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Test for Divergence and Other Theorems

The following test is very valuable if the terms of your series do not converge to zero.

Test for divergence

If $\displaystyle\sum_{n=1}^\infty a_n$ converges, then it must be true that $\displaystyle\lim_{n \to \infty} a_n = 0$.
Equivalently, if $\displaystyle\lim_{n \to \infty} a_n$ is not zero, then $\displaystyle\sum_{n=1}^\infty a_n$ diverges.

Warning: The converse is not true. Just because $\lim a_n=0$ it is not necessarily true that $\sum a_n$ converges. It is necessary that your terms go to zero in order to have a convergent series, but this is not enough to ensure convergence. In other words, this test for divergence can only be used to test for divergence; it does not help us determine convergence.

For example, the divergence test tells us that $\sum 2^n$ diverges, since our terms $2^n$ certainly do not converge to 0. The test does not help with the series $\sum\tfrac 1 n$, since the terms go to zero. It turns out that this series, called the harmonic series, does not converge. Similarly, we cannot use the test with $\sum\tfrac 1{n^2}$ since its terms go to 0, but it turns out that this series does converge.

Examples: Try to use the test for divergence for $\displaystyle\sum_{n=0}^\infty\frac{n^2}{2n^2-5}$ and $\displaystyle\sum_{n=1}^\infty\ln\left(\frac{n}{n-2}\right)$. DO before looking at the solutions.

Solutions: $\displaystyle\lim_{n\to\infty}\frac{n^2}{2n^2-5}=\frac{1}{2}\not = 0$, so the series diverges by the test for divergence. $\displaystyle\lim_{n\to\infty}\ln\left(\frac{n}{n-2}\right)=\ln(1)=0$. Since our terms go to zero, the divergence test does not help, and we don't know if the series converges or diverges without doing more work.

Because the definition of a convergent series is a limit, we have the following theorems about convergent series
$\displaystyle\sum_{n=1}^\infty a_n = L$ and $\displaystyle\sum_{n=1}^\infty b_n = M$. Warning: If either series is divergent, we do not have these facts.

$If \displaystyle\sum_{n=1}^\infty a_n = L$ and $\displaystyle\sum_{n=1}^\infty b_n = M,$
$\displaystyle\sum_{n=1}^\infty \left(a_n+b_n\right) = L+M$, $\displaystyle\sum_{n=1}^\infty \left(a_n-b_n\right) = L-M$, and
$\displaystyle\sum_{n=1}^\infty \left(c\, a_n\right) = cL$.
There is no similar information for $\displaystyle\sum_{n=1}^\infty\left( a_nb_n\right)$ or $\displaystyle\sum_{n=1}^\infty \left(\frac{a_n}{b_n}\right)$.

This video justifies these theorems.

The Fundamental Theorem of Calculus

The Indefinite Integral and the Net Change

Substitution

Area Between Curves

Volumes

Integration by Parts

Integrals of Trig Functions

Trig Substitutions

Partial Fractions

Strategies of Integration

Improper Integrals

Differential Equations

Models of Growth

Infinite Sequences

Infinite Series

Integral Test

Comparison Tests

Convergence of Series with Negative Terms

The Ratio and Root Tests

Strategies for testing Series

Power Series

Representing Functions as Power Series

Taylor and Maclaurin Series

Applications of Taylor Polynomials

Partial Derivatives

Multiple Integrals

Iterated Integrals over Rectangles

Double Integrals over General Regions

Test for Divergence and Other Theorems

Test for divergence