ich zeige es mal mit dem cosh.
$$ a_n:=\cosh(n)=\frac{e^n+e^{-n}}{2} $$
$$ \sum_{n=0}^\infty a_n=\sum_{n=0}^\infty\frac{e^n+e^{-n}}{2} =\sum_{n=0}^\infty\frac{e^{2n}+1}{2e^n}$$
Mit dem Quotientenkriterium folgt dann:
$$ \limsup_{n \to \infty} \Bigg| \frac{a_{n+1}}{a_n} \Bigg|=\limsup_{n \to \infty} \Bigg| \frac{\frac{e^{2n+2}+1}{2e^{n+1}}}{\frac{e^{2n}+1}{2e^n}} \Bigg|\\=\limsup_{n \to \infty} \Bigg| \frac{e^{2n+2}+1}{2e^{n+1}}\cdot \frac{2e^n}{e^{2n}+1} \Bigg|\\=\limsup_{n \to \infty} \Bigg| \frac{e^{2n+2}+1}{e}\cdot \frac{1}{e^{2n}+1} \Bigg|\\= \limsup_{n \to \infty} \Bigg| \frac{e^{2n+2}+1}{e^{2n+1}+e} \Bigg|\\=\limsup_{n \to \infty} \Bigg| \frac{\frac{e^{2n+2}}{e^{2n+2}}+\frac{1}{e^{2n+2}}}{\frac{e^{2n+1}}{e^{2n+2}}+\frac{e}{e^{2n+2}}} \Bigg|\\=\limsup_{n \to \infty} \Bigg| \frac{1+\frac{1}{e^{2n+2}}}{\frac{1}{e}+\frac{e}{e^{2n+2}}} \Bigg|=e>1$$
Also divergent.
