The Divergence and Integral Tests: Learn It 1

  • Use the divergence test to check if a series might converge
  • Apply the integral test to determine if a series converges
  • Estimate how close a partial sum is to the actual sum of a series

In the previous section, we determined whether series converged by calculating limits of partial sums [latex]{S_k}[/latex]. While this direct approach works, it’s often difficult or impossible to find these limits explicitly.

Fortunately, mathematicians have developed several tests that let us determine convergence without finding explicit limits. We’ll start with two fundamental tests: the divergence test and the integral test.

Divergence Test

The divergence test comes from a basic requirement for convergence: if a series converges, its terms must approach zero.

Here’s the reasoning: for a series [latex]\displaystyle\sum_{n=1}^{\infty} a_n[/latex] to converge to sum [latex]S[/latex], we need [latex]\underset{k\to \infty}{\text{lim}} S_k = S[/latex].

Therefore, from the algebraic limit properties of sequences,

[latex]\underset{k\to \infty }{\text{lim}}{a}_{k}=\underset{k\to \infty }{\text{lim}}\left({S}_{k}-{S}_{k - 1}\right)=\underset{k\to \infty }{\text{lim}}{S}_{k}-\underset{k\to \infty }{\text{lim}}{S}_{k - 1}=S-S=0[/latex].

 

This gives us a powerful test for divergence.

divergence test

If [latex]\underset{n\to \infty }{\text{lim}}{a}_{n}=c\ne 0[/latex] or [latex]\underset{n\to \infty }{\text{lim}}{a}_{n}[/latex] does not exist, then the series [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] diverges.

CautionThe divergence test can only prove divergence—never convergence. Even if [latex]\underset{n\to \infty}{\text{lim}} a_n = 0[/latex], the series might still diverge.

The harmonic series [latex]\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}[/latex] has [latex]\underset{n\to \infty}{\text{lim}} \frac{1}{n} = 0[/latex], but it diverges. When [latex]a_n \to 0[/latex], the divergence test is inconclusive—we need other methods.

For each of the following series, apply the divergence test. If the divergence test proves that the series diverges, state so. Otherwise, indicate that the divergence test is inconclusive.

  1. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{n}{3n - 1}[/latex]
  2. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{3}}[/latex]
  3. [latex]\displaystyle\sum _{n=1}^{\infty }{e}^{\frac{1}{{n}^{2}}}[/latex]