sum_(n=2)^oo 1/(nsqrt(n^2-1)) Determine the convergence or divergence of the series.

To evaluate the given series sum_(n=2)^oo 1/(nsqrt(n^2-1)) , we may apply Integral test to determine the convergence or divergence of the series.
Recall Integral test is applicable if f is a positive and decreasing function on interval [k,oo) where kgt=1 and a_n=f(x) .
If int_k^oo f(x) dx is convergent then the series sum_(n=k)^oo a_n is also convergent.
If int_k^oo f(x) dx is divergent then the series sum_(n=k)^oo a_n is also divergent.
For the  series sum_(n=2)^oo 1/(nsqrt(n^2-1)) , we have a_n=1/(nsqrt(n^2-1)) then we may let the function:
f(x) =1/(xsqrt(x^2-1))
The graph of the function is: 
As shown on the graph, f(x) is positive and decreasing on the interval [2,oo) . This confirms we may apply the Integral test to determine the converge or divergence of a series as:
int_2^oo1/(xsqrt(x^2-1)) dx= lim_(t-gtoo)int_2^t1/(xsqrt(x^2-1))dx
To determine the indefinite integral of int_2^t1/(xsqrt(x^2-1))dx , we may apply the integral formula for rational function with root as:
int 1/(usqrt(u^2-a^2))du= 1/a *arcsec(u/a)+C .
By comparing " 1/(xsqrt(x^2-1)) " with "1/(usqrt(u^2-a^2)) ", we determine the corresponding values as: u=x and a=1.Applying the integral formula, we get:
int_2^t1/(xsqrt(x^2-1))dx =1/1 *arcsec(x/1)|_2^t          
                        =arcsec(x)|_2^t
Applying definite integral formula: F(x)|_a^b = F(b)-F(a) 
arcsec(x)|_2^t =arcsec(t) -arcsec(2)
Applying int_2^t1/(xsqrt(x^2-1))dx = arcsec(t) -arcsec(2) , we get:
lim_(t-gtoo)int_2^t1/(xsqrt(x^2-1))dx =lim_(t-gtoo)[arcsec(t) -arcsec(2)]
                                  =lim_(t-gtoo)arcsec(t) -lim_(t-gtoo)arcsec(2)
                                  = pi/2 -arcsec(2)
                                  =pi/6
The lim_(t->oo)int_2^t 1/(xsqrt(x^2-1))dx =pi/6 implies that the integral converges.
 Conclusion: The integral int_2^oo1/(xsqrt(x^2-1)) dx  is convergent therefore the series sum_(n=2)^oo 1/(nsqrt(n^2-1)) must also be convergent.