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Prove for $a\gt0$ that following series is holomorphic

$$ \sum_{n=1}^\infty \frac {1}{(a+n)^z} \quad \textrm{for} \quad \operatorname{Re}z \gt 1 $$

I'm trying to prove this given that $Re \quad z \gt 1$ implies that there exists an $\varepsilon \gt 0$ such that $ Re \quad z \gt 1+ \varepsilon $.

I've said that if $$\frac {1}{(a+n)^z} = f_n(z)$$,how do I prove the $$|f_n(z)| \le \frac {1}{n^{1+\varepsilon}} $$

Can anyone help with this? Thanks

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The past 15 minutes you have asked 3 different questions on the same subject yet you have not made any effort at all in any of them! Before asking a new question, try to make your post readable and show us your effort. I am downvoting. –  Nameless Dec 2 '12 at 16:17
    
If I knew how to start the question I would be able to at least attempt but since I do not know how to start I am asking others to help me –  CJC Dec 2 '12 at 16:50
    
Can you comment on the convergence of $\sum_{n=1}^\infty \frac{1}{n^{1 + \epsilon}}$, given $\epsilon > 0$? –  anonymous Dec 2 '12 at 16:57
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John, no one asked you to know how to start. But why 3 questions at the same time? Post a question, wait for an answer, discuss in the comments and if you still don't understand how to solve the problem, then ask a similar question. –  Nameless Dec 2 '12 at 16:59
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@CJC It doesn't work that way. Instead you irritate people who regularly visit this site like me, because the front page is full of the same questions. –  Nameless Dec 2 '12 at 17:05
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1 Answer

up vote 2 down vote accepted

The following theorem is relevant: let $\Omega$ a domain in $\mathbb{C}$ and $\{f_n\}$ a sequence of holomorphic functions in $\Omega$. If $\{f_n\}$ converges uniformly to a function $f$ in every compact subset of $\Omega$, then $f$ is holomorphic in $\Omega$ (taken from Stein and Shakarchi).

Does that help?

Edited based on your edits:

You want to show that given $a > 0$ and $\Re(z) \ge 1 + \epsilon$ we have that $|\frac{1}{(n+a)^z}| \le \frac{C}{n^{1 + \epsilon}}$. To do this, first note that if $c$ is a real number larger than $1$, using our branch of the logarithm we have that $|c^z| = |e^{z\log(c)}| = |e^{\Re(z)\log(c)}| = c^{\Re(z)} \ge c^{1 + \epsilon}$. Thus by the sub-triangle inequality, we have that $|\frac{1}{(n + a)^z}| \le \frac{1}{(n - a)^{1 + \epsilon}}$, for large $n$.

Next, observe that since $\frac{n}{n-a} \to 1$ as $n \to \infty$, for large $n$ we have the estimate $n \le \frac{3}{2}(n-a)$. So we may take $C > 0$ to have $n \le C(n-a)$, for large $n$.

Thus: for sufficiently large $n$, we obtain the estimate $|\frac{1}{(n-a)^z}| \le C\frac{1}{n^{1 + \epsilon}}$.

From here, we deduce uniform convergence.

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Hi, I've seen this theorem before from my notes, not really sure how to apply it though. Thanks anyway –  CJC Dec 2 '12 at 17:39
    
First, consider a branch of the exponential defined for $\{z: \Re(z) > 1\}$, say the principal branch. Next, consider a compact subset $K$ of $\{\Re(z) > 1\}$. Show that for some $\epsilon > 0$, we have that $K \subset \{z: \Re(z) \ge 1 + \epsilon\}$. Deduce that the series converges uniformly to a function on $K$ and apply the theorem. –  anonymous Dec 2 '12 at 17:45
    
Could it be done by taking the modulus of the function and showing that it is less than or equal to 1/n^(1+ε)? Form previous questions I know that this converges? –  CJC Dec 2 '12 at 18:09
    
Almost. You'll want to obtain the estimate $|\frac{1}{(a+n)^z}| \le \frac{C}{|a + n|^{1 + \epsilon}}$, for some constant $C > 0$ (the issue being that for large n, your estimate is more or less correct, but you do have that factor of a in the denominator; the constant accounts for this). –  anonymous Dec 2 '12 at 18:20
    
Ok I understand what you mean. Thanks for your help –  CJC Dec 2 '12 at 18:26
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