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Some questions about power series:

i) Why the series $\sum_{n\geq 1} \frac{(-1)^{n}}{z + n} $ converges uniformly on compacts in $\mathbb{C}\setminus { \{ -1, -2, -3, \ldots \} } \ $ ?

ii) Let f be holomorphic in some neighborhood of $ 0 \ $; I have to show that if the series

(*) $ \ \ \ \sum_{n\geq 1} f^{(n)}(z) $

converges absolutely at z=0, then f is an entire function and the series (*) converges normally throughout $ \mathbb{C}$

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Are these from Remmert like the other question, or some other source? –  Jonas Meyer Feb 12 '12 at 8:36
    
Yes they are from Remmert –  WLOG Feb 12 '12 at 8:39
1  
What do you know? What similar results/exercises can you cite? What did you try to solve this one? Why did these tries fail? –  Did Feb 12 '12 at 11:11

1 Answer 1

up vote 1 down vote accepted

For i)

Let $K$ be a compact of $\mathbb{C}$. Then there exists $\delta > 0$ such that $\forall z \in K$ $d( z , \mathbb{N}) > \delta$ and since $K$ is bounded the sequence of function $ z \mapsto \frac{1}{1 +\frac{z}{n}}$ converges uniformily to $1$ on $K$ (Ask me if you want details on the proof of this statment)

Now remind us that to show the power series converges uniformly on $K$ we only have to demonstrate that it is unifomrly Cauchy on $K$. Put $B_k(z) = \frac{1}{z+k+1} - \frac{1} {z+k}$ and $a_k = 0$ if $k$ is even and $1$ if $k$ is odd.

Let's write $\sum_p^{q}{\frac{(-1)^n}{z +n}} = \sum_p^q{a_kB_k(z)} + a_{p-1}\frac{1}{z+q} - a_q\frac{1}{z+q+1}$

$B_k(z) = \frac{-1}{(z+k+1)(z+k)} = \frac{-1}{k^2}\frac{1}{(1+\frac{z}{k})(1+\frac{z+1}{k})}$

Since the sequence of function $ z \mapsto \frac{1}{1 +\frac{z}{n}}$ converges uniformily to $1$ on $K$, for all $\epsilon > 0$ there exists $p$ such that $\forall k \geq p$ and $\forall z \in K$ we have $|B_k(z)| \leq \frac{2}{k^2}$ and $|\frac{1}{z+k}| \leq \epsilon$

Then $ |\sum_p^q{ \frac{(-1)^n}{z +n}}| \leq 2\sum_p^q{\frac{1}{n^2}} + 2\epsilon $

Since it doesn't depends on $z$, the convergence is uniform.

(The method used in this proof is a particular case of the Abel Transformation)

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