# Prove the convergence of a series.

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

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

converges.

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.

• "Since the partial sum of the series is bounded, the series is convergent." - This, in general, does not hold. – Belgi Mar 22 '14 at 10:07
• 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. – Manolito Pérez Mar 22 '14 at 10:16
• @user136266 Take $\sum_{i=0}^\infty (-1)^i$. The partial sums are all within $[0,1]$, yet the series is not convergent. – fgp Mar 22 '14 at 10:20
• @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. – user75930 Mar 22 '14 at 10:24
• @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". – fgp Mar 22 '14 at 10:27

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 $1$.