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