# Term test – Serlo

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In this chapter we will see an easy and useful criterion for divergence. This criterion is called term test or also null-sequence test or divergence test. It says that every series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$, where ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ is not a null sequence, is divergent. When you invert that, this means that for every convergent series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ we have that${\displaystyle \lim _{k\to \infty }a_{k}=0}$( ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ is a null sequence).

## Term test

Theorem (Term test)

If a series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ is convergent, then ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ is a null sequence. That means, that every series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ is divergent, if ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ is divergent or ${\displaystyle \lim _{k\to \infty }a_{k}\neq 0}$.

Example (Term test)

The series ${\displaystyle \sum _{k=1}^{\infty }(-1)^{k}}$ is divergent, because ${\displaystyle \left((-1)^{k}\right)_{k\in \mathbb {N} }}$ is divergent (${\displaystyle 1}$ and ${\displaystyle -1}$ are both accumulation points and thus there is no limit).

Also the series ${\displaystyle \sum _{k=1}^{\infty }{\sqrt[{k}]{4}}}$ diverges, because ${\displaystyle \lim _{k\to \infty }{\sqrt[{k}]{4}}=1\neq 0}$.

Warning

The criterion that ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ is a null sequence is only a necessary but no sufficient criterion for the convergences of the series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$.

This means: from the fact that ${\displaystyle \lim _{k\to \infty }a_{k}=0}$ we cannot follow that ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ converges. For instance the harmonic series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k}}}$ is divergent although ${\displaystyle \lim _{k\to \infty }{\tfrac {1}{k}}=0}$.

## Proof with telescoping sums

Proof (Term test)

Our premise is that ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ is a convergent series. We want to conclude that ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ is a null sequence.

A member of the sequence ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ can be written as the difference between two consecutive partial sums ${\displaystyle S_{n}=\sum _{k=1}^{n}a_{k}}$ and ${\displaystyle S_{n-1}=\sum _{k=1}^{n-1}a_{k}}$:

{\displaystyle {\begin{aligned}a_{n}&=a_{n}+0\\&=a_{n}+a_{n-1}+a_{n-2}+\ldots +a_{2}+a_{1}-a_{n-1}-a_{n-2}-\ldots -a_{2}-a_{1}\\&={\color {Teal}(a_{n}+a_{n-1}+a_{n-2}+\ldots +a_{2}+a_{1})}-{\color {Indigo}(a_{n-1}+a_{n-2}+\ldots +a_{2}+a_{1})}\\&={\color {Teal}\sum _{k=1}^{n}a_{k}}-{\color {Indigo}\sum _{k=1}^{n-1}a_{k}}=S_{n}-S_{n-1}\end{aligned}}}

From our premise we know that ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ has a limit ${\displaystyle s\in \mathbb {R} }$. Thus we have

${\displaystyle \lim _{n\to \infty }S_{n}=\lim _{n\to \infty }\sum _{k=1}^{n}a_{k}=s}$

But also ${\displaystyle \lim _{n\to \infty }S_{n-1}=s}$, because the limit will not change if we simply shift indexes. Put together we obtain:

{\displaystyle {\begin{aligned}&\lim _{n\to \infty }a_{n}\\[0.3em]&\quad {\color {OliveGreen}\left\downarrow \ a_{n}=S_{n}-S_{n-1}{\text{ (see above)}}\right.}\\[0.3em]=&\lim _{n\to \infty }(S_{n}-S_{n-1})\\[0.3em]&\quad {\color {OliveGreen}\left\downarrow \ \lim _{n\to \infty }(a_{n}-b_{n})=\lim _{n\to \infty }a_{n}-\lim _{n\to \infty }b_{n}\right.}\\[0.3em]=&\lim _{n\to \infty }S_{n}-\lim _{n\to \infty }S_{n-1}=s-s=0\end{aligned}}}

We can conclude that ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ must be null sequence.

## Proof using Cauchy criterion

Proof (Term test)

We can proof the same result using the Cauchy criterion. As a reminder, every convergent series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ satisfies the Cauchy criterion:

${\displaystyle \forall \epsilon >0\,\exists N\in \mathbb {N} \,\forall n\geq m\geq N:\left|\sum _{k=m}^{n}a_{k}\right|<\epsilon }$

We don't consider all ${\displaystyle n\geq m}$, but only the case ${\displaystyle m=n}$:

{\displaystyle {\begin{aligned}&\forall \epsilon >0\,\exists N\in \mathbb {N} \,\forall n=m\geq N:\left|\sum _{k=m}^{n}a_{k}\right|<\epsilon \\&\quad {\color {OliveGreen}\left\downarrow \ n=m\right.}\\\Rightarrow \ &\forall \epsilon >0\,\exists N\in \mathbb {N} \,\forall n\geq N:\left|\sum _{k=n}^{n}a_{k}\right|<\epsilon \\&\quad {\color {OliveGreen}\left\downarrow \ \sum _{k=n}^{n}a_{k}=a_{n}\right.}\\[0.5em]\Rightarrow \ &\forall \epsilon >0\,\exists N\in \mathbb {N} \,\forall n\geq N:\left|a_{n}\right|<\epsilon \\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ |a_{n}|=|a_{n}-0|\right.}\\[0.5em]\Rightarrow \ &\forall \epsilon >0\,\exists N\in \mathbb {N} \,\forall n\geq N:\left|a_{n}-0\right|<\epsilon \end{aligned}}}

The last formula is exactly the ${\displaystyle \epsilon }$-definition for what it means that ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ is a null sequences. In other words we have showed that ${\displaystyle \lim _{n\to \infty }a_{n}=0}$.

## Example

Exercise

Show that ${\displaystyle \sum _{n=1}^{\infty }{\frac {n}{n+1}}}$ is divergent.

Solution

We have

{\displaystyle {\begin{aligned}\lim _{n\to \infty }{\frac {n}{n+1}}&=\lim _{n\to \infty }{\frac {1}{1+{\frac {1}{n}}}}\\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ {\text{limit theorems}}\right.}\\[0.5em]&={\frac {\lim _{n\to \infty }1}{\lim _{n\to \infty }1+\lim _{n\to \infty }{\frac {1}{n}}}}={\frac {1}{1+0}}=1\end{aligned}}}

From the above we see that ${\displaystyle a_{n}={\tfrac {n}{n+1}}}$ is not a null sequence. So the series ${\displaystyle \sum _{n=1}^{\infty }{\frac {n}{n+1}}}$ is divergent according to the term test.

## Outlook: Stronger version of term test

If we demand that ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ is monotonically decreasing, then we can show that even ${\displaystyle (na_{n})_{n\in \mathbb {N} }}$ is a null sequence. See the respective exercise.