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.
