`ln2/sqrt(2)+ln3/sqrt(3)+ln4/sqrt(4)+ln5/sqrt(5)+ln6/sqrt(6)+...` 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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For the series: `ln(2)/sqrt(2) + ln(3)/sqrt(3)+ ln(4)/sqrt(4)+ ln(5)/sqrt(5)+ ln(6)/sqrt(6) +...`, it follows the formula `sum_(n=2)^oo ln(n)/sqrt(n)` where `a_n = ln(n)/sqrt(n)` . To confirm if the Integral test will be applicable, we let `f(x) = ln(x)/sqrt(x)` .

Graph of the function `f(x)` :

Maximize view: 

As shown on the graphs,...

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For the series: `ln(2)/sqrt(2) + ln(3)/sqrt(3)+ ln(4)/sqrt(4)+ ln(5)/sqrt(5)+ ln(6)/sqrt(6) +...`, it follows the formula `sum_(n=2)^oo ln(n)/sqrt(n)` where `a_n = ln(n)/sqrt(n)` . To confirm if the Integral test will be applicable, we let `f(x) = ln(x)/sqrt(x)` .

Graph of the function `f(x)` :

 

Maximize view: 

As shown on the graphs, `f` is positive and continuous on the finite interval `[1,oo)` . To verify if the function will eventually decreases on the given interval, we may consider derivative of the function.

Apply Quotient rule for derivative: `d/dx(u/v) = (u'* v- v'*u)/v^2` .

Let `u = ln(x)` then `u' = 1/x`

      `v = sqrt(x)` or `x^(1/2)` then `v' = 1/(2sqrt(x))`

Applying the Quotient rule, we get:

`f'(x) = (1/x*sqrt(x)-1/(2sqrt(x))*ln(x))/(sqrt(x))^2`

           `= (1/sqrt(x) - ln(x)/(2sqrt(x)))/x`

           `= ((2-ln(x))/sqrt(x))/x`

          ` =((2-ln(x))/sqrt(x))* 1/x`

          `=(2-ln(x))/(xsqrt(x)) `

 or `(2-ln(x))/x^(3/2)`

Note that `2-ln(x) lt0` for higher values of x which means ` f'(x) lt0`.

Aside from this, we may verify by solving critical values of x .

Apply First derivative test: f'(c) =0 such that x =c as critical values.

`(2-ln(x))/x^(3/2)=0`

`2-ln(x)=0`

`ln(x) =2`

`x = e^2`

`x~~7.389`

Using `f'(7) ~~0.0015` , it satisfy `f'(x) gt0` therefore the function is increasing on the left side of `x=e^2` .

Using `f'(8) ~~-0.0018` , it satisfy `f'(x) lt0 ` therefore the function is decreasing on the right side of `x=e^2` .

Then, we may conclude that the function  `f(x)` is decreasing for an interval `[8,oo)` .

This confirms that the function is ultimately positive, continuous, and decreasing for an interval `[8,oo)`  . Therefore, we may apply the Integral test. 

Note: Integral test is applicable if f is positive, continuous , and decreasing function on interval `[k, oo)` 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.

To determine the convergence or divergence of the given series, we may apply improper integral as:

`int_8^oo ln(x)/sqrt(x)dx = lim_(t-gtoo)int_8^tln(x)/sqrt(x)dx`

                                  or `lim_(t-gtoo)int_8^tln(x)/x^(1/2)dx`

 To determine the indefinite integral of `int_8^tln(x)/x^(1/2)dx` , we may apply integration by parts: `int u dv = uv - int v du`

`u = ln(x)` then `du = 1/x dx` . 

`dv = 1/x^(1/2) dx` then `v= int 1/x^(1/2)dx = 2sqrt(x)`

Note: To determine v, apply Power rule for integration `int x^n dx = x^(n+1)/(n+1).`

`int 1/x^(1/2)dx =int x^(-1/2)dx`

                ` =x^(-1/2+1)/(-1/2+1)`

                `=x^(1/2)/(1/2)`

                `=x^(1/2)*2/1`

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

The integral becomes: 

`int_8^t ln(x)/sqrt(x) dx=ln(x) * 2 sqrt(x) - int 2sqrt(x) *1/x dx`

                    `=2sqrt(x)ln(x) - int 2x^(1/2) *x^(-1) dx`

                    `=2sqrt(x)ln(x) - int 2x^(-1/2) dx`

                   `=2sqrt(x)ln(x) - 2int x^(-1/2) dx`

                   `= [ 2sqrt(x)ln(x)- 2(2sqrt(x))]|_8^t`

                    `= [2sqrt(x)ln(x) - 4sqrt(x)]|_8^t`

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

`[2sqrt(x)ln(x) - 4sqrt(x)]|_8^t =[2sqrt(t)ln(t) - 4sqrt(t)] - [2sqrt(8)ln(8) - 4sqrt(8)]`

                                        ` =2sqrt(t)ln(t) - 4sqrt(t) - 2sqrt(8)ln(8) + 4sqrt(8)`

                                        ` =2sqrt(t)ln(t) - 4sqrt(t) - 4sqrt(2)ln(8) + 8sqrt(2)`

Note: `sqrt(8) = 2sqrt(2)`

Applying `int_8^t ln(x)/sqrt(x) dx=2sqrt(t)ln(t) - 4sqrt(t) - 4sqrt(2)ln(8) + 8sqrt(2)` , we get:

`lim_(t-gtoo)int_2^tln(x)/sqrt(x)dx =lim_(t-gtoo) [2sqrt(t)ln(t) - 4sqrt(t) - 4sqrt(2)ln(8) + 8sqrt(2)]`

         `=lim_(t-gtoo) 2sqrt(t)ln(t) - lim_(t-gtoo)4sqrt(t) - lim_(t-gtoo)4sqrt(2)ln(8) + lim_(t-gtoo) 8sqrt(2)`

         ` = oo-oo -4sqrt(2)ln(8) +8sqrt(2)`

        `=oo`

The `lim_(t-gtoo)int_8^tln(x)/sqrt(x)dx=oo`  implies that the integral diverges.

Conclusion:

The integral `int_8^ooln(x)/sqrt(x)dx` is divergent therefore the series`sum_(n=2)^ooln(n)/sqrt(n)` must also be divergent

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