Answer
Verified
411.9k+ views
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 \[\sum {{a_n}} \].
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 \[\sum {{a_n}} \].
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Change the following sentences into negative and interrogative class 10 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE