Question

How do you do the Comparison Test to see if \[\dfrac{1}{4{{n}^{2}}-1}\] converges, n is going to infinity?

Answer 1

Hint: These types of problems are based on the concept of limits. Let us consider \[{{a}_{n}}\] to be \[\dfrac{1}{4{{n}^{2}}-1}\]. Since n tends to infinity, \[4{{n}^{2}}\] also tends to infinity. \[4{{n}^{2}}-1\] can be approximated as \[4{{n}^{2}}\]. And let \[{{b}_{n}}\] be \[\dfrac{1}{{{n}^{2}}}\]. Substitute the values of \[{{a}_{n}}\] and \[{{b}_{n}}\] values in \[\displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}\] and find the value. Since \[\dfrac{1}{4}\] is positive and finite, then \[{{a}_{n}}\] and \[{{b}_{n}}\] are either convergent or divergent. But, we know \[\dfrac{1}{{{n}^{2}}}\] is convergent, therefore, \[\dfrac{1}{4{{n}^{2}}-1}\] is also convergent.

Complete step by step solution:
According to the question, we are asked to show \[\dfrac{1}{4{{n}^{2}}-1}\] is convergent when n tends to infinity.
We have been given the function is \[\dfrac{1}{4{{n}^{2}}-1}\]. --------(1)
First, we have to know the Comparison Test. It states that
“if \[{{a}_{n}}\] and \[{{b}_{n}}\] are series with positive terms and if \[\displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}\] is positive and finite, then either both the series converges or diverges”.
Let us assume \[{{a}_{n}}\] to be \[\dfrac{1}{4{{n}^{2}}-1}\].
\[\Rightarrow {{a}_{n}}=\dfrac{1}{4{{n}^{2}}-1}\]
Here, n tends to infinity.
Consider the denominator of \[{{a}_{n}}\].
Since n tends to infinity, \[{{n}^{2}}\] also tends to infinity and thus \[4{{n}^{2}}\] tends to infinity.
On subtracting 1 from infinity, we still consider the term as infinity.
Therefore, \[4{{n}^{2}}-1\simeq 4{{n}^{2}}\].
So, let us consider \[{{b}_{n}}\] to be \[\dfrac{1}{{{n}^{2}}}\].
According to the comparison test, we have to find \[\displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}\].
On substituting the values of \[{{a}_{n}}\] and \[{{b}_{n}}\], we get
\[\displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}=\displaystyle \lim_{n \to \infty }\dfrac{\dfrac{1}{4{{n}^{2}}-1}}{\dfrac{1}{{{n}^{2}}}}\]
We know that \[\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}\]. Using this property of division in the above expression, we get
\[\displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}=\displaystyle \lim_{n \to \infty }\dfrac{1}{4{{n}^{2}}-1}\times \dfrac{{{n}^{2}}}{1}\]
\[\Rightarrow \displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}=\displaystyle \lim_{n \to \infty }\dfrac{{{n}^{2}}}{4{{n}^{2}}-1}\]
Let us divide the numerator and denominator by \[{{n}^{2}}\].
\[\Rightarrow \displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}=\displaystyle \lim_{n \to \infty }\dfrac{\dfrac{{{n}^{2}}}{{{n}^{2}}}}{\dfrac{4{{n}^{2}}-1}{{{n}^{2}}}}\]
On further simplification, we get
\[\displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}=\displaystyle \lim_{n \to \infty }\dfrac{\dfrac{{{n}^{2}}}{{{n}^{2}}}}{\dfrac{4{{n}^{2}}}{{{n}^{2}}}-\dfrac{1}{{{n}^{2}}}}\]
Now, we find that \[{{n}^{2}}\] is common in both the numerator and denominator.
Let us cancel \[{{n}^{2}}\] form the numerator and denominator.
Therefore, we get
\[\displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}=\displaystyle \lim_{n \to \infty }\dfrac{1}{4-\dfrac{1}{{{n}^{2}}}}\]
Now, let us substitute the limits in the simplified function.
\[\Rightarrow \displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}=\dfrac{1}{4-\dfrac{1}{{{\infty }^{2}}}}\]
We know that any term divided by infinity is 0.
Therefore, we get
\[\displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}=\dfrac{1}{4-0}\]
\[\therefore \displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}=\dfrac{1}{4}\]
Here, we get \[\dfrac{1}{4}>0\].
According to the comparison test, \[\displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}\] is positive and finite.
Therefore, \[{{a}_{n}}\] and \[{{b}_{n}}\] are either convergent or divergent.
But we know that \[\dfrac{1}{{{n}^{2}}}\] is always convergent.
Then, the function \[\dfrac{1}{4{{n}^{2}}-1}\] should also be convergent.
Therefore, the function \[\dfrac{1}{4{{n}^{2}}-1}\] is convergent.

Note: We should not substitute the limits directly to the given function which will result in not defined answers. Also \[\dfrac{1}{{{n}^{2}}}\] is not divergent and is convergent. Avoid calculation mistakes based on sign convention. Also, we should know the Comparison Test rule properly to solve this question.