# sum_(n=2)^oo 1/(nsqrt(ln(n))) 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.

The Integral test is applicable if f is positive and decreasing function on infinite interval [k, oo) where kgt= 1 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.

For the given series sum_(n=2)^oo 1/(nsqrt(ln(n))) , the a_n =1/(nsqrt(ln(n))) then applying a_n=f(x) , we consider:

f(x) =1/(xsqrt(ln(x))) .

The graph of f(x) is:

As shown on the graph above, the function f(x) is positive and decreasing on the interval [2,oo) . This implies we may apply the Integral test to confirm the convergence or divergence of the given series.

We may determine the convergence or divergence of the improper integral as:

int_2^oo 1/(xsqrt(ln(x)))= lim_(t-gtoo)int_2^t 1/(xsqrt(ln(x)))dx

To determine the indefinite integral of int_2^t1/(xsqrt(ln(x)))dx , we may apply u-substitution by letting:

u = ln(x) and du = 1/x dx .

The integral becomes:

int 1/(xsqrt(ln(x)))dx=int 1/sqrt(ln(x)) *1/x dx

=int 1/sqrt(u) du

Apply the radical property: sqrt(x)= x^(1/2) and 1/x^m = x^(-m) .

int 1/sqrt(u) du=int 1/u^(1/2) du

=int u^(-1/2) du

Apply the Power rule for integration: int x^n dx = x^(n+1)/(n+1) .

int u^(-1/2) du =u^(-1/2+1)/(-1/2+1)

=u^(1/2)/(1/2)

 =u^(1/2)*(2/1)

 = 2u^(1/2)       or     2sqrt(u)

Plug-in  u=ln(x) on 2sqrt(u) , we get:

int_2^t1/(xsqrt(ln(x)))dx=2sqrt(ln(x))|_2^t

Apply the definite integral formula: F(x)|_a^b = F(b)-F(a) .

2sqrt(ln(x))|_2^t =2sqrt(ln(t))-2sqrt(ln(2))

Applying int_1^t1/(xsqrt(ln(x)))dx=2sqrt(ln(t))-2sqrt(ln(2)) , we get:

lim_(t-gtoo)int_1^t 1/(xsqrt(ln(x)))dx=lim_(t-gtoo)[2sqrt(ln(t))-2sqrt(ln(2))]

= oo -2sqrt(ln(2))

 =oo

Note: lim_(t-gtoo)2sqrt(ln(2))=2sqrt(ln(2)) and

lim_(t-gtoo)2sqrt(ln(t))= 2lim_(t-gtoo)sqrt(ln(t))

 =2sqrt(lim_(t-gtoo)ln(t))

 =2sqrt(oo)

 =oo

The lim_(t-gtoo)int_2^t 1/(xsqrt(ln(x)))dx= oo implies that the integral diverges.

Conclusion: The integral int_2^oo 1/(xsqrt(ln(x)))    diverges therefore the series sum_(n=2)^oo 1/(xsqrt(ln(x)))  must also diverges.

Approved by eNotes Editorial Team