Science:Infinite Series Module/Units/Unit 2/2.1 The Divergence Test/2.1.02 A Useful Theorem

The following theorem will yield the divergence test.

Theorem 1
If the infinite series ${\begin{aligned}\sum _{k=1}^{\infty }a_{k}\end{aligned}}$ is convergent, then $\lim _{k\rightarrow \infty }a_{k}=0.$

Proof of Theorem 1

The proof of this theorem can be found in most introductory calculus textbooks that cover the divergence test and is supplied here for convenience. Let the partial sum $s_{n}$ be

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} s_n = \sum_{k=1}^{n} a_k = a_1 + a_2 + a_3 + \ldots +a_{n-1} +  a_n.\end{align} }

Then

${\begin{aligned}s_{n-1}=\sum _{k=1}^{n-1}a_{k}=a_{1}+a_{2}+a_{3}+\ldots +a_{n-1}\end{aligned}}$

and $s_{n}-s_{n-1}=a_{n}.$

By assumption, a_n is convergent, so the sequence {s_n} is convergent (using the definition of a convergent infinite series). Let the number S be given by

$S=\lim _{n\rightarrow \infty }s_{n}.$

Since n-1 also tends to infinity as n tends to infinity, we also have

$S=\lim _{n\rightarrow \infty }s_{n-1}.$

Finally,

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \lim_{n \rightarrow \infty} a_n &= \lim_{n \rightarrow \infty} (s_n - s_{n-1}) \\  &= S - S \\ &=0. \end{align}}

Thus, if

$\sum _{k=1}^{\infty }a_{k}$

is convergent, then

$\lim _{k\rightarrow \infty }a_{k}=0,$

as required.