# Proving ln(x) using infinite series

1. Mar 31, 2008

### darewinder

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

Well we are given a series of steps done with the number "x" and in the end the end value is ln(x). Basically we are asked to prove why it isnt a coincedience

2. Relevant equations

I put the steps into an equation, but i cant prove it.

ln(x) =$$^{lim }_{n->inf}$$ $$(x^\frac{1} {2^n} -1)*2^n$$

3. The attempt at a solution

Well plugging in gives me inf times 0 so i thought of solving it using Hopital's rule but i cant get to a form where it is 0/0. I tried factoring, rationalizing but i couldnt get anywhere. I would appreciate if you guys can help me do this little bit.

and I would also appreciate if anyone can show me how to come up with a sequence for this. Second part of the question askes me to come up with my own limit of a sequence to get ln(x). I just flipped changed the sine of - inside the brackets to make it + and added the minus sign to the -2n at the end, but thats like the same thing. So if you guys can give me some ideas it would be great!! :)

Thank you

Last edited: Mar 31, 2008
2. Mar 31, 2008

### HallsofIvy

Staff Emeritus
Since n only occurs in 2n, you can simplify by letting a= 2n. Then the problem becomes showing that
$$ln(x)= \lim_{a\rightarrow \infty}(x^(1/a)-1)(a)$$

Now I would be inclined to "reverse" the function: If y= (x1/a-1)(a), then x= (y/a+ 1)a. Do you recognize the limit of that as a common formula for ey? And if x= ey, then y= ln(x).

3. Mar 31, 2008

### darewinder

tanks a bunch ill get back on you on the reverse function let me work it out. Im eating right now :) But the limit makes sense (argg why didn't i think of that)

4. Mar 31, 2008