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I'm trying to find the limit of the function

$$\frac{1}{n} \text{ln}\left(1 - a \frac{ n}{(1 - z)^n} \right)$$

as $n \to \infty$, where a and z are complex and $|z| < 1$.

I plugged this expression into Mathematica and after some experimenting found that it is convergent to some complex number, and that the result is actually independent of a, but I haven't been able to find a rigorous reason why. Can anyone see how you would evaluate this?

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By $Ln$ do you mean the principal branch of the natural logarithm? If so, where did you put the branch cut? –  Antonio Vargas May 28 '13 at 16:54
    
I chose it to be along the negative real axis but I don't think it matters too much. –  Josh May 28 '13 at 17:06
    
Well unfortunately if $z$ is real and positive, you will be taking the logarithm of a negative number. There seems to be something missing. –  Cameron Williams May 28 '13 at 17:33
    
z is complex and both real and imaginary parts can be positive or negative. –  Josh May 28 '13 at 17:42
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1 Answer

up vote 2 down vote accepted

Very rough answer: of course it is independent of $a$:

$$\frac{1}{n} \log{\left ( 1-a \frac{n}{(1-z)^n}\right)} \sim \frac{\log{(-a)}}{n} + \frac{\log{n}}{n} - \log{(1-z)} \sim -\log{(1-z)}$$

as $n \to \infty$, for $|1-z| < 1$. Otherwise, if $|1-z|>1$, the expression goes as

$$-\frac{a}{(1-z)^n} \to 0$$

in that limit.

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This holds for $|1-z| < 1$. If $|1-z|>1$ then $n/(1-z)^n \to 0$ so the whole quantity goes to $0$ as well. (+1 btw) –  Antonio Vargas May 28 '13 at 19:01
    
@AntonioVargas: quite right. –  Ron Gordon May 28 '13 at 19:05
    
Thanks for the replies @RonGordon but theres one bit that still puzzles me when I work through the above - is the approximation $$ \frac{1}{n} \log{(1 - a \frac{n}{(1-z)^n})} \sim \frac{1}{n} \log{(- a \frac{n}{(1-z)^n})} $$ valid? How would I go about a more rigorous proof of this? Thanks –  Josh May 29 '13 at 11:53
    
@Josh: just determine values of $n$ and $z$ such that $$\left | \frac{a n}{(1-z)^n}\right|$$ exceeds $1$ by, say, a factor of $10$. –  Ron Gordon May 29 '13 at 13:07
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