Question

How do you use the integral test to determine if \[\dfrac{{\ln 2}}{2} + \dfrac{{\ln 3}}{3} + \dfrac{{\ln 4}}{4} + ...\] is divergent or convergent?

Answer 1

Hint: Here in this question they have given the series and we have determined the given series is a convergent series or divergent series by using the integral test. Here before applying the integral test we have to check \[{a_n} > 0\] , where \[{a_n}\]represents the last term or the general expression of the given series. The \[{a_n}\] can be written as function of ‘n’ and then we apply the integral \[\int\limits_N^\infty {f(x)dx} \] for the given function and hence we say convergent or divergent series.

Complete step by step solution:
In mathematics, the integral test for convergence is a method used to test an infinite series of non-negative terms for convergence.
Now consider the given question, the series is given as
\[\dfrac{{\ln 2}}{2} + \dfrac{{\ln 3}}{3} + \dfrac{{\ln 4}}{4} + ...\]
This can be written in generally as
\[ \Rightarrow \dfrac{{\ln 2}}{2} + \dfrac{{\ln 3}}{3} + \dfrac{{\ln 4}}{4} + ... = \sum\limits_{n = 2}^\infty {\dfrac{{\ln \,n}}{n}} \]
Before applying the integral test to \[\sum\limits_{n = N}^\infty {f(n)} \] we must have \[{a_n} > 0\] for the given interval and \[f(n)\] must be decreasing on the same interval . Both of these true since \[\ln (n) > 0\] and \[n > 0\]on \[n \in [2,\infty )\], so \[\dfrac{{\ln (n)}}{n} > 0\] on the same interval
Now we will apply the integral test to the given function
Here \[f(x) = \dfrac{{\ln \,x}}{x}\]
The integral test is \[\int\limits_N^\infty {f(x)dx} \]
If the integral converges to a real, finite value, then the series converges. If the integral diverges, then the series does too.
Here the value of \[N = 2\] and \[f(x) = \dfrac{{\ln \,x}}{x}\] so we have
\[ \Rightarrow \int\limits_2^\infty {\dfrac{{\ln \,x}}{x}dx} \]
Now we will take the limit for the above function
\[ \Rightarrow \int\limits_N^\infty {\dfrac{{\ln \,x}}{x}dx} = \mathop {\lim }\limits_{b \to \infty } \int\limits_N^b {\dfrac{{\ln \,x}}{x}dx} \]
Now substitute the \[u = \ln \,x\], on differentiating this function we have \[du = \dfrac{1}{x}dx\].The integral values will get change, therefore \[u = \ln \,(2)\] and \[u = \ln \,(b)\] we have Now the given integral is written as
\[ \Rightarrow \int\limits_N^\infty {f(x)dx} = \mathop {\lim }\limits_{b \to \infty } \int\limits_{\ln \,(2)}^{\ln (b)} {u\,du} \]
On integrating we have
\[ \Rightarrow \mathop {\lim }\limits_{b \to \infty } \int\limits_{\ln \,(2)}^{\ln (b)} {u\,du} = \mathop {\lim }\limits_{b \to \infty } \left[ {\dfrac{{{u^2}}}{2}} \right]_{\ln \,2}^{\ln \,b}\]
Applying the limit values we have
\[ \Rightarrow \mathop {\lim }\limits_{b \to \infty } \int\limits_{\ln \,(2)}^{\ln (b)} {u\,du} = \mathop {\lim }\limits_{b \to \infty } \left( {\dfrac{1}{2}{{\left( {\ln \,b} \right)}^2} - \dfrac{1}{2}{{\left( {\ln \,2} \right)}^2}} \right)\]
Now taking the limit we have
\[ \Rightarrow \int\limits_N^\infty {f(x)dx} = \infty \]
So when we apply the integral test we will get infinity, so the series is a divergent series.

Note: The integral test for the given function is defined as \[\int\limits_N^\infty {f(x)dx} \], this is a integration concept so we have to know the integration formulas. Sometimes the integration is applied directly to the function but sometimes we use the substitution method and hence on simplifying we get a solution for the given question.