How to use the nth term test for divergence
What is the nth term test for convergence?
When the terms of a series decrease toward ???0???, we say that the series is converging. Otherwise, the series is diverging.
The ???n???th term test is inspired by this idea, and we can use it to show that a series is diverging.
Hi! I'm krista.
I create online courses to help you rock your math class. Read more.
Ironically, even though the ???n???th term test is one of the convergence tests that we learn when we study sequences and series, it can only test for divergence, it can never confirm convergence.
The ???n???th term test says that
if ???\lim_{n\to\infty}a_n\ne0???
then ???\sum{a_n}??? diverges
In other words,
If we take the limit as ???n\to\infty??? and the result is non-zero, then the series diverges
If we take the limit as ???n\to\infty??? and the result is zero, then the test is inconclusive
Notice that the only conclusion we can draw is that the series diverges. It’s possible that the series we’re testing converges, but we can’t use the ???n???th term test to show convergence. It can only be used to show divergence, and if it doesn’t prove divergence, then the test is inconclusive.
How to use the nth term test to determine the convergence of a series
Take the course
Want to learn more about Calculus 2? I have a step-by-step course for that. :)
Using the nth term test to say whether the series diverges
Example
Use the ???n???th term test to show whether the series diverges.
???\sum^{\infty}_{n=1}\frac{4n^3-4}{3n^3+2}???
To use the ???n???th term test we’ll take the limit of the series as it approaches ???\infty???.
If the result is non-zero, then the series diverges
If the result is zero, then the test is inconclusive
Taking the limit, we get
???\lim_{n\to\infty}\frac{4n^3-4}{3n^3+2}???
Notice that the only conclusion we can draw is that the series diverges. It’s possible that the series we’re testing converges, but we can’t use the nth term test to show convergence.
We’ll simplify the limit by dividing each term in the fraction by the variable of the highest degree, ???n^3???.
???\lim_{n\to\infty}\frac{4n^3-4}{3n^3+2}\left(\frac{\frac{1}{n^3}}{\frac{1}{n^3}}\right)???
???\lim_{n\to\infty}\frac{\frac{4n^3}{n^3}-\frac{4}{n^3}}{\frac{3n^3}{n^3}+\frac{2}{n^3}}???
???\lim_{n\to\infty}\frac{4-\frac{4}{n^3}}{3+\frac{2}{n^3}}???
Evaluating the limit at ???\infty???, we get
???\frac{4-\frac{4}{{\infty}^3}}{3+\frac{2}{{\infty}^3}}???
When we have a fraction in which the numerator is constant and the denominator is infinite, the whole fraction approaches ???0???.
???\frac{4-0}{3+0}???
???\frac43???
Since our answer is non-zero, the ???n???th term test proves that the series diverges.
