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# How to choose the ${B_n}$ for limit comparison test?

Last updated date: 26th Feb 2024
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Hint: The above question is based on the concept of limit comparison test. The main approach towards solving this is to divide ${A_n}$ and ${B_n}$ where ${A_n}$ is the original series and ${B_n}$ is the second series. We can come to know with the help of the second series whether the original series is converging or diverging.

Complete step by step solution:
In mathematics, the limit comparison test is a method of testing for the convergence of infinite series.
Now suppose that we have two series given
$\sum {_n{A_n}}$ and $\sum {_n{B_n}}$ where ${A_n} \geqslant 0,{B_n} > 0$ for all the n.
Then if $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{A_n}}}{{{B_{_n}}}} = L$ with $0 < L < \infty$,
then either both series diverge or both series converge.
Now,
If $L < \infty$and $\sum\limits_{n = 0}^\infty {{B_n}}$ converges, then $\sum\limits_{n = 0}^\infty {{A_n}}$ also converges.
If $L > 0$and $\sum\limits_{n = 0}^\infty {{B_n}}$ diverges to $\infty$, then $\sum\limits_{n = 0}^\infty {{A_n}}$ also converges.
Now let’s consider a series ${A_n}$ for as $\sum\limits_{n = 1}^\infty {\dfrac{2}{{{n^3} - 4}}}$.Now if I compare with $\dfrac{1}{{{n^3}}}$ as another series i.e${B_n}$
but $\dfrac{1}{{{n^3} - 4}}$ is not $\leqslant \dfrac{1}{{{n^3}}}$ for all n.
So now dividing the series of ${A_n}$ and ${B_n}$
$\mathop {\lim }\limits_{n \to \infty } \dfrac{{\dfrac{2}{{{n^3} - 4}}}}{{\dfrac{1}{{{n^3}}}}} = \dfrac{2}{{{n^3} - 4}} \times \dfrac{{{n^3}}}{1} = \dfrac{{2{n^3}}}{{{n^3} - 4}}$
Now if the power of the numerator is the same as in the denominator then we take the ratio of coefficients $\dfrac{2}{1}$. Since it turns out to be a positive finite number then both series do the same thing.
We know that ${B_n} = \sum\limits_{n = 1}^\infty {\dfrac{1}{{{n^3}}}}$is converging then the original series is also converging.

Note: If we want to know something about the series $\sum {{a_n}}$ , the limit comparison test suggests that we should look for an appropriate series $\sum {{b_n}}$, where the underlying sequences ${a_n}$ and ${b_n}$ behave similarly in the sense that $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{a_n}}}{{{b_n}}}$ exists. That is, if we understand how $\sum {{b_n}}$behaves then we understand $\sum {{a_n}}$.