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

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 ;)

## marked as duplicate by JimmyK4542, Winther, jgon, Did real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 17 '18 at 18:37

• You shouldn't be using the ratio test at all. – user10354138 Dec 17 '18 at 18:28
• I suppose you must assume $a_0=0$. – Yadati Kiran Dec 17 '18 at 18:33
• It is not finished there – VirtualUser Dec 17 '18 at 18:34
• @user10354138 I know, but there it should works too – VirtualUser Dec 17 '18 at 18:35
• @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. – Yadati Kiran Dec 17 '18 at 18:42

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}

• I used your hint to solve my problem, can you check if I done this well? – VirtualUser Dec 17 '18 at 18:39
• 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. – G Cab Dec 17 '18 at 22:53

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?