Question

Determine if $ \sum {\dfrac{{\ln n}}{{{n^3}}}} $ from $ \left[ {2,\infty } \right) $ is convergent or divergent by using the integral test?

Answer 1

Hint: In this problem, we have to do the integral test. In the integral test, the series $ \sum\limits_{n = 1}^\infty {{a_n}} $ is convergent if the improper integral $ \int\limits_1^\infty {f\left( x \right)} dx $ is convergent, where f is a continuous, positive and decreasing function on $ \left[ {1,\infty } \right) $ and let us assume $ {a_n} = f\left( n \right) $ . This test is used when the function can easily be integrated.

Complete step-by-step answer:
In this problem, we have to check that the series given is convergent or divergent and we can check it by using the integral test and if our series starts with $ n = 2 $ then also we can use the integral test.
Let $ {a_n} = \dfrac{{\ln n}}{{{n^3}}} $ , and for $ n = 1 $ , $ {a_n} = 0 $ . For this, the series is, $ \sum\limits_{n = 1}^\infty {\dfrac{{\ln n}}{{{n^3}}}} $ and we know from the above integral theorem that the series is convergent if the integral is convergent. Now, we can check whether the integral is convergent or not. So, the integral is,
 $ \int\limits_1^\infty {\dfrac{{\ln x}}{{{x^3}}}dx} $
On further solving, we get,
 $
   \Rightarrow \left[ { - \dfrac{1}{2}\dfrac{{\ln x}}{{{x^2}}}} \right]_1^\infty + \dfrac{1}{2}\int\limits_1^\infty {\dfrac{1}{x} \times \dfrac{1}{{{x^2}}}dx} \\
   \Rightarrow 0 + \dfrac{1}{2}\int\limits_1^\infty {\dfrac{{dx}}{{{x^3}}}} \\
   \Rightarrow \left[ { - \dfrac{1}{{4{x^2}}}} \right]_1^{ + \infty } = \dfrac{1}{4} \;
  $
Here the term $ \left[ { - \dfrac{1}{2}\dfrac{{\ln x}}{{{x^2}}}} \right]_1^\infty $ is solved as,
 $
  \left[ { - \dfrac{1}{2}\dfrac{{\ln x}}{{{x^2}}}} \right]_1^\infty = - \dfrac{1}{2}\int\limits_0^1 {\dfrac{{\ln x}}{{{x^2}}}dx + \left( {\dfrac{{ - 1}}{2}} \right)\int\limits_1^\infty {\dfrac{{\ln x}}{{{x^2}}}} } dx \\
   \Rightarrow - \dfrac{1}{2}\int\limits_0^1 {\dfrac{{\ln x}}{{{x^2}}}dx} + \left( {\dfrac{{ - 1}}{2}} \right)\int\limits_0^1 {\dfrac{{\ln \dfrac{1}{x}}}{{\dfrac{1}{{{x^2}}}}} \times \dfrac{1}{{{x^2}}}dx} \\
   \Rightarrow - \dfrac{1}{2}\int\limits_0^1 {\dfrac{{\ln x}}{{{x^2}}}dx} - \left( {\dfrac{{ - 1}}{2}} \right)\int\limits_0^1 {\dfrac{{\ln x}}{{{x^2}}}} \\
   = 0 \;
  $
The value of $ \int\limits_1^\infty {\dfrac{{\ln x}}{{{x^3}}}dx} = \dfrac{1}{4} $ .Hence, the integral is convergent, so, the series is convergent as well and this can be said by the theorem of absolute convergence test.



Note: By the help of p-series, we can also check that the series given is convergent or divergent. The p-series is represented like $ \sum\limits_{n = 1}^\infty {\dfrac{1}{{{n^p}}}} $ and if in series, the p is greater than $ 1 $ , then the series is convergent and if the series is smaller or equal to $ 1 $ , then the series is divergent. For this problem, we can change the series in the form of p-series and after checking the value of p, we can easily say whether the series is convergent or divergent.