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Infinite Series  Ratio Test 
The Ratio Test is probably the most important test and the test you will use the most as you are learning infinite series. It is used A LOT in power series. I believe it is the most powerful test of all and, if you look at the Infinite Series Table, you will see that it is listed first in Group 3. So I suggest you master it from the start. It's not hard, and if your algebra skills are strong, you might even find it fun to use. Also, the more familiar you are with it and the more practice you have, the sooner you will start to be able to look at a series and see almost right away if the Ratio Test will tell you what you need to know. Cool, eh?


Ratio Test 
Let \( \sum{a_n} \) be a series with nonzero terms and let
\( \displaystyle{\lim_{n \to \infty}{\left \frac{a_{n+1}}{a_n} \right} = L} \) 
Three cases are possible depending on the value of L.
\( L < 1 \): The series converges absolutely.
\( L = 1 \): The Ratio Test is inconclusive.
\( L > 1 \): The series diverges.

Quick Details
used to prove convergence  yes 
used to prove divergence  yes 
can be inconclusive  yes 
if \(L=\infty\) then \(L>1\) and the series diverges 
\(a_n\) terms can be positive or negative or both 
requires the use of limits at infinity 



When To Use The Ratio Test 
The ratio test is best used when you have certain elements in the sum. The best way to get a feel for this is to build a set of sheets containing examples of tests that work as you are working practice problems. This is one technique listed in the
infinite series study techniques section. This is an extremely powerful technique that will help you really understand infinite series.
Here is a list of things to watch for.
1. Sums that include factorials.
2. Sums with exponents containing n.
How To Use The Ratio Test 
In general, the idea is to set up the ratio
\( \displaystyle{\lim_{n \to \infty}{\left \frac{a_{n+1}}{a_n} \right} = L} \)
and evaluate it.
In detail, you need to determine what \(a_n\) is and then build \(a_{n+1}\), set up the fraction, combine like terms and then take the limit of each term. Setting up the limit and combining like terms are the easy parts. The challenge comes in taking the limit.
Key  It is important to remember to use the absolute value signs unless you are absolutely convinced that the term will always be positive. This is critical to practice up front since, once you get to Taylor Series, you can't and don't want to drop the absolute value signs. They are critical to the result. It is never wrong to include them and, as you work more problems, you will get a feel for when you need them and when you don't. In the practice problems and examples, we will use them unless we explicitly state that they are not needed. Some instructors are less rigid about this than others. As always, check with your instructor to see what they require.
Things To Notice
1. If you get \( \infty \) for the limit, this indicates divergence since it fits the case where the limit is greater than one. Notice that the theorem says nothing about the limit needing to be finite.
2. In the fraction that you are taking the limit of, the \( n+1 \) term is in the numerator and the \(n\) term is in the denominator. In order for the ratio test to work, they must appear like this. Do you know why? [answer ]
3. If you get one for the result, you cannot say anything about convergence or divergence of the series. You need to use another test. Sometimes, a comparison test ( either direct or limit ) will be the best next step.
Okay, it's time for some videos. This first video clip is a great overview of the ratio test. Notice that he doesn't use absolute value signs, so he requires that the terms be positive.
 Dr Chris Tisdell: Series, Comparison + Ratio Tests 

The beginning of this next video has a good discussion about the ratio test. Then the instructor shows two examples when the ratio test is inconclusive to emphasize that a series may converge or diverge when the ratio test is inconclusive.
 MIT OpenCourseware: Ratio Test for Convergence 

Okay, time for some practice problems.
Instructions   Unless otherwise instructed, determine the convergence or divergence of the following series, using the ratio test, if possible. [These instructions imply that if the ratio test fails ( \(L = 1\) ), you need to use another test to prove convergence or divergence.]
Practice A01 
\(\displaystyle{\sum_{n=0}^{\infty}{\left[\frac{n!}{2^n}\right]}}\) 


The series \(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{n!}{2^n} \right] } }\) diverges by the Ratio Test. 
Whenever you have a factorial, the ratio test will often work. So, let's try it.
\(\displaystyle{ a_n = \frac{n!}{2^n} }\) and \(\displaystyle{ a_{n+1} = \frac{(n+1)!}{2^{n+1}} }\)
\(\displaystyle{
\lim_{n \to \infty}{ \left \frac{a_{n+1}}{a_n} \right} =
\lim_{n \to \infty}{ \left \frac{(n+1)!}{2^{n+1}} \div \frac{n!}{2^n} \right } =
\lim_{n \to \infty}{ \left \frac{(n+1)!}{2^{n+1}} \cdot \frac{2^n}{n!} \right } }\)
Now combine like terms.
\(\displaystyle{ \frac{2^n}{2^{n+1}} = \frac{2^n}{2(2^n)} = 1/2 }\)
\(\displaystyle{\frac{(n+1)!}{n!} = \frac{1\cdot2\cdot3\cdot4 . . . n\cdot(n+1)}{1\cdot2\cdot3\cdot4 . . . n} = n+1 }\)
So our limit is now \(\displaystyle{ \lim_{n \to \infty}{ \left \frac{n+1}{2} \right} = \frac{\infty}{2} = \infty > 1 \to }\) the series diverges.
Note: We could have left off the absolute value signs all the way through this problem. However, to follow the theorem exactly, we need them unless we explicitly state that the ratio is positive. It is a common practice (although not always good) to leave them off without explaining that teachers may do quite often. If you don't understand something, always ask. That includes anything that you don't understand or is not clear on this site.
Practice A01 Final Answer 
The series \(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{n!}{2^n} \right] } }\) diverges by the Ratio Test. 
Practice A02 
\(\displaystyle{\sum_{n=0}^{\infty}{\left[\frac{n+2}{2n+9}\right]}}\) 


\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{n+2}{2n+9} \right] } }\) diverges by the nthterm test. 
\(\displaystyle{ a_n = \frac{n+2}{2n+9} }\)
\(\displaystyle{ a_{n+1} = \frac{n+3}{2(n+1)+9} = \frac{n+3}{2n+11} }\)
Now we can set up the limit and evaluate it.
\(\displaystyle{
\begin{array}{rcl}
\lim_{n \to \infty}{\left \frac{a_{n+1}}{a_n} \right}
& = & \lim_{n \to \infty}{\left \frac{n+3}{2n+11} \frac{2n+9}{n+2} \right} \\\\
& = & \lim_{n \to \infty}{\left \frac{(n+3)(2n+9)}{(2n+11)(n+2)} \frac{1/n^2}{1/n^2} \right} \\\\
& = & \lim_{n \to \infty}{\left \frac{(1+3/n)(2+9/n)}{(2+11/n)(1+2/n)} \right} = \frac{2}{2} = 1
\end{array}
}\)
So the ratio test failed. When the limit is equal to one, we cannot tell if the series converges or diverges.
However, the nthterm test yields
\(\displaystyle{ \lim_{n \to \infty}{\frac{n+2}{2n+9}} = \frac{1}{2} \neq 0 }\)
Therefore, the series diverges by the nthterm test.
Notes:
 Notice that we did not just stop after the ratio test failed. We needed to use another test to complete the problem, since we were told to determine convergence or divergence in the problem statement.
 Once you get more experience with infinite series, you will realize that the ratio test is not the best test to use here. Whenever you have polynomials, one in the numerator and one in the denominator, check out the ratio of the highest power terms. If you get a constant, that is a good indication that maybe the series diverges and the nthterm test may be the quickest way to determine divergence.
Practice A02 Final Answer 
\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{n+2}{2n+9} \right] } }\) diverges by the nthterm test. 
Practice A03 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{4^n}{n \cdot 3^n } \right] } }\) 


The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{4^n}{n \cdot 3^n} \right] } }\) diverges by the ratio test. 
Practice A03 Final Answer 
The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{4^n}{n \cdot 3^n} \right] } }\) diverges by the ratio test. 
Practice A04 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^{10}}{10^n} \right] } }\) 


The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^{10}}{10^n} \right] } }\) converges by the ratio test. 
Practice A04 Final Answer 
The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^{10}}{10^n} \right] } }\) converges by the ratio test. 
Practice A05 
\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{2^n+5}{3^n} \right] } }\) 


The series \(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{2^n+5}{3^n} \right] } }\) converges by the ratio test. 
Practice A05 Final Answer 
The series \(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{2^n+5}{3^n} \right] } }\) converges by the ratio test. 
Practice A06 
\(\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{3^k}{k^2} \right] } }\) 


The series \(\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{3^k}{k^2} \right] } }\) diverges by the ratio test. 
Practice A06 Final Answer 
The series \(\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{3^k}{k^2} \right] } }\) diverges by the ratio test. 
Practice A07 
\(\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{2^k}{k!} \right] } }\) 


The series \(\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{2^k}{k!} \right] } }\) converges by the ratio test. 
We have included two solutions to this problem by two different instructors.
Practice A07 Final Answer 
The series \(\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{2^k}{k!} \right] } }\) converges by the ratio test. 
Practice A08 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ n(5/4)^n \right] } }\) 


The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ n(5/4)^n \right] } }\) diverges by the ratio test. 
Practice A08 Final Answer 
The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ n(5/4)^n \right] } }\) diverges by the ratio test. 
Practice A09 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(1)^n 2^{3n}}{(2n)!} \right] } }\) 


The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(1)^n 2^{3n}}{(2n)!} \right] } }\) converges by the ratio test. 
Practice A09 Final Answer 
The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(1)^n 2^{3n}}{(2n)!} \right] } }\) converges by the ratio test. 
Practice A10 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^3}{(\ln 3)^n} \right] } }\) 


The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^3}{(\ln 3)^n} \right] } }\) converges by the ratio test. 
Practice A10 Final Answer 
The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^3}{(\ln 3)^n} \right] } }\) converges by the ratio test. 
Practice A11 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^2}{2^n} \right] } }\) 


The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^2}{2^n} \right] } }\) converges by the ratio test. 
Practice A11 Final Answer 
The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^2}{2^n} \right] } }\) converges by the ratio test. 
Practice A12 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n(3)^n}{4^{n1}} \right] } }\) 


The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n(3)^n}{4^{n1}} \right] } }\) converges by the ratio test. 
Practice A12 Final Answer 
The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n(3)^n}{4^{n1}} \right] } }\) converges by the ratio test. 
Practice A13 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(1)^{n+1} (n^2)2^n}{n!} \right] } }\) 


The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(1)^{n+1} (n^2)2^n}{n!} \right] } }\) converges by the ratio test. 
Practice A13 Final Answer 
The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(1)^{n+1} (n^2)2^n}{n!} \right] } }\) converges by the ratio test. 
Practice A14 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{7^n}{e^n} \right] } }\) 


The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{7^n}{e^n} \right] } }\) diverges by the ratio test. 
Practice A14 Final Answer 
The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{7^n}{e^n} \right] } }\) diverges by the ratio test. 
Practice B01 
\(\displaystyle{\sum_{n=2}^{\infty}{\left[\frac{n^2 2^n}{5^n}\right]}}\) 


The series \(\displaystyle{ \sum_{n=2}^{\infty}{ \left[ \frac{n^2 2^n}{ 5^n } \right] } }\) converges by the Ratio Test. 
Since we have n's in the exponents, we might be tempted to use the Root Test. However, the Root Test is best used when the base also has an n in it. So, let's use the Ratio Test.
\(\displaystyle{
\begin{array}{rcl}
\lim_{n \to \infty}{\left \frac{a_{n+1}}{a_n} \right} & = & \lim_{n \to \infty}{\left \frac{(n+1)^2 2^{n+1}}{5^{n+1}} \frac{5^n}{n^2 2^n} \right} \\\\
& = & \lim_{n \to \infty}{\left \frac{(n+1)^2}{n^2} \frac{2^{n+1}}{2^n} \frac{5^n}{5^{n+1}} \right} \\\\
& = & \lim_{n \to \infty}{\left \left(\frac{n+1}{n}\right) \frac{2}{5} \right} = 2/5
\end{array}
}\)
Practice B01 Final Answer 
The series \(\displaystyle{ \sum_{n=2}^{\infty}{ \left[ \frac{n^2 2^n}{ 5^n } \right] } }\) converges by the Ratio Test. 
Practice B02 
\(\displaystyle{\sum_{n=0}^{\infty}{\left[\frac{3n^2}{(2n1)!}\right]}}\) 


\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{3n^2}{(2n1)!} \right] } }\) converges absolutely by the ratio test. 
Since we have a factorial, let's try the ratio test.
\(\displaystyle{ a_n = \frac{3n^2}{(2n1)!} }\)
\(\displaystyle{ a_{n+1} = \frac{3(n+1)^2}{(2(n+1)1)!} = \frac{3(n+1)^2}{(2n+1)!} }\)
Set up the limit and evaluate it.
\(\displaystyle{
\begin{array}{rcl}
\lim_{n \to \infty}{\left \frac{a_{n+1}}{a_n} \right}
& = & \lim_{n \to \infty}{\left \frac{3(n+1)^2}{(2n+1)!} \frac{(2n1)!}{3n^2} \right} \\\\
& = & \lim_{n \to \infty}{\left \frac{(n+1)^2}{n^2} \frac{(2n1)!}{(2n+1)!} \right} \\\\
& = & \lim_{n \to \infty}{\left \frac{(n+1)^2}{n^2} \frac{(2n1)!}{(2n+1)(2n)(2n1)!} \right} \\\\
& = & \lim_{n \to \infty}{\left \frac{(n+1)^2}{n^2} \frac{1}{(2n+1)(2n)} \frac{1/n^4}{1/n^4} \right} \\\\
& = & \lim_{n \to \infty}{\left \frac{1/n + 1/n^2}{1(2+1/n)(2)} \right} = \frac{0}{4} = 0 < 1
\end{array}
}\)
Since the limit \(0 < 1\) the series converges absolutely.
Practice B02 Final Answer 
\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{3n^2}{(2n1)!} \right] } }\) converges absolutely by the ratio test. 
Practice B03 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{9^n}{(3)^{n+1}n} \right] } }\) 


\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{9^n}{(3)^{n+1}n} \right] } }\) diverges by the ratio test. 
\(\displaystyle{ a_n = \frac{9^n}{(3)^{n+1}n} }\)
Let's rewrite this a little bit to make the limit simpler.
\(\displaystyle{
\begin{array}{rcl}
a_n & = & \frac{9^n}{(3)^{n+1}n} \\\\
& = & \frac{9^n}{(1)^{n+1}3^{n+1}n} \\\\
& = & \frac{(1)^{n+1} 9^n}{3^{n+1}n}
\end{array}
}\)
In the above work, we can move the \(1\) term to the numerator since the fraction is the same whether the sign is in the numerator or denominator. This is the only time we can do this. So, now that \(a_n\) is simpler, let's write \(a_{n+1}\).
\(\displaystyle{ a_{n+1} = \frac{(1)^{n+2} 9^{n+1}}{3^{n+2}(n+1)} }\)
And, now we can calculate the limit.
\(\displaystyle{
\begin{array}{rcl}
\lim_{n \to \infty}{\left \frac{a_{n+1}}{a_n} \right}
& = & \lim_{n \to \infty}{\left \frac{9^{n+1}}{3^{n+2}(n+1)} \frac{3^{n+1}n}{9^n} \right} \\\\
& = & \lim_{n \to \infty}{\left \frac{9^{n+1}}{9^n} \frac{3^{n+1}}{3^{n+2}} \frac{n}{n+1} \right} \\\\
& = & \lim_{n \to \infty}{\left \frac{9}{3} \frac{n}{n+1} \right} = 3 > 1
\end{array}
}\)
Since the limit \(3 > 1\) the series diverges.
Practice B03 Final Answer 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{9^n}{(3)^{n+1}n} \right] } }\) diverges by the ratio test. 
Practice B04 
\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{3(1)^n}{n^2+1} \right] } }\) 


\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{3(1)^n}{n^2+1} \right] } }\) converges by the alternating series test. 
\(\displaystyle{ a_n = \frac{3(1)^n}{n^2+1} }\)
\(\displaystyle{ a_{n+1} = \frac{3(1)^{n+1}}{(n+1)^2+1} }\)
Now we can set up and evaluate the limit.
\(\displaystyle{
\begin{array}{rcl}
\lim_{n \to \infty}{\left \frac{a_{n+1}}{a_n} \right}
& = & \lim_{n \to \infty}{\left \frac{3}{(n+1)^2+1} \frac{n^2+1}{3} \right} \\\\
& = & \lim_{n \to \infty}{\left \frac{n^2+1}{(n+1)^2+1} \frac{1/n^2}{1/n^2} \right} \\\\
& = & \lim_{n \to \infty}{\left \frac{1+1/n^2}{(1+1/n)^2+1/n^2} \right} = \frac{1}{1} = 1
\end{array}
}\)
So the ratio test failed. When the limit is equal to one, we cannot tell if the series converges or diverges.
However, if your first inclination was to use the alternating series test, you were right. We show in a practice problem on the alternating series page that this series converges absolutely.
Notes:
 Notice that we did not just stop after the ratio test failed. We needed to use another test to complete the problem, since we were told to determine convergence or divergence in the problem statement.
 Once you get more experience with infinite series, you will realize that the ratio test is not the best test to use here. With an alternating series, it is usually best to start with the alternating series test.
Practice B04 Final Answer 
\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{3(1)^n}{n^2+1} \right] } }\) converges by the alternating series test. 
Practice B05 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(2n)!}{n!n!} \right] } }\) 


The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(2n)!}{n!n!} \right] } }\) diverges by the ratio test. 
Practice B05 Final Answer 
The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(2n)!}{n!n!} \right] } }\) diverges by the ratio test. 
Practice B06 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^n}{3^{1+3n}} \right] } }\) 


The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^n}{3^{1+3n}} \right] } }\) diverges by the ratio test. 
Practice B06 Final Answer 
The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^n}{3^{1+3n}} \right] } }\) diverges by the ratio test. 
Practice B07 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{4^n(n!)^2}{(n+2)!} \right] } }\) 


The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{4^n(n!)^2}{(n+2)!} \right] } }\) diverges by the ratio test. 
Practice B07 Final Answer 
The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{4^n(n!)^2}{(n+2)!} \right] } }\) diverges by the ratio test. 
Practice B08 
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^n}{n!} } }\) 

Practice C01 
\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{(5)^n}{4^{2n+1}(n+1)} \right] } }\) 


\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{(5)^n}{4^{2n+1}(n+1)} \right] } }\) converges absolutely by the ratio test. 
Let's try the ratio test.
\(\displaystyle{ a_n = \frac{(5)^n}{4^{2n+1}(n+1)} }\)
Before we set up \(a_{n+1}\), let's simplify this a bit.
\(\displaystyle{
\begin{array}{rcl}
a_n & = & \frac{(5)^n}{4^{2n+1}(n+1)} \\\\
& = & \frac{(1)^n 5^n}{4 \cdot 4^{2n} (n+1)} \\\\
& = & \frac{(1)^n 5^n}{4 (16)^n (n+1)} \\\\
& = & \frac{1}{4} \left( \frac{5}{16} \right)^n \frac{(1)^n}{n+1}
\end{array}
}\)
Okay, so this \(a_n\) will be simpler to work with and, since \(1/4\) is just a constant, it does not affect convergence or divergence and can be taken outside the sum. This leaves
\(\displaystyle{ a_n = \frac{ (1)^n 5^n}{16^n (n+1)} ~~~ \to ~~~ a_{n+1} = \frac{ (1)^{n+1} 5^{n+1}}{16^{n+1} (n+2)} }\)
So now we set up the limit and evaluate it.
\(\displaystyle{
\begin{array}{rcl}
\lim_{n \to \infty}{\left \frac{a_{n+1}}{a_n} \right}
& = & \lim_{n \to \infty}{\left \frac{5^{n+1}}{16^{n+1} (n+2)} \frac{16^n (n+1)}{5^n} \right} \\\\
& = & \lim_{n \to \infty}{\left \frac{5^{n+1}}{5^n} \frac{16^n}{16^{n+1}} \frac{n+1}{n+2} \right} \\\\
& = & \lim_{n \to \infty}{\left \frac{5}{16} \frac{n+1}{n+2} \right} \\\\
& = & \frac{5}{16} < 1
\end{array}
}\)
Since the limit \(5/16 < 1\) the series converges absolutely.
Note: This is also an alternating series, so we could have used the alternating series test to determine that it converges absolutely.
Practice C01 Final Answer 
\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{(5)^n}{4^{2n+1}(n+1)} \right] } }\) converges absolutely by the ratio test. 