# Homework Help: Sum of an infinite series

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1. Nov 3, 2016

### Buffu

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

Find the sum of the given infinite series.
$$S = {1\over 1\times 3} + {2\over 1\times 3\times 5}+{3\over 1\times 3\times 5\times 7} \cdots$$

2. Relevant equations

3. The attempt at a solution
I try to reduce the denominator to closed form by converting it to a factorial.

$$\sum_{k \ge 1} {k\over {\prod^{k + 1}_{a = 1} 2a -1 }}$$
$$\sum_{k \ge 1} {k\times \prod^{k + 1}_{a = 1} 2a\over \prod^{k + 1}_{a = 1} 2a -1 \times \prod^{k + 1}_{a = 1} 2a}$$
$$\sum_{k \ge 1} {k\times 2^{k+1} \times (k+1)! \over (2(k+1))!}$$

I hit the dead end here. Although i can simply this a bit more but i still can't find a series that i can sum easily.
Please provide some hints as to how can i proceed further.

Last edited: Nov 3, 2016
2. Nov 3, 2016

### Staff: Mentor

What are you supposed to show in this problem? Your problem description is lacking this information.

3. Nov 3, 2016

### Fightfish

The trick is to write it as a telescopic sum:
$$\frac{n}{(2n+1)!!} = \frac{1}{2(2n-1)!!} - \frac{1}{2(2n+1)!!}$$
(in case you haven't seen it before, !! is the double factorial)

4. Nov 3, 2016

### Buffu

I am really sorry Mark. I have edited the question. Please see if my edit is sufficient enough.

5. Nov 3, 2016

### Staff: Mentor

Much better. I wasn't sure whether you were supposed to determine if the series converged or diverged, or if it converged, find the sum.

6. Nov 3, 2016

### Buffu

I don't know what double factorial means. Sorry.
Does double factorial means $n!! = n(n-2)(n-4)\times\cdots \times3 \times 1$ or $(n!)!$?

Also, can you tell how you know that this is a telescopic series ?
Have you done this question before ?

7. Nov 3, 2016

### Fightfish

This one.
Nope, haven't done it before - but its a standard technique for evaluating infinite summation of series. It probably helps to examine the structure of the first few terms. This can lead you to a good guess / ansatz that you can then prove in general for an arbitrary term.

8. Nov 3, 2016

### Buffu

Thank you very much. You are genius for sure.