_i\to 0$. This gives again a quotient of $1$ and by comparison $(2)$ converges.