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.
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