Infinite Series  nthTerm Test
You CAN Ace Calculus 

17trek > calculus > svc > infinite series > nthterm test 
Topics You Need To Understand For This Page
It also helps to know these topics but they are not required. 
Page Tools, Related Topics and Links
sections on this page 

external links you may find helpful 
Search 17calculus
The nthTerm Test is also called the Divergence Test.
17calculus is a 17trek subject 

quick links 
nthTerm Test
If \( \displaystyle{\lim_{n \to \infty}{a_n} \neq 0} \), then \( \displaystyle{\sum_{n=1}^{\infty}{a_n}}\) diverges.
Mathematical Logic Form
\[ \lim_{n \to \infty}{a_n} \neq 0 ~ \to ~ \sum_{n=1}^{\infty}{a_n} \text{ diverges} \] 17calculus Mathematical Logic
Alternate Form
The nthTerm Test can also be stated in this alternate form.
If \( \displaystyle{\sum_{n=1}^{\infty}{a_n}}\) converges, then
\( \displaystyle{\lim_{n \to \infty}{a_n} = 0} \).
Using Mathematical Logic this can be written
\(\displaystyle{ \sum_{n=1}^{\infty}{a_n} \text{ converges } ~ \to ~ \lim_{n \to \infty}{a_n} = 0 } \), which is the contrapositive of the form above.
This alternate form is not very useful for computation. So we use the theorem above for all discussion and practice problems here at 17calculus.
How To Use The nthTerm Test 

To use the nthterm test, just take the limit \( \displaystyle{\lim_{n \to \infty}{a_n} } \). If this limit turns out to be nonzero, the series diverges and you are done. If the limit is equal to zero, then the test is inconclusive and says nothing about the series. It may converge or it may diverge. You need to use another test to determine convergence or divergence.
Things To Watch Out For 

The nthTerm Test seems to be pretty straightforward. However, there is a trap that many students fall into.
Basically, this test says that, for a series
\( \displaystyle{ \sum_{n=1}^{\infty}{a_n}} \), if
\( \displaystyle{ \lim_{n \to \infty}{a_n} \neq 0 }, \) then the series diverges. Pretty simple, eh?
Trap   There is a tendency by almost every student to use this theorem to prove convergence. The statement says nothing about convergence. Let us give you a couple of examples that demonstrate what we mean. Let's compare the convergence or divergence of these two very similar series. (If you don't know about pseries yet, just take our word for the convergence/divergence conclusions. You will understand this soon enough.)
A.  \(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n}}}\)  \( \displaystyle{\lim_{n\to\infty}{\frac{1}{n}} = 0} \)  diverges 
pseries with \(p=1\) 

B. 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2}}}\)  \( \displaystyle{\lim_{n\to\infty}{\frac{1}{n^2}} = 0} \)  converges 
pseries with \(p=2\) 
It is important to notice that in both cases the limit of the terms goes to zero. However, series A diverges while series B converges.
So you can see that just because the limit goes to zero, this does not guarantee the series will converge. The only time you can apply this theorem is when the limit does not go to zero. This guarantees divergence. When the limit does go to zero, you still don't know if the series converges or diverges. You need to use another test to determine convergence.
Below is the nthterm test row from the infinite series table. Notice that the convergence line says that this test cannot be used. Think about this thoroughly and completely so that you get your head around it.
nthTerm Test 

convergence: cannot be used 
divergence: \(\displaystyle{ \lim_{n \to \infty}{ a_n \neq 0} }\) 
A Quick Video 

Here is a quick video that explains this test.  
Okay, so you are now ready for some practice problems. Once you go through those, the next logical step is the pseries.