Question

Show that product of (1+e^nz) from n=1,2,3,...
Diverges in the half-plane Re(z) >=0

Answer 1

Rewrite the expression in other form, using the Euler's formula.

$1+e^{nz}=1+e^{nx}*e^{iny}=$

$=1+e^{nx}*(cos(ny)+isin(ny))$

The real part of this expression $1+e^{nx}*cos(ny)\geqslant0$ by the condition of the question.

Consider the imaginary part of this expression $e^{nx}*sin(ny)$. The exponential function is monotonously increased, when $n->\infty$. So $e^{nx}->\infin$, then $e^{nx}*sin(ny)->\infin$.

Now, consider the real part of the expression $1+e^{nx}*cos(ny)$. It also has the exponential function that when $n->\infty$, $e^{nx}->\infty$. 1 is very small in comparison with exponential function. So $1+e^{nx}*cos(ny)->\infty$ also diverges.

Q.E.D.