<s>[INST] 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 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 ;)
 [/INST] 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)
$$
\eqalign{
  &  = {{a_1 } \over {\left( {a_1  + 1} \right)}} + {{a_2 } \over {\left( {a_1  + 1} \right)\left( {a_2  + 1} \right)}}
 + {{a_3 } \over {\left( {a_1  + 1} \right)\left( {a_2  + 1} \right)\left( {a_3  + 1} \right)}} +  \cdots  =   \cr 
  &  = {{a_1 } \over {\left( {a_1  + 1} \right)}} + {{a_2  + 1 - 1} \over {\left( {a_1  + 1} \right)\left( {a_2  + 1} \right)}}
 + {{a_3  + 1 - 1} \over {\left( {a_1  + 1} \right)\left( {a_2  + 1} \right)\left( {a_3  + 1} \right)}} +  \cdots  =   \cr 
  &  = {{a_1 } \over {\left( {a_1  + 1} \right)}} + {1 \over {\left( {a_1  + 1} \right)}} - {1 \over {\left( {a_1  + 1} \right)\left( {a_2  + 1} \right)}}
 + {1 \over {\left( {a_1  + 1} \right)\left( {a_2  + 1} \right)}} - {1 \over {\left( {a_1  + 1} \right)\left( {a_2  + 1} \right)\left( {a_3  + 1} \right)}} +  \cdots  =   \cr 
  &  = {{a_1 } \over {\left( {a_1  + 1} \right)}} + {1 \over {\left( {a_1  + 1} \right)}}
 - {1 \over {\left( {a_1  + 1} \right)\left( {a_2  + 1} \right)\left( {a_3  + 1} \right)}} +  \cdots  =   \cr 
  &  = 1 - {1 \over {\left( {a_1  + 1} \right)\left( {a_2  + 1} \right)\left( {a_3  + 1} \right)}} +  \cdots  =  \cdots  \cr} 
$$
</s>