# Show that series converges to 1 while another diverges

1. Nov 29, 2013

### Jaggis

1. The problem statement, all variables and given/known data

Let $a_n≥ 0$ when $n ≥ 1$.

Show that

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

when $\sum_{n=1}^{\infty}{a_n}$ diverges.

3. The attempt at a solution

I tried to do something with partial sums because they should approach 1 as n goes to infinity. I looked at the difference and quotient of partial sums for n and n+1 but it didn't go anywhere. I didn't find any use for the fact that the n:th term in the first series should go to zero when n goes to infinity.

2. Nov 29, 2013

### Staff: Mentor

The sum of the first n terms has a very simple expression (did you write down the first 2 or 3 terms and simplified?). You just have to show that the denominator of that sum diverges, and that is not hard.

3. Nov 30, 2013

### Jaggis

The expression I got for a partial sum would be

$S_n = \frac{a_1(1+ a_1)...(1+a_n) +...+ a_{n-1}(1+a_n) + a_n}{(1+a_1)(1+a_2)...(1+a_n)}$.

If the whole series converges to 1 then according to my knowledge also

$S_n→ 1$ as $n→ ∞$.

I don't know how to get this result, even if the denominator of $S_n$ diverged.

4. Nov 30, 2013

### Staff: Mentor

For S_n, it is tricky to see how to simplify the numerator. Try S_2 or S_3 and see how many factors of the denominator you can include somehow in the numerator.

The divergence of the denominator is the following step, it does not help you at this point.

5. Nov 30, 2013

### haruspex

That's not quite right. You should not have a1(1+ a1) at the start.

6. Dec 2, 2013

### Jaggis

Yes, you are right. Should be a1(1+ a2).

I'm thinking

$\frac{a_1}{1+a_1} + \frac{a_2}{(1+a_1)(1+a_2)} = \frac{(1+a_1)(1+a_2) -1}{(1+a_1)(1+a_2)} = 1-\frac{1}{(1+a_1)(1+a_2)}$.

And then the last term goes to zero?

7. Dec 2, 2013

### Staff: Mentor

If you generalize that to the sum of the first n fractions (which looks exactly as you can expect based on your result): yes.

8. Dec 2, 2013

### Jaggis

OK, thanks for your help. I return with more questions on the subject in case I find myself in trouble with something.