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Question: Using the integral test, how do you show whether \[\sum {\dfrac{1}{{n{{(\log n)}^p}}}} \] diverges o...

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

Explanation

Solution

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 an=f(n){a_n} = f(n) , f(x)f(x) is a non-negative non-increasing function, then the condition for ∑n∞an\sum\limits_n^\infty {{a_n}} to converge is that the integral ∫1∞f(x)dx\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 ∑1n(log⁡n)p\sum {\dfrac{1}{{n{{(\log n)}^p}}}} converges or diverges from n=3 to infinity.
Let p be a natural number, that is p⩾1p \geqslant 1
For p=1,
∫3∞1n(log⁡n)pdn=∫3∞1nlog⁡ndn\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 ⇒dlog⁡ndn=dudn ⇒1ndn=du   \log n = u \\\ \Rightarrow \dfrac{{d\log n}}{{dn}} = \dfrac{{du}}{{dn}} \\\ \Rightarrow \dfrac{1}{n}dn = du \;
So, we get –
∫3∞1log⁡n×1ndn=∫n=3n=∞1udu ⇒∫3∞1nlog⁡ndn=[log⁡u]n=3n=∞ ⇒∫3∞1nlog⁡ndn=[log⁡(log⁡n)]n=3n=∞ ⇒∫3∞1nlog⁡ndn=log⁡(log⁡∞)−log⁡(log⁡3)   \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⁡∞=∞\log \infty = \infty so log⁡(log⁡∞)=∞\log (\log \infty ) = \infty
⇒∫3∞1nlog⁡ndn=∞\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
∫3∞1n(log⁡n)pdn\int\limits_3^\infty {\dfrac{1}{{n{{(\log n)}^p}}}dn}
Let –
log⁡n=u ⇒du=1ndn   \log n = u \\\ \Rightarrow du = \dfrac{1}{n}dn \;

⇒∫3∞1n(log⁡n)pdn=∫n=3∞1updu ⇒∫3∞1n(log⁡n)pdn=[u−p+1−p+1]n=3∞ ⇒∫3∞1n(log⁡n)pdn=[1(1−p)(log⁡n)p−1]n=3∞ ⇒∫3∞1n(log⁡n)pdn=1(1−p)(log⁡∞)p−1−1(1−p)(log⁡3)p−1   \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⁡∞=0\log \infty = 0 so we get 1(1−p)(log⁡∞)p−1=0\dfrac{1}{{(1 - p){{(\log \infty )}^{p - 1}}}} = 0 .
⇒∫3∞1n(log⁡n)pdn=0−1(1−p)(log⁡3)p−1\Rightarrow \int\limits_3^\infty {\dfrac{1}{{n{{(\log n)}^p}}}dn} = 0 - \dfrac{1}{{(1 - p){{(\log 3)}^{p - 1}}}}
Clearly, 1(1−p)(log⁡3)p−1\dfrac{1}{{(1 - p){{(\log 3)}^{p - 1}}}} is a finite quantity, so the integral is finite for p>1 and thus converges.
Hence, ∑1n(log⁡n)p\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.