You should perform ratio test such that:

`lim_(n->oo) |((ln sqrt(n+1)) /sqrt (n+1))*((sqrt n)/(ln(sqrt n)))|` = `lim_(n->oo)sqrt(n/(n+1))*lim ((ln(sqrt(n+1)))/ (ln (sqrt n)))`

`lim_(n->oo) |((ln sqrt(n+1)) /sqrt (n+1))*((sqrt n)/(ln(sqrt n)))| = 1*lim ((ln(sqrt(n+1)))/ (ln (sqrt n)))` `= 1*oo/oo = oo/oo`

You need to use l'Hospital's theorem such that:

`lim_(n->oo) |((ln sqrt(n+1)) /sqrt (n+1))*((sqrt...

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You should perform ratio test such that:

`lim_(n->oo) |((ln sqrt(n+1)) /sqrt (n+1))*((sqrt n)/(ln(sqrt n)))|` = `lim_(n->oo)sqrt(n/(n+1))*lim ((ln(sqrt(n+1)))/ (ln (sqrt n)))`

`lim_(n->oo) |((ln sqrt(n+1)) /sqrt (n+1))*((sqrt n)/(ln(sqrt n)))| = 1*lim ((ln(sqrt(n+1)))/ (ln (sqrt n)))` `= 1*oo/oo = oo/oo`

You need to use l'Hospital's theorem such that:

`lim_(n->oo) |((ln sqrt(n+1)) /sqrt (n+1))*((sqrt n)/(ln(sqrt n)))| = lim_(n->oo) ((ln(sqrt(n+1)))')/ ((ln (sqrt n))')`

`lim_(n->oo) ((ln(sqrt(n+1)))')/ ((ln (sqrt n))') = lim_(n->oo) (1/(sqrt(n+1))*1/(2sqrt(n+1)))/(1/sqrt n*1/(2sqrt n))`

`lim_(n->oo) ((ln(sqrt(n+1)))')/ ((ln (sqrt n))') = lim_(n->oo) (2n)/(2(n+1))`

`lim_(n->oo) ((ln(sqrt(n+1)))')/ ((ln (sqrt n))') = lim_(n->oo) n/(n+1)`

`lim_(n->oo) ((ln(sqrt(n+1)))')/ ((ln (sqrt n))') = 1`

`lim_(n->oo) |((ln sqrt(n+1)) /sqrt (n+1))*((sqrt n)/(ln(sqrt n)))| = 1`

**Hence, performing the ratio test yields 1, meaning that the given series may be convergent, absolutely convergent or divergent.**

`sum_(n=2)^oo ln sqrtn/sqrtn`

To determine if the series is convergent or divergent, we may use Integral Test.

Let's re-write the series in integral form. We have:

`int_2^oo ln sqrtx/sqrtx dx = int_2^oo lnx^(1/2)/x^(1/2)dx = int_2^oo 1/2 lnx/x^(1/2) = 1/2 int_2^oo lnx/x^(1/2)`

Since the upper limit is infinity, replace it with a variable *t* to be able to evaluate the integral.

`1/2int_2^t ln x / x^(1/2) dx`

Then, use integration by parts. The formula is `int udv = uv - intv du` . So let,

`u = lnx` and `dv = 1/x^1/2 dx=x^(-1/2)dx`

`du=(dx)/x ` `v=intx^(-1/2)dx = 2x^1/2`

Substitute these to the formula.

`1/2 [lnx*(2x^(1/2)) - int(2x^(1/2))/x dx] = x^(1/2)lnx - int x^(-1/2)dx = x^(1/2)lnx -2x^(1/2)`

So we have:

`int_2^t lnx/x^(1/2 ) dx = x^(1/2 )lnx -2x^(1/2)` `|_2^t`

Note that *t* represents infinity. To evaluate further, take the limit of the integral as *t* approaches infinity.

`int_2^oo (lnsqrtx)/sqrtx = lim_(t->oo) int_2^t lnx/x^(1/2)dx = lim_(t->oo) (x^(1/2)lnx - 2x^(1/2)) |_2^t`

`=lim_(t->oo) [(t^(1/2)lnt-2t^(1/2) - (2^(1/2)ln2-2*2^(1/2))]`

`=lim_(t->oo) [t^(1/2)lnt - 2t^(1/2) + 1.85] = lim_(x->oo) t^(1/2)(lnt - 2) + lim _(x->oo) 1.85 `

Let's evaluate the limits separately.

For the first one, apply the rule `lim_(x->oo) ln x = oo` . And if we subtract a number from a large number, this results to a large value. Also, use the rule `lim_(x->oo) x^c = oo` . So we have,

`lim_(x->oo) t^(1/2)(lnt-2)= oo(oo-2)=oo*oo=oo`

Note that when we multiply two large numbers, the product would be a large value. Hence, the limit is infinite.

For the second limit, apply the rule `lim_(x-gtoo) c = c` where *c* is a constant.

`lim_(t->oo) 1.85 = 1.85`

So we have:

`lim_(x->oo) t^(1/2)(ln-2)+ lim_(x->oo) 1.85 = oo+1.85 = oo`

The resulting limit is infinite since adding a large number with a small one results to large sum.

**Hence, `int_2^oo (ln sqrtx) / sqrtx = oo` , which indicates that the integral is divergent. **

**Therefore, the series `sum_(n=2)^oo (ln sqrtx)/sqrtx` is divergent.**