Question

How do you use the integral test to determine if $\sum\limits_{n = 3}^\infty {\dfrac{1}{{n\ln n\ln (\ln n)}}} $ is convergent or divergent?

Answer 1

Hint: To solve this question use the Integral Test to check given series is convergent or divergent.
According to the integral test We determine the convergence of $\sum\limits_{n = 3}^\infty {\dfrac{1}{{n\ln n\ln (\ln n)}}} $ by comparing it with
$\int\limits_3^\infty {\dfrac{1}{{x\ln x\ln (\ln x)}}} dx$. Then the series $\sum\limits_{n = 3}^\infty {\dfrac{1}{{n\ln n\ln (\ln n)}}} $ and the integral $\int\limits_3^\infty {\dfrac{1}{{x\ln x\ln (\ln x)}}} dx$ both converge or both diverge.
If the integral gives finite value then it converges otherwise it will diverge.
To solve the integral $\int\limits_3^\infty {\dfrac{1}{{x\ln x\ln (\ln x)}}} dx$, make substitution $u = \ln (\ln x)$.
Which gives the derivative of $u$ as $du = \dfrac{1}{{x\ln x}}dx$. Then change the limit integration of $u$ from $\ln (\ln 3)$ to $\infty $.
Solve the integration using formula $\int {\dfrac{1}{x}dx} = \ln x + C$ and remember $\mathop {\lim }\limits_{u \to \infty } \ln u \to \infty $.

Complete step-by-step answer:
Apply the integral test to find convergence of the series $\sum\limits_{n = 3}^\infty {\dfrac{1}{{n\ln n\ln (\ln n)}}} $.
The Integral Test: Suppose the sequence ${a_n} = f(n)$ is of positive terms, where $f$ is a continuous, positive decreasing function. Then the series $\sum\limits_{n = N}^\infty {{a_n}} $ , N is positive integer and the integral $\int\limits_N^\infty {f(x)} dx$ both converge or both diverge.
Consider the function $f(n) = \dfrac{1}{{n\ln n\ln (\ln n)}}$ so, $f(x) = \dfrac{1}{{x\ln x\ln (\ln x)}}$.
The function $f(x) = \dfrac{1}{{x\ln x\ln (\ln x)}}$is continuous, positive, and decreasing for $x \geqslant 3$ so we can apply the integral test to the series.
Now, we have to find$\int\limits_3^\infty {\dfrac{1}{{x\ln x\ln (\ln x)}}} dx$.
Here we use the substitution$u = \ln (\ln x)$.
Use the chain rule to find the derivative of $u = \ln (\ln x)$.
$\therefore $ $du = \dfrac{1}{{\ln x}}\dfrac{d}{{dx}}(\ln x)$
$ \Rightarrow du = \dfrac{1}{{\ln x}} \times \dfrac{1}{x}dx$
$ \Rightarrow du = \dfrac{1}{{x\ln x}}dx$
Apply the limits of $x$.
Put $x = 3$into $u$
 $ \Rightarrow u = \ln (\ln 3)$
Put $x = \infty $into $u$
 $ \Rightarrow u = \ln (\ln \infty )$
$ \Rightarrow u = \infty $($\because \ln \infty = \infty $ )
So the integral becomes,
$\int\limits_3^\infty {\dfrac{1}{{x\ln x\ln (\ln x)}}} dx = \int\limits_3^\infty {\dfrac{1}{{x\ln x}}} \times \dfrac{1}{{\ln (\ln x)}}dx$
$ \Rightarrow \int\limits_3^\infty {\dfrac{1}{{x\ln x\ln (\ln x)}}} dx = \int\limits_{\ln (\ln 3)}^\infty {\dfrac{1}{u}} du$
Apply the formula of integration$\int {\dfrac{1}{x}dx} = \ln x + C$.
$ \Rightarrow \int\limits_{\ln (\ln 3)}^\infty {\dfrac{1}{u}} du = \left[ {\ln u} \right]_{\ln (\ln 3)}^\infty $
Here, as \[u\] approaches to $\infty $$\ln u$ also approaches to $\infty $that implies $\mathop {\lim }\limits_{u \to \infty } \ln u \to \infty $.
$\therefore \left[ {\ln u} \right]_{\ln (\ln 3)}^\infty = \infty - \ln (\ln (\ln 3))$
$\therefore \left[ {\ln u} \right]_{\ln (\ln 3)}^\infty = \infty $
So the integral $\int\limits_3^\infty {\dfrac{1}{{x\ln x\ln (\ln x)}}} dx$ does not have finite values hence $\sum\limits_{n = 3}^\infty {\dfrac{1}{{n\ln n\ln (\ln n)}}} $is divergent.

The series $\sum\limits_{n = 3}^\infty {\dfrac{1}{{n\ln n\ln (\ln n)}}} $ is divergent.

Note:
Students always forget to change the limit of integration whenever we substitute the value in the definite integral we have to find the limits for the substitution.
Integration of $\int {\dfrac{1}{x}dx} = \ln x + C$, don’t use power formulas to integrate.
The chain rule: If $f(t)$ and $g(t)$ are the function of $t$ then the derivative of its composite function is given by $\dfrac{d}{{dx}}[f(g(t)] = g'(t)f'(g(t))$