root test

Root test

Let ${\sum }_{k=1}^{\infty }{a}_k$ be a series.
i) if there exists some $q<1$ such that

$$\sqrt[k]{|a_k|}\le q\text{for almost all }k,$$

then ${\sum }_{k=1}^{\infty }{a}_k$ is absolutely convergent.
ii) Conversely, if

$$\sqrt[k]{|a_k|}\ge 1\text{for infinitely many }k,$$

then ${\sum }_{k=1}^{\infty }{a}_k$ is divergent.

Proof:
… by comparison test with a geometric series.

Assume there is some $k_0$ s.t. $|a_k|\le q^k$ for $k\ge k_0$.

First split the sum, then, for $k\ge k_0$, we have $|a_k|\le q^k$:

$$\begin{array}{rl}\sum _{k=1}^{\infty }|a_k|& =\sum _{k=1}^{k_0-1}|a_k|+\sum _{k=k_0}^{\infty }|a_k|\\ & \le \sum _{k=1}^{k_0-1}|a_k|+\sum _{k=k_0}^{\infty }{q}^k\end{array}$$

${\sum }_{k=1}^{k_0-1}|a_k|$ is finite, ${\sum }_{k=k_0}^{\infty }{q}^k$ is also finite, for $q\in [0,1)$, so ${\sum }_{k=1}^{\infty }|a_k|$ is finite, which means ${\sum }_{k=1}^{\infty }{a}_k$ is absolutely convergent.

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Equivalenty, via limsup, the root test says:

$$a:=\underset{k\to \infty }{\mathrm{limsup}}\sqrt[k]{|a_k|}$$

i) $a<1⟹$ absolutely convergent
ii) $a>1⟹$ divergent
iii) $a=1⟹$ inconclusive

Note: $a>0$ says nothing about the terms failing to vanish (that’d be the case if you remove the root). The root goes to $1$ for any fixed $x>0$, as $k\to \infty$.
$a=0$ … terms vanish faster than every geometric sequence, e.g. $a_k=1/k^k$ (root: $1/k\to 0$)
$0<a<1$ … terms vanish like $a^k$ (geometric decay)
$a=1$ … terms decay subexponentially (slower than every geometric sequence) or not at all
$a>1$ … terms grow geometrically along an infinite subsequence

EXAMPLE

Take $a=\frac{1}{2}$: that means $\sqrt[k]{|a_k|}\approx \frac{1}{2}$, i.e. $|a_k|\approx (\frac{1}{2}{)}^k\to 0$. Infinitely many positive terms, yes, but a sum of infinitely many positive terms is finite whenever they shrink fast enough, as the geometric series $\sum (\frac{1}{2}{)}^k=1$ shows.

The root test only detects exponential decay.

$\sum \frac{1}{k}$ diverges. $a=\sqrt[k]{1/k}=\frac{1}{k^{1/k}}\to 1$
$\sum \frac{1}{k^2}$ converges. $a=\sqrt[k]{1/k^2}=\frac{1}{(k^{1/k}{)}^2}\to 1$

It asks “do the terms behave like $q^k$ for some $q$?” Polynomial decay like $\sum k^{-c}$ for some $c>0$ converges for some $c$, but since ${\lim }_{k\to \infty }\sqrt[k]{k^{-c}}=1$ independent of $c$, the root test sees both as “no exponential decay” ($a=1$) and can’t tell them apart.

Give the definition of the root test.

Let ${\sum }_{k=1}^{\infty }{a}_k$ be a series.
i) if there exists some $q<1$ such that

$$\sqrt[k]{|a_k|}\le q\text{for almost all }k,$$

then ${\sum }_{k=1}^{\infty }{a}_k$ is absolutely convergent. Conversely, if that value is $\ge 1$ for infinitely many $k$, the series is divergent.

Equivalently: $a={\mathrm{limsup}}_{k\to \infty }\sqrt[k]{|a_k|}$
$<1⟹$ convergent,
$>1⟹$ divergent,
$=1⟹$ inconclusive.

What is the idea behind the root test?

The root test asks whether the terms behave like $q^k$, i.e. if they decay geometrically (exponentially).

Determine whether ${\sum }_{k=1}^{\infty }\sin (k)k^{42}{2}^{-k}$ converges. Since sin is [0, 1], it only shrinks the terms, so we can easily drop/split it for a comparison test:

$$\sum |\sin (x)k^{42}{2}^{-k}|\le \sum |k^{42}{2}^{-k}|$$

Apply the root test to the latter:

$$\sqrt[k]{k^{42}{2}^{-k}}=\sqrt[k]{k^{42}}\cdot \sqrt[k]{2^{-k}}=(k^{1/k}{)}^{42}\cdot 2^{-1}\to 1^{42}\cdot \frac{1}{2}=\frac{1}{2}<1$$

The series converges absolutely, hence converges.