The Integral test is applicable if f is positive and a 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/(n(ln(n))^3) , then a_n =1/(n(ln(n))^3) .
Then applying a_n=f(x) , we consider:f(x) =1/(x(ln(x))^3) .
The graph of f(x) is:
As shown on the graph above, the function f(x) is positive and decreasing on the finite 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/(x(ln(x))^3)= lim_(t-gtoo)int_2^t 1/(x(ln(x))^3)dx
To determine the indefinite integral of int_2^t 1/(x(ln(x))^3)dx , we may apply u-substitution by letting:
u = ln(x) and du = 1/x dx .
The integral becomes:
int 1/(x(ln(x))^3)dx=int 1/(ln(x))^3 *1/x dx
=int 1/u^3 du
Apply Law of exponent: 1/x^m = x^(-m) .
int 1/u^3 du=int u^(-3) du
Apply Power rule for integration: int x^n dx = x^(n+1)/(n+1) .
int u^(-3) du =u^(-3+1)/(-3+1)
=u^(-2)/(-2)
= - 1/(2u^2)
Plug-in u=ln(x) on - 1/(2u^2) , we get:
int_2^t 1/(x(ln(x))^3)dx=- 1/(2(ln(x))^2)|_2^t
Apply definite integral formula: F(x)|_a^b = F(b)-F(a) .
- 1/(2(ln(x))^2)|_2^t=- 1/(2(ln(t))^2)-(- 1/(2(ln(2))^2))
=- 1/(2(ln(t))^2)+ 1/(2(ln(2))^2)
Applying int_1^t 1/(x(ln(x))^3)dx=- 1/(2(ln(t))^2)+ 1/(2(ln(2))^2) , we get:
lim_(t-gtoo)int_1^t 1/(x(ln(x))^3)dx=lim_(t-gtoo)[- 1/(2(ln(t))^2)+ 1/(2(ln(2))^2)]
= 0+1/(2(ln(2))^2)
=1/(2(ln(2))^2)
Note: lim_(t-gtoo)1/(2(ln(2))^2)=1/(2(ln(2))^2) and
lim_(t-gtoo)- 1/(2(ln(t))^2)= [lim_(t-gtoo) 1]/[lim_(t-gtoo)2(ln(t))^2]
=-1/oo
=-0 or 0
Thelim_(t-gtoo)int_2^t 1/(x(ln(x))^3)dx= 1/(2(ln(2))^2) implies that the integral converges.
Conclusion: The integral int_2^oo 1/(x(ln(x))^3) is convergent therefore the series sum_(n=2)^oo 1/(n(ln(n))^3) must also be convergent.
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