`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))`...

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.