The limit comparison test for convergence

Determining the convergence of a series by comparing it to a similar comparison series

The limit comparison test for convergence lets us determine the convergence or divergence of the given series $a_n$ by comparing it to a similar, but simpler comparison series $b_n$

We’re usually trying to find a comparison series that’s a geometric or p-series, since it’s very easy to determine the convergence of a geometric or p-series.

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We can use the limit comparison test to show that

the original series $a_n$ is diverging if

$a_n\geq 0$ and $b_n>0$,

$L=\lim_{n\to\infty}\frac{a_n}{b_n}$ and $0<L<\infty$, and

the comparison series $b_n$ is diverging

the original series $a_n$ is converging if

$a_n\geq 0$ and $b_n>0$,

$L=\lim_{n\to\infty}\frac{a_n}{b_n}$ and $0<L<\infty$, and

the comparison series $b_n$ is converging

How to apply the limit comparison test

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Determining convergence with the limit comparison test

Example

Use the limit comparison test to say whether or not the series is converging.

$\sum^{\infty}_{n=1}\frac{6n}{2n^3+3}$

We need to find a series that’s similar to the original series, but simpler. The original series is

$a_n=\frac{6n}{2n^3+3}$

For the comparison series, we’ll use the same numerator as the original series, since it’s already pretty simple, but we’ll drop the $6$ since it has little effect on the series as $n\to\infty$. Looking at the denominator, we can see that the first term $2n^3$ carries more weight and will affect our series more than the second term $3$, so we’ll just use the first term from the original denominator for the denominator of our comparison series, but drop the $2$, and the comparison series is

$b_n=\frac{n}{n^3}$

$b_n=\frac{1}{n^2}$

We can see that this simplified version of $b_n$ is just a p-series, where $p=2$. We’ll use the p-series test for convergence to say whether or not $b_n$ converges. Remember, the p-series test says that the series will

converge when $p>1$

diverge when $p\le1$

Since $p=2$ in $b_n$, we know that $b_n$ converges.

That means we need to show that $a_n>0$ and $b_n>0$ and that

$\lim_{n\to\infty}\frac{a_n}{b_n}=L>0$

in order to prove that the original series $a_n$ is also converging.

Let’s try to verify that $a_n>0$ and $b_n>0$ by checking a few points for both $a_n$ and $b_n$, like $n=1$, $n=2$ and $n=3$.

Looking at these three terms, we can see that $a_n>0$ and $b_n>0$. There’s no positive value of $n$ that will make a term in $a_n$ or $b_n$ negative.

We’re usually trying to find a comparison series that’s a geometric or p-series, since it’s very easy to determine the convergence of a geometric or p-series.

The last thing we need to verify is

$\lim_{n\to\infty}\frac{a_n}{b_n}=L>0$

Plugging $a_n$ and $b_n$ into the limit formula gives

$L=\lim_{n\to\infty}\frac{\frac{6n}{2n^3+3}}{\frac{1}{n^2}}$

$L=\lim_{n\to\infty}\frac{6n}{2n^3+3}\left(\frac{n^2}{1}\right)$

$L=\lim_{n\to\infty}\frac{6n^3}{2n^3+3}$

$L=\lim_{n\to\infty}\frac{6n^3}{2n^3+3}\left(\frac{\frac{1}{n^3}}{\frac{1}{n^3}}\right)$

$L=\lim_{n\to\infty}\frac{\frac{6n^3}{n^3}}{\frac{2n^3}{n^3}+\frac{3}{n^3}}$

$L=\lim_{n\to\infty}\frac{6}{2+\frac{3}{n^3}}$

$L=\frac{6}{2+\frac{3}{\infty}}$

$L=\frac{6}{2+0}$

$L=3$

Since

$L=3>0$,

$a_n>0$ and $b_n>0$, and

the comparison series $b_n$ is converging,

we can say the the original series $a_n$ is also converging.

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