If $a_n > 0$ prove that $\sum_{n=1}^{\infty} \frac{a_n}{(a_1+1)(a_2+1)\cdots(a_n+1)}$ converges

Question

I have an interesting task: If $a_n > 0$, prove that $$\sum_{n=1}^{\infty} \frac{a_n}{(a_1+1)(a_2+1)\cdots(a_n+1)}$$ converges.

I thought that it will be simple because ratio test gives me: $$\frac{u_{n+1}}{u_n}= \frac{a_{n+1}}{a_{n+1}+1}\cdot a_n^{-1} < 1 \cdot a_n^{-1} = \frac{1}{a_n}$$ and $a_n$ should be in $[0,1]$. But... In my opinion it can be over that... why need I assume that $ a_n \rightarrow g \in [0,1] $? There is similar topic on this forum, but It was not solved there...
@edit I saw that: $$\sum_{n=1}^{N}\frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)} = 1-\frac{1}{(1+a_1)(1+a_2)...(1+a_N)} < 1 $$ So if series of partial sum is bounded from up, the sum converges, that is right? @edit2 but It is good? Look at that: $$ \sum_{n=1}^{N}\frac{a_n+1-1}{(1+a_1)(1+a_2)...(1+a_n)} = \sum_{n=1}^{N}\frac{1}{(1+a_1)(1+a_2)...(1+a_{n-1})}-\frac{1}{(1+a_1)(1+a_2)...(1+a_n)} $$ why somebody changed first part into $1$? @edit3 Ok, I think that I have understood, thanks for your time ;)

Answer 1

Hint:

$$ \eqalign{ & \sum\limits_{1\, \le \,n\,} {{{a_n } \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right) \cdots \left( {a_n + 1} \right)}}} = \cr & = \sum\limits_{1\, \le \,n\,} {{{\left( {a_n + 1} \right) - 1} \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right) \cdots \left( {a_n + 1} \right)}}} = \cdots \cr} $$

(continuing)

$$ \eqalign{ & = {{a_1 } \over {\left( {a_1 + 1} \right)}} + {{a_2 } \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right)}} + {{a_3 } \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right)\left( {a_3 + 1} \right)}} + \cdots = \cr & = {{a_1 } \over {\left( {a_1 + 1} \right)}} + {{a_2 + 1 - 1} \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right)}} + {{a_3 + 1 - 1} \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right)\left( {a_3 + 1} \right)}} + \cdots = \cr & = {{a_1 } \over {\left( {a_1 + 1} \right)}} + {1 \over {\left( {a_1 + 1} \right)}} - {1 \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right)}} + {1 \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right)}} - {1 \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right)\left( {a_3 + 1} \right)}} + \cdots = \cr & = {{a_1 } \over {\left( {a_1 + 1} \right)}} + {1 \over {\left( {a_1 + 1} \right)}} - {1 \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right)\left( {a_3 + 1} \right)}} + \cdots = \cr & = 1 - {1 \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right)\left( {a_3 + 1} \right)}} + \cdots = \cdots \cr} $$

Answer 2

Here, the ratio test is useless because you have zero information on $a_n$.

May I suggest that you compute the first partial sums to “get a feeling” about what happens?