Using the ratio test to determine whether or not a series converges
What is the ratio test, and what does it show?
The ratio test for convergence lets us determine the convergence or divergence of a series ???a_n??? using the limit
???L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|???
Once we find a value for ???L???, we can say that
the series converges absolutely if ???L<1???.
the series diverges if ???L>1??? or if ???L??? is infinite.
the test is inconclusive if ???L=1???.
The ratio test is used most often when our series includes a factorial or something raised to the ???n???th power.
Using the ratio test to determine whether the series converges absolutely or diverges
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Proving absolute convergence with the ratio test
Example
Use the ratio test to say whether the series converges or diverges.
???\sum^{\infty}_{n=1}\frac{n^3}{4^n}???
To use the ratio test, we need to solve for the limit
???L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|???
and then evaluate the value of ???L???.
???L=\lim_{n\to\infty}\left|\frac{\frac{(n+1)^3}{4^{n+1}}}{\frac{n^3}{4^n}}\right|???
We can drop the absolute value bars since all of our terms will be positive.
???L=\lim_{n\to\infty}\frac{\frac{(n+1)^3}{4^{n+1}}}{\frac{n^3}{4^n}}???
???L=\lim_{n\to\infty}\frac{(n+1)^3}{4^{n+1}}\left(\frac{4^n}{n^3}\right)???
Grouping like bases together, we get
???L=\lim_{n\to\infty}\frac{(n+1)^3}{n^3}\left(\frac{4^n}{4^{n+1}}\right)???
???L=\lim_{n\to\infty}\frac{(n+1)^3}{n^3}\left(4^{n-(n+1)}\right)???
???L=\lim_{n\to\infty}\frac{(n+1)^3}{n^3}\left(4^{-1}\right)???
???L=\lim_{n\to\infty}\frac{(n+1)^3}{n^3}\left(\frac14\right)???
???L=\frac14\lim_{n\to\infty}\frac{(n+1)^3}{n^3}???
???L=\frac14\lim_{n\to\infty}\frac{n^3+3n^2+3n+1}{n^3}???
???L=\frac14\lim_{n\to\infty}\frac{n^3+3n^2+3n+1}{n^3}\left(\frac{\frac{1}{n^3}}{\frac{1}{n^3}}\right)???
???L=\frac14\lim_{n\to\infty}\frac{\frac{n^3}{n^3}+\frac{3n^2}{n^3}+\frac{3n}{n^3}+\frac{1}{n^3}}{\frac{n^3}{n^3}}???
???L=\frac14\lim_{n\to\infty}\frac{1+\frac{3}{n}+\frac{3}{n^2}+\frac{1}{n^3}}{1}???
???L=\left(\frac14\right)\frac{1+\frac{3}{\infty}+\frac{3}{\infty}+\frac{1}{\infty}}{1}???
???L=\left(\frac14\right)\frac{1+0+0+0}{1}???
???L=\frac14???
Since ???L<1???, we can say that the original series ???a_n??? converges absolutely.