Help with convergence sequence by cases I have to study the converge or diverge or this sequence:
\begin{eqnarray}
c_{n}=\left\{\left(\cos (\frac{a}{n})\right)^{\frac{1}{\ln(|\cos(\frac{b}{n})|)}}\right\} \hspace{0.3cm} \text{for a,b > 0}
\end{eqnarray}

So I noticed that:
\begin{eqnarray}
(\cos (\frac{a}{n}))^{\frac{1}{\ln(|\cos(\frac{b}{n})|)}} &=& e^{\frac{1}{\ln(|\cos(\frac{b}{n})|)}\ln(\cos(\frac{a}{n}))}\\
&=& e^{\frac{\ln(\cos(\frac{a}{n}))}{\ln|(\cos(\frac{b}{n})|}}\\
&=& e^{\log_{|\cos(\frac{b}{n})|}{\cos}(\frac{a}{n})}\\
&=& e^{\log_{\cos(\frac{b}{n})}{\cos(\frac{a}{n})}}
\end{eqnarray}
So, if $a=b$ :
\begin{eqnarray}
\lim_{n\rightarrow \infty} c_{n}= e
\end{eqnarray}
If $b>a$
\begin{eqnarray}
e>e^{\log_{\cos(\frac{b}{n})}{\cos(\frac{a}{n})}}>1
\end{eqnarray}
Can I find a specific value of this limit in this case? and If $a > b$ diverges.

HINT
Since $\cos \frac{b}{n}\to 1$ we can eliminate absolute value and we can use that
$$\frac{\ln\left(\cos \frac{a}{n}\right)}{\ln\left(\cos \frac{b}{n}\right)}=$$
$$= \frac{\ln(1+(\cos \frac{a}{n}-1))}{\cos \frac{a}{n}-1} \frac{\cos \frac{b}{n}-1}{\ln(1+(\cos \frac{b}{n}-1))}\frac{\cos \frac{a}{n}-1} {\left(\frac{a}{n}\right)^2}\frac{\left(\frac{b}{n}\right)^2}{\cos \frac{b}{n}-1}\frac{a^2}{b^2}$$
then apply standard limits.
