`sum_(n=2)^oo 1/(nsqrt(ln(n)))` Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.

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The Integral test is applicable if `f` is positive and decreasing function on infinite interval `[k, oo)` where `kgt= 1` and `a_n=f(x)` . Then the series `sum_(n=k)^oo a_n` converges if and only if the improper integral` int_k^oo f(x) dx ` converges. If the integral diverges then the series also diverges.

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The Integral test is applicable if `f` is positive and decreasing function on infinite interval `[k, oo)` where `kgt= 1` and `a_n=f(x)` . Then the series `sum_(n=k)^oo a_n` converges if and only if the improper integral` int_k^oo f(x) dx ` converges. If the integral diverges then the series also diverges.

For the given series `sum_(n=2)^oo 1/(nsqrt(ln(n)))` , the `a_n =1/(nsqrt(ln(n)))` then applying` a_n=f(x)` , we consider:

`f(x) =1/(xsqrt(ln(x)))` .  

The graph of f(x) is:

As shown on the graph above, the function `f(x)` is positive and decreasing on the interval `[2,oo)` . This implies we may apply the Integral test to confirm the convergence or divergence of the given series.

We may determine the convergence or divergence of the improper integral as:

`int_2^oo 1/(xsqrt(ln(x)))= lim_(t-gtoo)int_2^t 1/(xsqrt(ln(x)))dx`

To determine the indefinite integral of `int_2^t1/(xsqrt(ln(x)))dx` , we may apply u-substitution by letting:

`u = ln(x)` and `du = 1/x dx` . 

The integral becomes: 

`int 1/(xsqrt(ln(x)))dx=int 1/sqrt(ln(x)) *1/x dx`

                        `=int 1/sqrt(u) du`

Apply the radical property: `sqrt(x)= x^(1/2)` and `1/x^m = x^(-m)` .

`int 1/sqrt(u) du=int 1/u^(1/2) du`

                 `=int u^(-1/2) du`

Apply the Power rule for integration: `int x^n dx = x^(n+1)/(n+1)` .

`int u^(-1/2) du =u^(-1/2+1)/(-1/2+1)`

                   `=u^(1/2)/(1/2)`

                   ` =u^(1/2)*(2/1)`

                   ` = 2u^(1/2)`       or     `2sqrt(u)`

Plug-in  `u=ln(x)` on `2sqrt(u)` , we get:

`int_2^t1/(xsqrt(ln(x)))dx=2sqrt(ln(x))|_2^t`

Apply the definite integral formula: `F(x)|_a^b = F(b)-F(a)` .

`2sqrt(ln(x))|_2^t =2sqrt(ln(t))-2sqrt(ln(2))`

Applying `int_1^t1/(xsqrt(ln(x)))dx=2sqrt(ln(t))-2sqrt(ln(2))` , we get:

`lim_(t-gtoo)int_1^t 1/(xsqrt(ln(x)))dx=lim_(t-gtoo)[2sqrt(ln(t))-2sqrt(ln(2))]`

                                     `= oo -2sqrt(ln(2))`

                                     ` =oo`

Note: `lim_(t-gtoo)2sqrt(ln(2))=2sqrt(ln(2))` and

`lim_(t-gtoo)2sqrt(ln(t))= 2lim_(t-gtoo)sqrt(ln(t))`

                           ` =2sqrt(lim_(t-gtoo)ln(t))`

                            ` =2sqrt(oo)`

                            ` =oo`

The `lim_(t-gtoo)int_2^t 1/(xsqrt(ln(x)))dx= oo` implies that the integral diverges.

Conclusion: The integral `int_2^oo 1/(xsqrt(ln(x)))`    diverges therefore the series `sum_(n=2)^oo 1/(xsqrt(ln(x)))`  must also diverges. 

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