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I'm trying to do this problem for complex variable:

Study the convergence of this sequence of analytic functions in $D(0, 1)$. \begin{equation} i) \left\{ f_n (x) = \frac{z^n}{n + z^n} \right\}_{n\geq1} \\ ii) \left\{ f_n (x) = \frac{nz^n}{1 + z^n} \right\}_{n\geq1} \end{equation}

The answer from my professor is :

i) From this estimation $$ \lvert f_n(z) \rvert \leq \frac{1}{n-1} \quad in\, D(0, 1) $$ we deduct the convergency of $f_n(z)$ in $D(0, 1)$ when $n\rightarrow\infty .$

The answer to the second one is pretty much the same but with

$$\lvert f_n(z) \rvert \leq \frac{nr^n}{1-r^n} \quad in\,D(0,r), r<1 $$

But I still don't know why $$\lvert f_n(z) \rvert \leq \frac{1}{n-1}$$ or why $$\lvert f_n(z) \rvert \leq \frac{nr^n}{1-r^n}$$

I've tried using $z^n = \lvert z \rvert^n e^{in\theta} = \lvert z \rvert^n (\cos n\theta + i \sin n\theta)$ or writing $z$ as $z = x + i y$, but I don't get to anything. My professor never explains things completely, and don't worry, it is not so we can learn more by trying hard.

Thanks in advance.

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2 Answers 2

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The priniciple is that you can bound a fraction above by bounding its numerator above and its denominator below.

Since $|z|<1$, we get $|z^2|=|z|\cdot|z|<1\cdot1=1$. Extending by induction yields $|z^n|<1$.

On the other hand, $|n+z^n|>||n|-|z^n||$ by the triangle inequality. Since $|z^n|<1$, we can drop the outside absolute value bars. Furthermore, this means that $n-|z^n|>n-1$. Thus, $$|n+z^n|>||n|-|z^n||=n-|z^n|>n-1.$$ Taking reciprocals yields $\frac{1}{|n+z^n|}<\frac{1}{n-1}$.

We put this together as: $$\left|\frac{z^n}{n+z^n}\right|=\frac{|z^n|}{|n+z^n|}<\frac{1}{|n+z^n|}<\frac{1}{n-1}.$$

Less formally, the power of $z$ in the numerator is less than 1 because $z$ is in the disc. The denominator is bounded below by $n-1$ because the most $z^n$ could take away from $n$, while staying in the unit disc, is 1. If this is a post-proofs course, then I'd take it for granted too (just draw a picture).

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  • $\begingroup$ Thanks. I wish my professor were half as explicit as you are. $\endgroup$
    – Danowsky
    Commented May 12, 2016 at 5:14
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For the second one, use that $\left|nz^n\right|\leq nr^n$ and that $$\left | 1+ z^n \right | \geq \left | 1 - {\left | z\right |}^n \right | \geq 1 - {\left | z\right |}^n=1-r^n$$

so that $$\frac{1}{\left | 1+ z^n \right |}\leq \frac{1}{1-r^n}$$

The first one can be dealt with similarly: $\left|z^n\right|\leq r^n=1$ (because $r=1$) and $\left | n+ z^n \right | \geq n-r^n = n-1$.

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