1

Does the converse of $n$th term test for divergence hold?

I am new to this topic so I don't have much idea about the proof. Please help.

I only know that the series converge if limit of the $n$th term of that sequence approaches $0$ as $n$ approaches $\infty$.

user
  • 154,566
The Learner
  • 170
  • Well no. The last line is not the $n$th term test. The $n$th term test says that if a series converges, its summands (terms) tend to zero as $n$ approaches infinity. What the "last line" in the question says is the converse, which does not necessarily hold. The harmonic series is the classical counterexample. – kamills Nov 23 '19 at 10:11
  • Sorry for the wrong statement – The Learner Nov 23 '19 at 12:32

1 Answers1

3

Let consider a convergent series

$$\lim_{n\to \infty} S_n=\lim_{n\to \infty}\sum_{k=1}^n a_k = L \in \mathbb R$$

then

$$\lim_{n\to \infty} \left(S_n-S_{n-1}\right)=\lim_{n\to \infty}\left( \sum_{k=1}^na_k-\sum_{k=1}^{n-1}a_k\right)=\lim_{n\to \infty} a_n =L-L=0$$

therefore we can conclude that

$$\sum_{k=1}^\infty a_k=L \in\mathbb R \implies a_n \to 0$$

Therefore $a_n \to 0$ represents a necessary condition for the convergence of the series.

To see that the converse implication doesn't hold let consider the counter example $a_n=\frac1n$ which leads to the divergent harmonic series.

Therefore $a_n \to 0$ doesn't represent a sufficient condition for the convergence of the series.

user
  • 154,566