Question

How do you determine if $\sum {\dfrac{{{n^3}}}{{({n^4}) - 1}}} $ from $n = 2$ to $n = \infty $ is convergent ?

Answer 1

Hint: In this question, we need to find whether the given series is convergent. Firstly, we will see the behaviour of the series given in the question and denote it by ${x_n} = \dfrac{{{n^3}}}{{({n^4}) - 1}}$. Then we consider one more series and denote it by ${y_n} = \dfrac{1}{n}$ . Then we apply the limit comparison test, and check whether the series is convergent or divergent.

Complete step-by-step answer:
Given the series of the form $\sum\limits_{n = 2}^{n = \infty } {\dfrac{{{n^3}}}{{({n^4}) - 1}}} $
We are asked to determine whether the given series is convergent or divergent.
We will apply the limit comparison test (LCT) to check the convergence.
The limit comparison test states that if ${x_n}$ and ${y_n}$ are series with positive terms and if $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{x_n}}}{{{y_n}}}$ is positive and finite. Then either both the series converge or diverge.
Let us consider the series inside the summation and denote it by ${x_n}$.
i.e. ${x_n} = \dfrac{{{n^3}}}{{({n^4}) - 1}}$
Let us now examine the behaviour of the series ${x_n}$.
For larger values of $n$, the denominator ${n^4} - 1$ of the series ${x_n}$ behaves like ${n^4}$.
Hence, for larger $n$, the series ${x_n}$ acts like,
$\dfrac{{{n^3}}}{{{n^4}}} = \dfrac{1}{n}$
Let us consider ${y_n} = \dfrac{1}{n}$.
Now we apply the limit comparison test, to examine whether the series is convergent or not.
Let us consider,
$\mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{{x_n}}}{{{y_n}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{\dfrac{{{n^3}}}{{{n^4} - 1}}}}{{\dfrac{1}{n}}}} \right)$
Now multiplying the numerator by the reciprocal of the denominator, we get,
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{{x_n}}}{{{y_n}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{{n^3}}}{{{n^4} - 1}} \times \dfrac{n}{1}} \right)$
This can also be written as,
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{{x_n}}}{{{y_n}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{{n^3}({n^1})}}{{{n^4} - 1}}} \right)$
We have the multiplication rule of exponential which is given by, ${a^c} \cdot {a^d} = {a^{c + d}}$
We have in the numerator $a = n$, $c = 3$ and $d = 1$.
Hence we have,
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{{x_n}}}{{{y_n}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{{n^{3 + 1}}}}{{{n^4} - 1}}} \right)$
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{{x_n}}}{{{y_n}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{{n^4}}}{{{n^4} - 1}}} \right)$
We write the expression inside the parenthesis as,
$\dfrac{{{n^4}}}{{{n^4} - 1}} = \dfrac{1}{{1 - \dfrac{1}{{{n^4}}}}}$
So we have now,
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{{x_n}}}{{{y_n}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{1}{{1 - \dfrac{1}{{{n^4}}}}}} \right)$
Note that as $n \to \infty $, we have $\dfrac{1}{{{n^4}}} \to 0$
So we obtain the limit as,
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{{x_n}}}{{{y_n}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{1}{1}} \right)$
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\dfrac{{{x_n}}}{{{y_n}}}} \right) = 1$.
Note that the limit obtained above is positive and finite. Hence, by the limit comparison test, either both the series ${x_n}$ and ${y_n}$ are convergent or both of them divergent.
But we have the series ${y_n} = \dfrac{1}{n}$ which is divergent.
So by LCT, the series ${x_n} = \dfrac{{{n^3}}}{{({n^4}) - 1}}$ is also divergent.
Hence the series $\sum\limits_{n = 2}^{n = \infty } {\dfrac{{{n^3}}}{{({n^4}) - 1}}} $ is divergent by limit comparison test.

Note:
Students must know the meaning of convergence and divergence of a series.
The nth partial sum of the series $\sum\limits_{n = 1}^\infty {{a_n}} $ is given by ${S_n} = {a_1} + {a_2} + .... + {a_n}$. If the sequence of these partial sums $\{ {S_n}\} $ converges to L, then the sum of the series converges to L. If $\{ {S_n}\} $ diverges, then the sum of the series diverges.
We must know the limit comparison test, which states that if ${x_n}$ and ${y_n}$ are series with positive terms and if $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{x_n}}}{{{y_n}}}$ is positive and finite. Then either both the series converge or diverge.