Root test – "Math for Non-Geeks"

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 ${\displaystyle \sum _{k=1}^{\infty }q^{k}}$ with ${\displaystyle 0\leq q<1}$.

We have already learned how to use the direct comparison test with a majorant. In summary, a series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ with ${\displaystyle a_{k}\geq 0}$ is convergent, if there exists a convergent series ${\displaystyle \sum _{k=1}^{\infty }b_{k}}$ with ${\displaystyle a_{k}\leq b_{k}}$.

Furthermore we know that every geometric series ${\displaystyle \sum _{k=1}^{\infty }q^{k}}$ with ${\displaystyle q\in [0,1)}$ is convergent.

Let ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ be a series. Also let ${\displaystyle a_{k}\geq 0}$ for all ${\displaystyle k\in \mathbb {N} }$, because this is a constraint we need for the direct comparison test. For our majorant we need a ${\displaystyle q\in [0,1)}$ with ${\displaystyle a_{k}\leq q^{k}}$. Then we have

${\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }a_{k}&\leq \sum _{k=1}^{\infty }q^{k}=\sum _{k=0}^{\infty }q^{k}-q^{0}\\[0.3em]&={\frac {1}{1-q}}-1={\frac {q}{1-q}}<\infty \end{aligned}}}$

The series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ is convergent by our direct comparison test. The inequality ${\displaystyle a_{k}\leq q^{k}}$ can be transformed:

${\displaystyle a_{k}\leq q^{k}\iff {\sqrt[{k}]{a_{k}}}\leq q}$

Thus if there exists a ${\displaystyle q}$ with ${\displaystyle 0\leq q<1}$, so that ${\displaystyle {\sqrt[{k}]{a_{k}}}\leq q}$, then ${\displaystyle a_{k}\leq q^{k}}$ and the series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ is convergent.

For the convergence behaviour we can ignore the value of finitely many summands. Thus ${\displaystyle {\sqrt[{k}]{a_{k}}}\leq q}$ must not be satisfied for all ${\displaystyle k\in \mathbb {N} }$, but for all ${\displaystyle k\in \mathbb {N} }$, with finitely many exceptions. So the inequality ${\displaystyle {\sqrt[{k}]{a_{k}}}\leq q}$ must be satisfied for almost all ${\displaystyle k\in \mathbb {N} }$.

We can restate the requirement, that there must exist a ${\displaystyle q\in [0,1)}$ with ${\displaystyle {\sqrt[{k}]{a_{k}}}\leq q}$ for almost all ${\displaystyle k\in \mathbb {N} }$, using the limit superior:

${\displaystyle \exists {\tilde {q}}\in [0,1):\limsup _{k\to \infty }{\sqrt[{k}]{a_{k}}}\leq {\tilde {q}}}$

Or in other terms:

${\displaystyle \limsup _{k\to \infty }{\sqrt[{k}]{a_{k}}}<1}$

If ${\displaystyle {\sqrt[{k}]{a_{k}}}\leq q}$ for almost all ${\displaystyle k}$, the succession ${\displaystyle \left({\sqrt[{k}]{a_{k}}}\right)_{k\in \mathbb {N} }}$ is bounded above and must have an accumulation point less than or equal to ${\displaystyle q}$. This accumulation point is equal to ${\displaystyle \limsup _{k\to \infty }{\sqrt[{k}]{a_{k}}}}$ and ${\displaystyle \limsup _{k\to \infty }{\sqrt[{k}]{a_{k}}}\leq q}$.

Conversely, let ${\displaystyle \limsup _{k\to \infty }{\sqrt[{k}]{a_{k}}}\leq q}$ for a ${\displaystyle q\in [0,1)}$. Then for all all ${\displaystyle \epsilon >0}$ the inequality ${\displaystyle {\sqrt[{k}]{a_{k}}}\leq q+\epsilon }$ is satisfied for almost all ${\displaystyle k\in \mathbb {N} }$. Because ${\displaystyle q<1}$ there exists ${\displaystyle \epsilon >0}$, that is small enough so that ${\displaystyle q+\epsilon <1}$. Set ${\displaystyle {\tilde {q}}=q+\epsilon }$. We have ${\displaystyle {\tilde {q}}<1}$ and the inequality ${\displaystyle {\sqrt[{k}]{a_{k}}}\leq {\tilde {q}}}$ holds for almost all ${\displaystyle k}$.

Summary: Instead of ${\displaystyle {\sqrt[{k}]{a_{k}}}\leq q}$ for almost all ${\displaystyle k\in \mathbb {N} }$ it suffices to show ${\displaystyle \limsup _{k\to \infty }{\sqrt[{k}]{a_{k}}}<1}$, to prove convergence the series.

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

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

In case ${\displaystyle \limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}=1}$ 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 ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k}}}$, which is divergent. We have

${\displaystyle {\begin{aligned}\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}&=\limsup _{k\to \infty }{\sqrt[{k}]{\left|{\frac {1}{k}}\right|}}\\[0.5em]&=\limsup _{k\to \infty }{\sqrt[{k}]{\frac {1}{k}}}\\[0.5em]&=\limsup _{k\to \infty }{\frac {1}{\sqrt[{k}]{k}}}\\[0.5em]&\,{\color {OliveGreen}\left\downarrow \ \lim _{k\to \infty }{\sqrt[{k}]{k}}=1\right.}\\[0.5em]&={\frac {1}{1}}=1\end{aligned}}}$

But also the convergent series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k^{2}}}}$ satisfies this equation:

${\displaystyle {\begin{aligned}\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}&=\limsup _{k\to \infty }{\sqrt[{k}]{\left|{\frac {1}{k^{2}}}\right|}}\\[0.5em]&=\limsup _{k\to \infty }{\sqrt[{k}]{\frac {1}{k^{2}}}}\\[0.5em]&=\limsup _{k\to \infty }{\frac {1}{\sqrt[{k}]{k^{2}}}}\\[0.5em]&=\limsup _{k\to \infty }\left({\frac {1}{\sqrt[{k}]{k}}}\right)^{2}\\[0.5em]&\,{\color {OliveGreen}\left\downarrow \ \lim _{k\to \infty }{\sqrt[{k}]{k}}=1\right.}\\[0.5em]&=\left({\frac {1}{1}}\right)^{2}=1\end{aligned}}}$

These examples show that we cannot conclude convergence or divergence from ${\displaystyle \limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}=1}$. In this case we have to employ another convergence criterion!

To apply the root test on a series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ we can proceed as follows: We compute ${\displaystyle {\sqrt[{k}]{|a_{k}|}}}$ and find the limit (if the limit exists) or the limit superior.

1. If ${\displaystyle \limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}<1}$, then the series converges absolutely.
2. If ${\displaystyle \limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}|}}>1}$, then the series diverges.
3. If ${\displaystyle {\sqrt[{k}]{|a_{k}|}}\geq 1}$ for infinitely many ${\displaystyle k}$, then the series diverges.
4. Else, if none of the above is true, we cannot make a statement about convergence behaviour using the root test.

Math for Non-Geeks: Template:Aufgabe

Math for Non-Geeks: Template:Aufgabe

1. ↑ Siehe the answer to the question „Where is the root test first proved“ of Q&A website „History of Science and Mathematics“