For What positive values of x ; the below mentioned series is convergent and divergent? $$ \sum \frac{1}{x^{n}+x^{-n}}$$
My attempt
$$
\begin{aligned}
&\begin{aligned}
\therefore & u_{n+1}=\frac{x^{n+1}}{x^{2 n+2}+1} \\
\therefore & \frac{u_{n+1}}{u_{n}}=\frac{x^{n+1}}{x^{2 n+2}+1} \cdot \frac{x^{2 n}+1}{x^{n}} \\
& \frac{u_{n+1}}{u_{n}}=x \frac{1+\left(\frac{1}{x}\right)^{2 n}}{x^{2}+\left(\frac{1}{x}\right)^{2 n}}
\end{aligned}\\
&  i f x>1 \quad \therefore \quad \frac{1}{x}<1 \quad \therefore \text { when } n \rightarrow \infty,\left(\frac{1}{x}\right)^{2 n} \rightarrow 0
\end{aligned}\\
$$
What to do next?

In your attempt you have shown that $\lim \sup \frac {u_{n+1}} {u_n} \leq x \frac {1+0} {x^{2}+0}=\frac  1 x <1$. Hence Ratio Test tells you that the series is convergent for $x>1$. But there is a bettter approach as shown below:
For $x>1$ it is dominated by $\sum \frac 1 {x^{n}}$ which is convergent.
For $0<x<1$ it is dominated by $\sum x^{n}$ which is convergent.
For $x=1$ it is divergent.
