# Help with series

1. Feb 10, 2009

### djuiceholder

Help with "series"

so I have learned how to do different problems of series.
but theres this problem that I spent hours last night but could not come up with anything.
which is-

n/2^(n-1)
n=1

so from there i was able to take out 2 and end up with: 2 n /(2^n)

i dont know how to do after that. however, i thought about doing (1/2)^n but then again it does not make sense to me.

help needed. i will appreciate that. thanks

2. Feb 10, 2009

### Tom Mattson

Staff Emeritus
Re: Help with "series"

What are you supposed to do with it? I would assume you are supposed to determine if it converges, but unless you say I can't be sure. Also, please show us an attempt at the problem. Then we can help you with where you get stuck.

3. Feb 11, 2009

### Delphi51

Re: Help with "series"

What an interesting piece of work! I wonder who thought that up. It's hypnotic.
Of course it is easy to find the sum on a spreadsheet . . . approximately, I suppose.
I decided it just HAD to be a telescopic series and played with it for a good long time before I found it. Is that enough of a hint?

I started with the nth term written as the difference 4n over 2^n minus 2n over 2^n and then tried to write the 2nd term in the form of the first term with n replaced by n+1 as is required to collapse a telescopic series. Unfortunately that wasn't quite right and nothing I guessed was either. Finally I resorted to adding x to that first term and subtracting x from the second. To find x, I replaced n with n+1 in the first term and set it equal to the second. It came out very nicely and converges to 4 as the spreadsheet predicted, so I expect it is correct. I'm no mathematician, so this may not be the conventional way of summing this series.

I hope I have not spoiled your fun with this intriguing series!

4. Feb 11, 2009

### Delphi51

Re: Help with "series"

Oh, a nicer way to do it is to guess that the nth term is an+b over 2^n minus a(n+1)+b over 2^(n+1), where a and b are two parameters to be found. This setup guarantees that the series will be telescoping. Set that expression equal to 2n over 2^n and solve for a and b.

5. Feb 11, 2009

### Dick

Re: Help with "series"

There's another way to think about it. Can you sum the series 1+x+x^2+x^3+...? Now think what happens if you take the derivative and substitute a special value of x.

6. Feb 11, 2009

### Delphi51

Re: Help with "series"

That's quite a find, Dick! Glad you didn't mention it before I had all that fun finding the telescoping series.

7. Feb 11, 2009

### Dick

Re: Help with "series"

It's just the 'standard' way to do it. Glad you had fun!