Show that $\sum\limits_{n=1}^\infty \frac{1}{n^z}$ converges absolutely

Question

I want to show that $\large\sum\limits_{n=1}^\infty \frac{1}{n^z}$ converges absolutely for $\Re(z) > 1$, so

I want to show that $\large\sum\limits_{n=1}^\infty |\frac{1}{n^z}|$ converges, or

$\large\sum\limits_{n=1}^\infty |\frac{1}{\exp(z\log n)}|$

or

$\sum\limits_{n=1}^\infty |\exp(-z\log n)|$

any ideas?

Answer 1

First, note that $$\sum_{n = 1}^\infty \frac{1}{n^z}$$ is the zeta function $\zeta(z)$. Now, the modulus of a complex number $z = x + iy$ is defined by $|z| = \sqrt{x^2 + y^2}$. So, in particular, $|e^{iy}| = |\cos y + i\sin y| = \sqrt{\cos^2 y + \sin^2 y} = 1$.

Using Riemann's notation $z = \sigma + it$ for a complex number, we see that: $$|\zeta(z)| \leq \sum_{n = 1}^\infty \left|\frac{1}{n^z}\right| = \sum_{n = 1}^\infty \frac{1}{|n^{\sigma + it}|} = \sum_{n = 1}^\infty \frac{1}{|n^\sigma||n^{it}|} = \sum_{n = 1}^\infty \frac{1}{|n^\sigma||e^{i t \log n}|} = \sum_{n = 1}^\infty \frac{1}{|n^\sigma|}.$$ Observe that we set $y = t \log n$ and used the fact that $|e^{iy}| = 1$.

Next, we see that for $\sigma \ne 0$, $$\int_1^\infty x^{-\sigma} dx = \lim_{M \to \infty} \left[\frac{1}{1 - \sigma} x^{1 - \sigma}\right]_1^M.$$ However, $\lim_{x \to \infty} x^{-\sigma} = 0$ and $x^{-\sigma}$ is decreasing, which means $\zeta(z)$ converges for $\sigma = \Re[z] > 1$ by the Cauchy Integral Test.

Answer 2

$\frac{1}{|n^z|} = \frac{1}{n^x}$ where $z = x +iy$, now just observe that the series $\sum_{n = 1}^{\infty} \frac{1}{n^x}$ converges only if $x > 1$ by the integral test.