`ln2/2+ln3/3+ln4/4+ln5/5+ln6/6+...` 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.
Integral test is applicable if f is positive, continuous and decreasing function on infinite interval `[k,oo)` where `k>=1` and `f(n)=a_n` . Then the series `sum_(n=k)^ooa_n` converges or diverges if and only if the improper integral `int_k^oof(x)dx` converges or diverges.
Given series is `ln2/2+ln3/3+ln4/4+ln5/5+ln6/6+........`
The series can be written as `sum_(n=1)^ooln(n+1)/(n+1)`
Refer the attached graph for f(x),
From the graph, we observe that the function is positive, continuous and decreasing for `x>=2`
We can also determine whether f(x) is decreasing by finding the derivative f'(x), such that `f'(x)<0` for `x>1`
Since the function satisfies the conditions for the integral test , we can apply the same.
Now let's determine the convergence or divergence of the integral `int_1^ooln(x+1)/(x+1)dx`
Let's first evaluate the indefinite integral,`intln(x+1)/(x+1)dx`
Apply integral substitution:`u=ln(x+1)`
`=u^2/2+C` , where C is a constant
Substitute back `u=ln(x+1)`
Since the integral `int_1^ooln(x+1)/(x+1)dx` diverges, so the given series also diverges as per the integral test.