# 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

{\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }a_{k}\end{aligned}}}

is convergent, then

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

### 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 ${\displaystyle s_{n}}$ be

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

{\displaystyle {\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 ${\displaystyle s_{n}-s_{n-1}=a_{n}.}$

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

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

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

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

Finally,

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

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

is convergent, then

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

as required.