f(x + iy)=∑(k = 1 to ∞)1/k^(x + iy)

Explanation:

Step1: Identify the series

This is a complex - valued Dirichlet series. The general form of a Dirichlet series is $\sum_{k = 1}^{\infty}\frac{a_k}{k^s}$, where in this case $a_k = 1$ and $s=x+iy$.

Step2: Analyze convergence

For $x>1$, the series $\sum_{k = 1}^{\infty}\frac{1}{k^{x+iy}}=\sum_{k = 1}^{\infty}\frac{1}{k^x}\cdot\frac{1}{k^{iy}}$ converges absolutely since $\sum_{k = 1}^{\infty}\left|\frac{1}{k^{x+iy}}\right|=\sum_{k = 1}^{\infty}\frac{1}{k^x}$ is a p - series with $p=x>1$. When $x = 1$, the series $\sum_{k = 1}^{\infty}\frac{1}{k^{1+iy}}=\sum_{k = 1}^{\infty}\frac{\cos(y\ln k)-i\sin(y\ln k)}{k}$ converges conditionally for $y\neq0$ by the Dirichlet's test for series. For $x<1$, the series diverges.

Answer:

The series $\sum_{k = 1}^{\infty}\frac{1}{k^{x+iy}}$ converges absolutely for $x > 1$, converges conditionally for $x = 1,y\neq0$ and diverges for $x<1$.