If $a_n >0$ for all $n\geq1$, show that the series

$$\sum_{n=1}^{\infty} \frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)}$$


Can someone check if my solution is correct:

$$\sum_{n=1}^{M}\frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)} = 1-\frac{1}{(1+a_1)(1+a_2)...(1+a_M)} < 1 $$

Since the partial sum of the series is bounded, the series is convergent.

  • $\begingroup$ "Since the partial sum of the series is bounded, the series is convergent." - This, in general, does not hold. $\endgroup$ – Belgi Mar 22 '14 at 10:07
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    $\begingroup$ Well, given that the partial sum is bounded above, if you additionally prove that it is a non-decreasing sequence, then it is convergent. Nevertheless, I haven't checked your equality yet. It doesn't look straightforward. $\endgroup$ – Manolito Pérez Mar 22 '14 at 10:16
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    $\begingroup$ @user136266 Take $\sum_{i=0}^\infty (-1)^i$. The partial sums are all within $[0,1]$, yet the series is not convergent. $\endgroup$ – fgp Mar 22 '14 at 10:20
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    $\begingroup$ @fgp, isn't it true that a series of non-negative terms is convergent iff its partial sums form a bounded sequence. But in your example, the terms can be negative. $\endgroup$ – user75930 Mar 22 '14 at 10:24
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    $\begingroup$ @SabyasachiMukherjee Yes. But the OP's answer fails to state that explicitly. You can't just say "The partial sums form a bounded sequence, hence the sum converges". You have to say "The terms are all non-negative, hence the partial sums form a monotone sequence, and since the partial sums are also bounded, the series converges". $\endgroup$ – fgp Mar 22 '14 at 10:27


Indeed, your formula is correct and it can be proved inductively.

The series converges, because its partial sums form an increasing sequence which is upper bounded by $1$. Hence it converges, and its limit (i.e., the sum of the series) is also less or equal to $1$.

  • $\begingroup$ It should be pointed out that in this case, the ratio test cannot be used, because the limit of the ratio between two consecutive terms of the sequence (i.e. two consecutive partial sums) is one. $\endgroup$ – Manolito Pérez Mar 22 '14 at 10:31

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