Using the comparison test to determine convergence or divergence

Comparison test for convergence

The 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 comparison test to show that

the original series $a_n$ is diverging if

the original series $a_n$ is greater than or equal to the comparison series $b_n$ and both series are positive, $a_n\geq b_n\geq 0$, and

the comparison series $b_n$ is diverging

Note: If $a_n<b_n$, the test is inconclusive

the original series is converging if

the original series $a_n$ is less than or equal to the comparison series $b_n$ and both series are positive, $0\leq a_n\leq b_n$, and

the comparison series $b_n$ is converging

Note: If $b_n<a_n$, the test is inconclusive

How to find the comparison series and use the comparison test to say whether the series converges or diverges

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Comparison test for a rational function

Example

Use the comparison test to say whether or not the series converges.

$\sum^{\infty}_{n=1}\frac{n}{\sqrt{n^5}+n}$

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

$a_n=\frac{n}{\sqrt{n^5}+n}$

For the comparison series, we’ll use the same numerator as the original series, since it’s already pretty simple. Looking at the denominator, we can see that the first term $\sqrt{n^5}$ carries more weight and will affect our series more than the second term $n$, so we’ll just use the first term from the original denominator for the denominator of our comparison series, and the comparison series is

$b_n=\frac{n}{\sqrt{n^5}}$

$b_n=\frac{n}{n^{\frac52}}$

$b_n=n^{1-\frac52}$

$b_n=n^{-\frac32}$

$b_n=\frac{1}{n^{\frac32}}$

We can see that this simplified version of $b_n$ is just a p-series, where $p=3/2$. 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 $p=3/2$ in $b_n$, we know that $b_n$ converges.

That means we need to show that $0\leq a_n\leq b_n$ to prove that the original series $a_n$ is also converging. If we can’t show that $0\leq a_n\leq b_n$, then the test is inconclusive with this particular comparison series.

We’re usually trying to find a comparison series that’s a geometric or p-series.

Let’s try to verify that $0\leq a_n\leq b_n$ by checking a few points for both $a_n$ and $b_n$, like $n=1$, $n=4$ and $n=9$.

Looking at these three terms, we can see that $0\leq a_n\leq b_n$, since $a_n$ is always positive and always smaller than $b_n$.

Therefore, we can say that the original series $a_n$ converges.

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