# Convergence of complex series

• Apr 5th 2011, 05:13 AM
ychana
Convergence of complex series
Hi guys, I think this is easy, but I keep getting stuck!

"Prove the series $\displaystyle \sum_{n=1}^{\infty} (1+n)^{-z}$ converges for all complex z with real part strictly greater than 1."

I assume we take the absolute value $\displaystyle \sum_{n=1}^{\infty} |(1+n)^{-z}|$

and then $\displaystyle \sum_{n=1}^{\infty} |(1+n)^{-Re(z)}||(1+n)^{-Im(z)}|$

$\displaystyle \leq \sum_{n=1}^{\infty} |(1+n)^{-Re(z)}|$

But where to now, I assume the comparison test, but what would one compare this too?

Many thanks,

• Apr 5th 2011, 07:07 AM
chisigma
If we consider the series...

$\displaystyle \displaystyle S=\sum_{n=2}^{\infty} \frac{1}{n^{\text{Re} (z)}}$ (1)

... setting $\displaystyle \text{Re} (z) = 1+\varepsilon\ ,\ \varepsilon>0$ we have that for n 'large enough' is...

$\displaystyle \displaystyle \frac{1}{n^{1 + \varepsilon}}} < \frac{1}{n\ \ln^{2} n}$ (2)

Now the series...

$\displaystyle \displaystyle \sum_{n=2}^{\infty} \frac{1}{n\ \ln^{2} n}$ (3)

... converges so that the series (1) for $\displaystyle \text{Re} (z)>1$ converges too...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$