logo

If an Infinite Series Converges, Then the Infinite Sequence Converges to 0 📂Calculus

If an Infinite Series Converges, Then the Infinite Sequence Converges to 0

Theorem

If $\displaystyle \sum _{ n=1 }^{ \infty }{ { a }_{ n }}$ converges, then $\displaystyle \lim _{ n\to \infty }{ { a }_{ n }}=0$

Explanation

This theorem might be a bit surprising and counterintuitive at first. You might wonder why the converse doesn’t hold. A classic counterexample involves considering the following sequences:

$$ \begin{align*} { a }_{ n }&=\frac { 1 }{ n } \\ { b }_{ n }&=\sqrt { n }-\sqrt { n-1 } \end{align*} $$

Both sequences converge to 0, but their sum diverges to infinity. For the first case, refer to Orem’s proof. Thinking about the contrapositive, it states ‘if an infinite sequence does not converge to 0, then the infinite series diverges’, which is used to demonstrate divergence in infinite series and is known as the Divergence Test.

Proof

Let $$ \begin{align*} S:=\sum _{ n=1 }^{ \infty }{ { a }_{ n }} \end{align*} $$

and express $a_n$ as follows:

$$ \begin{align*} { a }_{ n }=\sum _{ k=1 }^{ n }{ { a }_{ k }}-\sum _{ k=1 }^{ n-1 }{ { a }_{ k }} \end{align*} $$

Taking limits on both sides yields:

$$ \begin{align*} \lim _{ n\to \infty }{ { a }_{ n }} =& \lim _{ n\to \infty }{ \left( \sum _{ k=1 }^{ n }{ { a }_{ k }}-\sum _{ k=1 }^{ n-1 }{ { a }_{ k }} \right) } \\ =& \sum _{ n=1 }^{ \infty }{ { a }_{ n }}-\sum _{ n=1 }^{ \infty }{ { a }_{ n }} \\ &=S-S \\ &=0 \end{align*} $$