Student Question

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.

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 seriessum_(n=2)^ooln(n)/sqrt(n) must also be divergent