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You CAN Ace Calculus

17calculus > infinite series > nth-term test

precalculus
differential equations

Infinite Series - nth-Term Test

nth-Term 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 nth-Term 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.

The nth-Term Test is also called the Divergence Test.

How To Use The nth-Term Test

To use the nth-term test, just take the limit \( \displaystyle{\lim_{n \to \infty}{a_n} } \). If this limit turns out to be non-zero, 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 nth-Term Test seems to be pretty straight-forward. 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 p-series 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

p-series with \(p=1\)

B.

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2}}}\) \( \displaystyle{\lim_{n\to\infty}{\frac{1}{n^2}} = 0} \)

converges

p-series 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. [ See the Infinite Series - Integral Test page for a video showing a proof of the convergence/divergence of these two series. ]

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 nth-term 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.

nth-Term Test
\(\displaystyle{ \sum_{n=1}^{\infty}{a_n} }\)

convergence:

this test cannot be used

divergence:

\(\displaystyle{ \lim_{n \to \infty}{ a_n \neq 0} }\)

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 p-series.

next: p-series →

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Practice Problems

Instructions - - Unless otherwise instructed, determine whether these series converge or diverge.

Level A - Basic

Practice A01

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{ 2 } } }\)

answer

solution

Practice A02

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5n}{n^3} } }\)

answer

solution

Practice A03

\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^2+3n+1}{2n^2+4}} }\)

answer

solution

Practice A04

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{n^2+2}{n^2+1}}}\)

answer

solution


Level B - Intermediate

Practice B01

\(\displaystyle{\sum_{n=2}^{\infty}{\frac{(-1)^n n}{(\ln(n))^2}}}\)

answer

solution

Practice B02

\(\displaystyle{ \sum_{n=1}^{\infty}{ \ln(\cos n) } }\)

answer

solution


Level C - Advanced

Practice C01

\(\displaystyle{\sum_{n=1}^{\infty}{\sin(n\pi/2)}}\)

answer

solution

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