# Sum to infinity problem again

1. Jan 24, 2014

### lionely

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

Prove the following result:

$$\frac{1}{2.2} + \frac{ 1}{3.2^2} + \frac{1}{4.2^3} ..... = 2ln2 -1$$

2. Relevant equations

3. The attempt at a solution

I tried writing down the nth term of the series which is 1/(n+1)2^n

But don't know where to move after this.

2. Jan 24, 2014

### Office_Shredder

Staff Emeritus
The series you wrote down has as its first denominator 2.2, but the general term that you wrote down would imply the first denominator is 4. Can you confirm that those decimal points are in fact supposed to be multiplications?

You can tex multiplication with \times, or by just wrapping the thing to be squared in parentheses and putting no multiplication sign at all (which is probably the best way to do things).

Have you covered Taylor series in your class? Problems like this almost always come down to knowing the Taylor series of an appropriate function.

3. Jan 24, 2014

### lionely

The dot means multiply and , I've done Maclaurin's series.

4. Jan 24, 2014

### lionely

I basically know I need to try and split up the nth term into separate series and sum each one and then they'll maybe start looking similar to other series, but splitting them up is the problem.

5. Jan 24, 2014

### Office_Shredder

Staff Emeritus
Don't worry about splitting things up. There are two ways that you can approach this problem. One is that you can look at the series on the left and say 'gee that looks a lot like a McLaurin series of a function that I know the value of', and the other is to look at the right side and say 'if I want to express this as an infinite series I should take a certain function here and write it out as a McLaurin series'. Either way is valid and will work on this problem. So the first question is which function/McLaurin series is being highlighted in the equation that you are given?

6. Jan 24, 2014

### lionely

I would say the McLaurin series of ln(1+x)? and x =1 ?

7. Jan 24, 2014

### Office_Shredder

Staff Emeritus
That sounds like a pretty good idea. What happens when you follow through on the algebra?

8. Jan 24, 2014

### lionely

well ln(1+1) would = 1 - (1^2)/2 +(1^3)/3 ................

9. Jan 24, 2014

### Office_Shredder

Staff Emeritus
Oops I was wrong in my last post. As you can see ln(1+1) doesn't give you what you want. You are going to want to use the Taylor series of ln(1+x) evaluated at some point x, just not x=1

10. Jan 24, 2014

### lionely

But my teacher didn't really teach me about that, so I don't think I should use it here. Is there no way to do this using Mclaurin series?

11. Jan 24, 2014

### Office_Shredder

Staff Emeritus
No, I mean you want the McLaurin series of ln(1+x), but the value of x you should plug in is not x=1.

What can you plug in to get a series that looks a lot like the one on the left? It might not be immediately obvious that you are also getting the thing on the right, but a nice property of logarithms will save the day.

12. Jan 24, 2014

### lionely

since it says 2ln2 which is ln4 should i try x=3 ? But isn't this expansion only valid for -1<x</ 1?

13. Jan 24, 2014

### ehild

Do it with ln (1/2) = ln(1+(-1/2)), that is, with x=-1/2.

ehild