Using the comparison test to determine convergence or divergence

 
 
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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.

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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???.

calculating the first few terms of the series and comparison series

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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