Is the series $\sum^\infty_{n=1}\left( a^{\frac{1}{n}}+a^{-\frac{1}{n}}-2 \right)$ converging? I've been trying to solve this for limit comparison test with $a_n=a^\frac{1}{n}+a^{-\frac{1}{n}}-2 , b_n= \frac{1}{n}$, 
but
$\frac{a_n}{b_n}\rightarrow\ln{a}(a^{\frac{1}{x}}-a^{-\frac{1}{x}})\rightarrow 0$.
Any help appreciated.
 A: Assume $a>0$. Let $x_n=\frac 1{2n}.$ Then $$\sum_{n=1}^\infty \left(a^{1/n}+a^{-1/n}-2\right)=\sum_{n=1}^\infty(a^{x_n}-a^{-x_n})^2$$ $$=\sum_{n=1}^\infty a^{2x_n}(1-a^{-2x_n})^2=\sum_{n=1}^\infty a^{2x_n}(1-e^{-2x_n\ln a})^2,$$ which by mean value theorem equals 
$$\sum_{n=1}^\infty a^{2x_n}(e^{-2y_n\ln a}\cdot 2x_n\ln a)^2,$$ where $0<y_n<x_n$. Clearly the series is bounded by $$C\cdot \sum_{n=1}^\infty (2x_n)^2=C\sum_{n=1}^\infty\frac 1{n^2}$$ where $C$ is a bounded constant. It follows that the original series is convergent.
Note: $1-e^x=e^0-e^x=e^{\xi}(0-x),$ where $\xi$ is between $0$ and $x$.
A: Let be $a>0$. Since
$$
a^x  + a^{ - x}  - 2 = 2\left[ {\cosh (x\ln a) - 1} \right]
$$
and since
$$
\mathop {\lim }\limits_{t \to 0} \left[ {\frac{{\cosh (t) - 1}}
{{t^2 }}} \right] = \frac{(\ln a)^2}
{2}
$$
by setting $
t = \frac{{\ln a}}
{n}
$
you get
$$
\begin{gathered}
  \mathop {\lim }\limits_{n \to  + \infty } n^{ - 2} \left( {a^{\frac{{\text{1}}}
{n}}  + a^{ - \frac{{\text{1}}}
{n}}  - 2} \right) =  \hfill \\
   = \mathop {\lim }\limits_{n \to  + \infty } 2n^{ - 2} \left[ {\cosh \left( {\frac{{\ln a}}
{n}} \right) - 1} \right] = \frac{(\ln a)^2}
{2} \hfill \\
   \hfill \\ 
\end{gathered} 
$$
and your series is convergent by asymptotic comparison test.
