# Math for Non-Geeks/ Root test

Let us turn our attention to the root test, which is a powerful tool for proving the convergence or divergence of a given series. It is based off the direct comparison test, in fact we will compare the series with the geometric series $\sum _{k=1}^{\infty }q^{k}$ with $0\leq q<1$.
The root test was first published in 1821 in the textbook „Cours d'analyse“ from French mathematician Augustin Louis Cauchy.

## Derivation
### Recap

We have already learned how to use the direct comparison test with a majorant. In summary, a series $\sum _{k=1}^{\infty }a_{k}$ with $a_{k}\geq 0$ is convergent, if there exists a convergent series $\sum _{k=1}^{\infty }b_{k}$ with $a_{k}\leq b_{k}$.
Furthermore we know that every geometric series $\sum _{k=1}^{\infty }q^{k}$ with $q\in [0,1)$ is convergent.

### First derivation

Let $\sum _{k=1}^{\infty }a_{k}$ be a series. Also let $a_{k}\geq 0$ for all $k\in \mathbb {N}$, because this is a constraint we need for the direct comparison test. For our majorant we need a $q\in [0,1)$ with $a_{k}\leq q^{k}$. Then we have
${\begin{aligned}\sum _{k=1}^{\infty }a_{k}&\leq \sum _{k=1}^{\infty }q^{k}=\sum _{k=0}^{\infty }q^{k}-q^{0}\\[0.3em]&={\frac {1}{1-q}}-1={\frac {q}{1-q}}<\infty \end{aligned}}$
The series $\sum _{k=1}^{\infty }a_{k}$ is convergent by our direct comparison test. The inequality $a_{k}\leq q^{k}$ can be transformed:
$a_{k}\leq q^{k}\iff {\sqrt[{k}]{a_{k}}}\leq q$
Thus if there exists a $q$ with $0\leq q<1$, so that ${\sqrt[{k}]{a_{k}}}\leq q$, then $a_{k}\leq q^{k}$ and the series $\sum _{k=1}^{\infty }a_{k}$ is convergent.
Comprehension question: We start the geometric series $\sum _{k=1}^{\infty }q^{k}$ at $k=1$ and not (as usual) at $k=0$. Why does this make sense in the above derivation?
In the step $a_{k}\leq q^{k}\iff {\sqrt[{k}]{a_{k}}}\leq q$ we take the $k$-th root, but the $0$-th root is not defined, because that would mean $q^{\frac {1}{0}}$. This is why $k$ should start at $1$.

### Bringing limit superior into play

For the convergence behaviour we can ignore the value of finitely many summands. Thus ${\sqrt[{k}]{a_{k}}}\leq q$ must not be satisfied for all $k\in \mathbb {N}$, but for all $k\in \mathbb {N}$, with finitely many exceptions. So the inequality ${\sqrt[{k}]{a_{k}}}\leq q$ must be satisfied for almost all $k\in \mathbb {N}$.
We can restate the requirement, that there must exist a $q\in [0,1)$ with ${\sqrt[{k}]{a_{k}}}\leq q$ for almost all $k\in \mathbb {N}$, using the limit superior:
$\exists {\tilde {q}}\in [0,1):\limsup _{k\to \infty }{\sqrt[{k}]{a_{k}}}\leq {\tilde {q}}$
Or in other terms:
$\limsup _{k\to \infty }{\sqrt[{k}]{a_{k}}}<1$
If ${\sqrt[{k}]{a_{k}}}\leq q$ for almost all $k$, the succession $\left({\sqrt[{k}]{a_{k}}}\right)_{k\in \mathbb {N} }$ is bounded above and must have an accumulation point less than or equal to $q$. This accumulation point is equal to $\limsup _{k\to \infty }{\sqrt[{k}]{a_{k}}}$ and $\limsup _{k\to \infty }{\sqrt[{k}]{a_{k}}}\leq q$.
Conversely, let $\limsup _{k\to \infty }{\sqrt[{k}]{a_{k}}}\leq q$ for a $q\in [0,1)$. Then for all all $\epsilon >0$ the inequality ${\sqrt[{k}]{a_{k}}}\leq q+\epsilon$ is satisfied for almost all $k\in \mathbb {N}$. Because $q<1$ there exists $\epsilon >0$, that is small enough so that $q+\epsilon <1$. Set ${\tilde {q}}=q+\epsilon$. We have ${\tilde {q}}<1$ and the inequality ${\sqrt[{k}]{a_{k}}}\leq {\tilde {q}}$ holds for almost all $k$.
Summary: Instead of ${\sqrt[{k}]{a_{k}}}\leq q$ for almost all $k\in \mathbb {N}$ it suffices to show $\limsup _{k\to \infty }{\sqrt[{k}]{a_{k}}}<1$, to prove convergence the series.

### What about absolute convergence?

What happens if not all $a_{k}\geq 0$? Then we cannot use the above argumentation, because for even $k$ and $a_{k}<0$ the values ${\sqrt[{k}]{a_{k}}}$ is not defined. But we still have a valid argumentation if we talk about $\sum _{k=1}^{\infty }|a_{k}|$, to show the absolute convergence of the series (from which normal convergence follows immediately). Thus for series with $a_{k}\geq 0$ for all $k$ we have $|a_{k}|=a_{k}$. In this case nothing changes because for $a_{k}$ we can also use $|a_{k}|$. The conclusion is:

### Root test for divergence

We have derived the root test to show the convergence of a series. Can we also derive a criterion for divergence? Let us assume that $\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}>1$. Then for infinitely many $k\in \mathbb {N}$ we have the inequality ${\sqrt[{k}]{|a_{k}|}}\geq 1$. For these $k$ we have $|a_{k}|\geq 1^{k}=1$, thus $\left(|a_{k}|\right)_{k\in \mathbb {N} }$ cannot be a null sequence. But then $\left(a_{k}\right)_{k\in \mathbb {N} }$ is also no null sequence. From the term test it follows immediately that $\sum _{k=1}^{\infty }a_{k}$ is divergent. We can generalize this case if instead of $\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}>1$ we require the inequality ${\sqrt[{k}]{|a_{k}|}}\geq 1$ for almost all $k\in \mathbb {N}$.

## Theorem

Theorem (Root test)
Let $\sum _{k=1}^{\infty }a_{k}$ be a series. If $\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}<1$, then the series is absolutely convergent. If $\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}>1$, then it is divergent. Even if ${\sqrt[{k}]{|a_{k}|}}\geq 1$ for infinitely many $k\in \mathbb {N}$, the series diverges.
The proof uses the same ideas as the derivation above:
Proof (Root test)
Proof step: From $\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}<1$ follows the absolute convergence of $\sum _{k=1}^{\infty }a_{k}$.
Let $\sum _{k=1}^{\infty }a_{k}$ be a series. If $\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}<1$, then $\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}\leq q$ for a $0\leq q<1$ (we can choose $q=\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}$ for example).
Choose $\epsilon >0$ small enough, so that $q+\epsilon <1$. From the definition of limit superior it follows that for almost all $k\in \mathbb {N}$ the inequality ${\sqrt[{k}]{|a_{k}|}}\leq q+\epsilon$ holds. But that means $|a_{k}|\leq (q+\epsilon )^{k}$ for almost all $k$. Since $\sum _{k=1}^{\infty }(q+\epsilon )^{k}$ is convergent (it is a geometric series with $q+\epsilon <1$) also the series $\sum _{k=1}^{\infty }|a_{k}|$ is convergent by direct comparison. So the series $\sum _{k=1}^{\infty }a_{k}$ converges absolutely.
Proof step: From $\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}>1$ or ${\sqrt[{k}]{|a_{k}|}}\geq 1$ for infinitely many $k$ follows the divergence of $\sum _{k=1}^{\infty }a_{k}$.
Let $\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}>1$ or ${\sqrt[{k}]{|a_{k}|}}\geq 1$ for infinitely many $k$. This implies $|a_{k}|\geq 1^{k}=1$ for infinitely many $k$. But then $\left(|a_{k}|\right)_{k\in \mathbb {N} }$ cannot be a null sequence and thus $\left(a_{k}\right)_{k\in \mathbb {N} }$ is also not a null sequence. From the term test we know that $\sum _{k=1}^{\infty }a_{k}$ is divergent.
Hint
If $\left({\sqrt[{k}]{|a_{k}|}}\right)_{k\in \mathbb {N} }$ is convergent, then $\lim _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}=\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}$. Therefore we could also consider the limit $\lim _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}$ if it exists. This is usually done when proving convergence using the root test.

## Limitations of the root test

In case $\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}=1$ we cannot say whether we have convergence or divergence. In fact there exist both convergent and divergent series that satisfy this equation. First consider the harmonic series $\sum _{k=1}^{\infty }{\frac {1}{k}}$, which is divergent. We have
${\begin{aligned}\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}&=\limsup _{k\to \infty }{\sqrt[{k}]{\left|{\frac {1}{k}}\right|}}\\[0.5em]&=\limsup _{k\to \infty }{\sqrt[{k}]{\frac {1}{k}}}\\[0.5em]&=\limsup _{k\to \infty }{\frac {1}{\sqrt[{k}]{k}}}\\[0.5em]&\,{\color {OliveGreen}\left\downarrow \ \lim _{k\to \infty }{\sqrt[{k}]{k}}=1\right.}\\[0.5em]&={\frac {1}{1}}=1\end{aligned}}$
But also the convergent series $\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}$ satisfies this equation:
${\begin{aligned}\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}&=\limsup _{k\to \infty }{\sqrt[{k}]{\left|{\frac {1}{k^{2}}}\right|}}\\[0.5em]&=\limsup _{k\to \infty }{\sqrt[{k}]{\frac {1}{k^{2}}}}\\[0.5em]&=\limsup _{k\to \infty }{\frac {1}{\sqrt[{k}]{k^{2}}}}\\[0.5em]&=\limsup _{k\to \infty }\left({\frac {1}{\sqrt[{k}]{k}}}\right)^{2}\\[0.5em]&\,{\color {OliveGreen}\left\downarrow \ \lim _{k\to \infty }{\sqrt[{k}]{k}}=1\right.}\\[0.5em]&=\left({\frac {1}{1}}\right)^{2}=1\end{aligned}}$
These examples show that we cannot conclude convergence or divergence from $\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}=1$. In this case we have to employ another convergence criterion!

## How to apply the root test

To apply the root test on a series $\sum _{k=1}^{\infty }a_{k}$ we can proceed as follows: We compute ${\sqrt[{k}]{|a_{k}|}}$ and find the limit (if the limit exists) or the limit superior.
* If $\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}<1$, then the series converges absolutely.
* If $\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}>1$, then the series diverges.
* If ${\sqrt[{k}]{|a_{k}|}}\geq 1$ for infinitely many $k$, then the series diverges.
* Else, if none of the above is true, we cannot make a statement about convergence behaviour using the root test.

## Exercises
### Exercise 1

Math for Non-Geeks: Template:Aufgabe

### Exercise 2

Math for Non-Geeks: Template:Aufgabe
* ↑ Siehe the answer to the question „Where is the root test first proved“ of Q&A website „History of Science and Mathematics“