use the integral test to determine whether the series is convergent or divergent
∑ n=9 1/n(ln n)^3
To use the integral test to determine the convergence or divergence of the series ∑ n=9 1/n(ln n)^3, we need to compare it with the integral of the function f(x) = 1/(x(ln x)^3).
The integral test states that if the integral of a function f(x) from a to infinity is convergent or divergent, then the corresponding series ∑ f(n) from a to infinity has the same convergence behavior.
Let's perform the integral to determine the convergence of the series.
∫[9,∞] 1/(x(ln x)^3) dx
First, let's substitute u = ln x, which gives us du = 1/x dx.
∫(1/u^3) du
Integrating, we get:
-(1/(2u^2)) + C
Substituting back u = ln x:
-(1/(2(ln x)^2)) + C
Now, let's evaluate the improper integral:
lim [b→∞] -(1/(2(ln x)^2)) + C - [9, b]
= lim [b→∞] -(1/(2(ln b)^2)) + C - (-(1/(2(ln 9)^2)) + C)
Since C - C cancels out, we are left with:
lim [b→∞] -(1/(2(ln b)^2)) - -(1/(2(ln 9)^2))
To determine if the integral converges or diverges, we need to evaluate this limit.
The integral converges if this limit is finite and diverges if the limit is infinite or undefined.
Let's evaluate the limit:
lim [b→∞] -(1/(2(ln b)^2)) - -(1/(2(ln 9)^2))
As b approaches infinity, the term -(1/(2(ln b)^2)) approaches zero, since ln b increases without bound, making the denominator very large.
Thus, the limit becomes:
-(1/(2(ln 9)^2))
Since this limit is a finite value (not infinity or undefined), the integral converges.
By the integral test, the series ∑ n=9 1/n(ln n)^3 also converges.
Therefore, the series is convergent.