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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 ;)

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marked as duplicate by JimmyK4542, Winther, jgon, Did real-analysis Dec 17 '18 at 18:37

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    $\begingroup$ You shouldn't be using the ratio test at all. $\endgroup$ – user10354138 Dec 17 '18 at 18:28
  • $\begingroup$ I suppose you must assume $a_0=0$. $\endgroup$ – Yadati Kiran Dec 17 '18 at 18:33
  • $\begingroup$ It is not finished there $\endgroup$ – VirtualUser Dec 17 '18 at 18:34
  • $\begingroup$ @user10354138 I know, but there it should works too $\endgroup$ – VirtualUser Dec 17 '18 at 18:35
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    $\begingroup$ @VirtualUser: Since sequence of partial sums are monotonic and bounded (from above), so the partial sums converge to the supremum as a consequence of Monotone convergence theorem. Hence the summation converges to the supremum. $\endgroup$ – Yadati Kiran Dec 17 '18 at 18:42
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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)

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

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  • $\begingroup$ I used your hint to solve my problem, can you check if I done this well? $\endgroup$ – VirtualUser Dec 17 '18 at 18:39
  • $\begingroup$ yes, you got the idea (it's a telescoping sum), but you shall pay attention to the starting point (there is not an $a_0$ to subtract): I continued for some further steps .. now you shall be able to conclude. $\endgroup$ – G Cab Dec 17 '18 at 22:53
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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?

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