Question

How do you test the series `Sigma lnn/(nsqrtn)` from n is `[1,oo)` for convergence?

Answer 1

The series:

`sum_(n=1)^oo lnn/(nsqrtn)`

is convergent.

Explanation:

You can use the integral test, choosing as test function:

`f(x) = lnx/(xsqrtx)`

We can verify that:

(i) `f(x) > 0` for `x in (1,+oo)`

(ii) `lim_(x->oo) lnx/(xsqrtx) = 0`

The limit is in the indeterminate form `oo/oo` so we can use l'Hospital's rule to solve it:

`lim_(x->oo) lnx/(xsqrtx) = lim_(x->oo) (d/(dx) lnx )/(d/(dx) x^(3/2)) = lim_(x->oo) (1/x) 1/(3/2x^(1/2)) = lim_(x->oo) 2/(3x^(3/2)) = 0`

(iii) `f(x)` is monotone decreasing for `x in (1+oo)`

We can calculate the first derivative:

`d/(dx) lnx/(xsqrtx) = d/(dx) lnx *x^(-3/2) = (1/x)x^(-3/2) -3/2x^(-5/2)lnx = x^(-5/2) (1-3/2lnx) = (2-3lnx)/(2x^(5/2))`

and we can see that `x > 1 => f'(x) < 0`

(iv) `f(n) = lnn/(nsqrt(n))`

Thus all the hypotheses of the integral test theorem are satisfied and the convergence of the series is equivalent to the convergence of the improper integral:

`int_1^oo lnx/(xsqrtx)dx`

We solve the indefinite integral by parts:

`int lnx * x^(-3/2) dx = int lnx d(-2x^(-1/2)) =-2lnx/sqrt(x) +2 int x^(-1/2)/xdx = -2lnx/sqrt(x) +2int x^(-3/2)dx = -2lnx/sqrt(x) -4 x^(-1/2) +C = - (2(lnx+2))/sqrt(x)+C`

and we have:

`int_1^oo lnx/(xsqrtx)dx = [- (2(lnx+2))/sqrt(x)]_1^oo`

`int_1^oo lnx/(xsqrtx)dx = 4 - lim_(x->oo) lnx/(xsqrtx)`

We have already calculated at point (ii) above that such limit is zero, then:

`int_1^oo lnx/(xsqrtx)dx = 4`

proving that the series is convergent.