Question

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

Answer 1

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 \[\sum {{a_n}} \].