`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.