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Question: How to choose the \({B_n}\) for limit comparison test?...

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

Explanation

Solution

The above question is based on the concept of limit comparison test. The main approach towards solving this is to divide An{A_n} and Bn{B_n} where An{A_n} is the original series and Bn{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
∑nAn\sum {_n{A_n}} and ∑nBn\sum {_n{B_n}} where An⩾0,Bn>0{A_n} \geqslant 0,{B_n} > 0 for all the n.
Then if lim⁡n→∞AnBn=L\mathop {\lim }\limits_{n \to \infty } \dfrac{{{A_n}}}{{{B_{_n}}}} = L with 0<L<∞0 < L < \infty ,
then either both series diverge or both series converge.
Now,
If L<∞L < \infty and ∑n=0∞Bn\sum\limits_{n = 0}^\infty {{B_n}} converges, then ∑n=0∞An\sum\limits_{n = 0}^\infty {{A_n}} also converges.
If L>0L > 0and ∑n=0∞Bn\sum\limits_{n = 0}^\infty {{B_n}} diverges to ∞\infty , then ∑n=0∞An\sum\limits_{n = 0}^\infty {{A_n}} also converges.
Now let’s consider a series An{A_n} for as ∑n=1∞2n3−4\sum\limits_{n = 1}^\infty {\dfrac{2}{{{n^3} - 4}}} .Now if I compare with 1n3\dfrac{1}{{{n^3}}} as another series i.eBn{B_n}
but 1n3−4\dfrac{1}{{{n^3} - 4}} is not ⩽1n3 \leqslant \dfrac{1}{{{n^3}}} for all n.
So now dividing the series of An{A_n} and Bn{B_n}

\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}} $$.