determin whether the given series converges or diverges Sum(upper^infinity,lower n=2) (ln sqrt(n)) /sqrt n

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.

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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.

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