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

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.