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I need some help with a homework problem.

This is Ahlfors exercise 1 p. 178:

Using Taylor's Theorem applied to a branch of $\log (1 + \frac{z}{n})$ prove that $\lim (1 + \frac{z}{n})^n=e^z$ uniformly on all compact sets.

What I did:

Taking the principal branch we have by Taylor's:

$$\log \left(1 + \frac{z}{n}\right) = z -\frac{z^2}{n}+\frac{2z^3}{n^2}- \ldots +f_m(z)z^m$$ Where $f_m(z)$ is a analytic function in the region where the branch is defined, hence: $$1 + \frac{z}{n} = e^{z -\frac{z^2}{n}+\frac{2z^3}{n^2}- \ldots +f_m(z)z^m}$$ $$\Rightarrow \left(1 + \frac{z}{n}\right)^n = e^{n\left(z -\frac{z^2}{n}+\frac{2z^3}{n^2}- \ldots +f_m(z)z^m\right)}$$ then I got stuck, I really apreciate your help. Thanks.

DonAntonio
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user58848
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2 Answers2

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You want to show that, equivalently,

$$\lim_{n \rightarrow \infty} n \log{ \left ( 1+\frac{z}{n} \right )} = z $$

uniformly, on all compact sets. You have made a mistake above in writing out you $\log$ term; you should write, rather:

$$n \log{\left ( 1 + \frac{z}{n} \right )} = z - \frac{z^2}{2 n} + \frac{z^3}{3 n^2} + \ldots + R \left ( \frac{z}{n} \right ) z $$

where $\lim_{n \rightarrow \infty} R \left ( \frac{z}{n} \right ) = 0 $ uniformly on all compact sets.

Ron Gordon
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  • Why $R= R(\frac{z}{n})$ and why it tends to zero? – user58848 Jan 19 '13 at 08:18
  • It must tend to zero as the Taylor expansion for $\log(1+z/n)$ converges $\forall |z/n| < 1$. – Ron Gordon Jan 19 '13 at 08:21
  • the section of the homework is a section before the taylor expansion theorem, so we are suppose to use just the version I've mentioned, but thank you anyway. – user58848 Jan 19 '13 at 08:29
  • Actually, no. Taylor's theorem is stated on Page 125, as Theorem 8, as a finite series in the form I have stated. Have a look at this series and reconsider. – Ron Gordon Jan 19 '13 at 08:38
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My answer is to a large extent in the same spirit as Ron Gordon's. I simply try to make it a bit clearer and remove some unnecessary redundancy.

Take compact set $C_r$ to be the closed disc centered at 1 of radius $r$.

For any sufficiently large $n > n_r$, $\log{\left ( 1 + \frac{z}{n} \right )}$ is well defined and analytic on $C_r$.

Using Theorem 8 on Page 125 in Ahlfors' book, we have

$$\log{\left ( 1 + \frac{z}{n} \right )} = \frac{z}{n} + f_2\left(1+\frac{z}{n}\right) \left(\frac{z}{n}\right)^2 $$

where $f_2\left (1 + \frac{z}{n} \right ) $ is analytic on $C_r$. Since $C_r$ is compact, $f_2\left (1 + \frac{z}{n} \right ) \leq M_r $ for all $n > n_r$ and $z \in C_r$.

Multiply both sides by $n$, we have

$$n\log{\left ( 1 + \frac{z}{n} \right )} = z + f_2\left(1+\frac{z}{n}\right) \frac{z^2}{n} $$

where the $\Big|f_2\left(1+\frac{z}{n}\right) \frac{z^2}{n}\Big| \leq \frac{M_r(1+r)^2}{n}$ tends uniformly to 0 on $C_r$ as $n \rightarrow \infty$. Equivalently, we have

$$\lim_{n\rightarrow\infty} \left(1+\frac{z}{n}\right)^n = e^z$$

uniformly on $C_r$.

Since any compact set is contained in a $C_r$ for a sufficiently large $r$, we have what we intend to prove.

Petra Axolotl
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