Question

Using the integral test, how do you show whether \[\sum {\dfrac{1}{{n{{(\log n)}^p}}}} \] diverges or converges from n=3 to infinity?

Answer 1

Hint: In this question, we have to show whether the given series converges or diverges by using the integral test. According to integral test, if ${a_n} = f(n)$ , $f(x)$ is a non-negative non-increasing function, then the condition for $\sum\limits_n^\infty {{a_n}} $ to converge is that the integral $\int\limits_1^\infty {f(x)} dx$ is finite and the condition for it to diverge is that the integral is infinite. So, we can easily solve this question by using the above information.

Complete step-by-step answer:
We have to integral test to show whether \[\sum {\dfrac{1}{{n{{(\log n)}^p}}}} \] converges or diverges from n=3 to infinity.
Let p be a natural number, that is $p \geqslant 1$
For p=1,
$\int\limits_3^\infty {\dfrac{1}{{n{{(\log n)}^p}}}} dn = \int\limits_3^\infty {\dfrac{1}{{n\log n}}} dn$
Let
$
  \log n = u \\
   \Rightarrow \dfrac{{d\log n}}{{dn}} = \dfrac{{du}}{{dn}} \\
   \Rightarrow \dfrac{1}{n}dn = du \;
 $
So, we get –
$
  \int\limits_3^\infty {\dfrac{1}{{\log n}} \times \dfrac{1}{n}} dn = \int\limits_{n = 3}^{n = \infty } {\dfrac{1}{u}du} \\
   \Rightarrow \int\limits_3^\infty {\dfrac{1}{{n\log n}}dn} = [\log u]_{n = 3}^{n = \infty } \\
   \Rightarrow \int\limits_3^\infty {\dfrac{1}{{n\log n}}dn} = [\log (\log n)]_{n = 3}^{n = \infty } \\
   \Rightarrow \int\limits_3^\infty {\dfrac{1}{{n\log n}}dn} = \log (\log \infty ) - \log (\log 3) \;
 $
We know that $\log \infty = \infty $ so $\log (\log \infty ) = \infty $
$ \Rightarrow \int\limits_3^\infty {\dfrac{1}{{n\log n}}dn} = \infty $
The integral for p=1 is infinite and thus diverges.
At p>1
$\int\limits_3^\infty {\dfrac{1}{{n{{(\log n)}^p}}}dn} $
Let –
$
  \log n = u \\
   \Rightarrow du = \dfrac{1}{n}dn \;
 $
\[
   \Rightarrow \int\limits_3^\infty {\dfrac{1}{{n{{(\log n)}^p}}}dn} = \int\limits_{n = 3}^\infty {\dfrac{1}{{{u^p}}}du} \\
   \Rightarrow \int\limits_3^\infty {\dfrac{1}{{n{{(\log n)}^p}}}dn} = [\dfrac{{{u^{ - p + 1}}}}{{ - p + 1}}]_{n = 3}^\infty \\
   \Rightarrow \int\limits_3^\infty {\dfrac{1}{{n{{(\log n)}^p}}}dn} = [\dfrac{1}{{(1 - p){{(\log n)}^{p - 1}}}}]_{n = 3}^\infty \\
   \Rightarrow \int\limits_3^\infty {\dfrac{1}{{n{{(\log n)}^p}}}dn} = \dfrac{1}{{(1 - p){{(\log \infty )}^{p - 1}}}} - \dfrac{1}{{(1 - p){{(\log 3)}^{p - 1}}}} \;
 \]
We know that $\log \infty = 0$ so we get $\dfrac{1}{{(1 - p){{(\log \infty )}^{p - 1}}}} = 0$ .
$ \Rightarrow \int\limits_3^\infty {\dfrac{1}{{n{{(\log n)}^p}}}dn} = 0 - \dfrac{1}{{(1 - p){{(\log 3)}^{p - 1}}}}$
Clearly, $\dfrac{1}{{(1 - p){{(\log 3)}^{p - 1}}}}$ is a finite quantity, so the integral is finite for p>1 and thus converges.
Hence, \[\sum {\dfrac{1}{{n{{(\log n)}^p}}}} \] diverges for p=1 and converges for p>1.

Note: A series is defined as an expression in which infinitely many terms are added one after the other to a given starting quantity. When we get further and further in a sequence, the terms get closer and closer to a specific limit, that is, when adding the terms one after the other, if we get partial sums that become closer and closer to a given number, then the series converges and it is known as the convergence of the series, otherwise it is a divergent series.