# Term test – Serlo

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• ↳ Project "Serlo"
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• Introduction • Complex numbers • Supremum and infimum • Sequences • Convergence and divergence • Subsequences, Accumulation points and Cauchy sequences • Series • Convergence criteria for series • Overview: convergence criteria • Cauchy criterion • Term test • Bounded series and convergence • Direct comparison test • Root test • Ratio test • Alternating series test • Cauchy condensation test • Application of convergence criteria • Exercises • Exponential and Logarithm functions • Trigonometric and Hyperbolic functions • Continuity • Differential Calculus 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 $\sum _{k=1}^{\infty }a_{k}$ , where $(a_{k})_{k\in \mathbb {N} }$ is not a null sequence, is divergent. When you invert that, this means that for every convergent series $\sum _{k=1}^{\infty }a_{k}$ we have that$\lim _{k\to \infty }a_{k}=0$ ( $(a_{k})_{k\in \mathbb {N} }$ is a null sequence).

## Term test

Theorem (Term test)

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

Example (Term test)

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

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

Warning

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

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

## Proof with telescoping sums

Proof (Term test)

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

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

{\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 $\sum _{k=1}^{\infty }a_{k}$ has a limit $s\in \mathbb {R}$ . Thus we have

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

{\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 $(a_{n})_{n\in \mathbb {N} }$ 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 $\sum _{k=1}^{\infty }a_{k}$ satisfies the Cauchy criterion:

$\forall \epsilon >0\,\exists N\in \mathbb {N} \,\forall n\geq m\geq N:\left|\sum _{k=m}^{n}a_{k}\right|<\epsilon$ We don't consider all $n\geq m$ , but only the case $m=n$ :

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

## Example

Exercise

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

Solution

We have

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

## Outlook: Stronger version of term test

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