Divergence Test

The divergence test is one of the easiest ways to test whether a Series converges or not, by using the Limit (Calculus) of it’s sequence.

Definition §

Let $a_n$ be a sequence of real numbers, and ${\sum }_{n=1}^{\infty }{a}_n$ be it’s respective series.

$$\underset{n\to \infty }{\lim }{a}_n\ne 0\to \sum _{n=1}^{\infty }{a}_{n}diverges$$

• If the limit does not exist, it still satisfies the inequality, and hence the divergence test cannot be used.
• If ${\lim }_{n\to \infty }{a}_n=0$ then ${\sum }_{n=1}^{\infty }{a}_n$ may converge or diverge, other tests are needed.

The principal of this test works by looking at the values that the sequence approaches as $n\to \infty$. If the values does not equal 0, then the sum of the terms (the series) will keeping increasing (or decreasing if the sequence contains negative terms) , and hence will diverge.

Proof §

Proof:

The proof of the divergence test can be done by proving the contrapositive, which is:

$$\sum _{k=1}^{\infty }(a_k)\text{ is convergent}\to \underset{k\to \infty }{\lim }(a_k)=0$$

Note, we are talking about the sequence ($a_k$) not it’s partial sums ($s_k$)!

If ${\sum }_{k=1}^{\infty }$ is convergent, then so is the sequence of partial sums, $(s_k{)}_{k\in \mathbb{N}}$. Then, because the sequence of partial sums is convergent, it is Cauchy. This means for all $ϵ>0$, there exists a $N>0$ such that for all $n,m>N$ (with $n>m$):

$$|s_n-s_m|=\left|\sum _{k=m+1}^n{a}_k\right|<ϵ$$

For any $m$ we can take $n=m+1$, giving us ${\sum }_{k=m+1}^n=a_n$ (summing on a single term). Hence, we have:

$$|a_n-0|<ϵ$$

Which implies the sequence converges to 0.