Exercises — Root test

The whole term is raised to the $n$-th power $(\cdots {)}^n$. That is the trigger for the root test: the $\sqrt[n]{\cdot }$ will cancel the outer power exactly. $\sqrt[n]{{\left(\frac{n}{2n+1}\right)}^n}=\frac{n}{2n+1}.$ What we did / why: taking the $n$-th root undoes "raise to the $n$-th," leaving the base. Now divide top and bottom by $n$: $\frac{n}{2n+1}=\frac{1}{2+1/n}\stackrel{n\to \infty }{\to }\frac{1}{2}.$ So $L=\frac{1}{2}<1$ $\Rightarrow$ converges absolutely.

Root each factor separately: $\sqrt[n]{\frac{5^n}{n^4}}=\frac{\sqrt[n]{5^n}}{\sqrt[n]{n^4}}=\frac{5}{(\sqrt[n]{n}{)}^4}.$ Why this step: $\sqrt[n]{5^n}=5$ (an $n$-th root of an $n$-th power), and $\sqrt[n]{n^4}=(n^{1/n}{)}^4$. Since $\sqrt[n]{n}\to 1$, the denominator $\to 1^4=1$, so $L=\frac{5}{1}=5>1\Rightarrow \text{diverges}.$ The exponential $5^n$ on top beats the polynomial $n^4$ — the polynomial factor is invisible to the root test.

Perfect $n$-th power in the denominator: $\sqrt[n]{\frac{1}{(\arctan n{)}^n}}=\frac{1}{\arctan n}.$ Why this step: $\sqrt[n]{(\arctan n{)}^n}=\arctan n$ because the base is fixed relative to the exponent. As $n\to \infty$, $\arctan n\to \frac{\pi }{2}\approx 1.5708$, so $L=\frac{1}{\pi /2}=\frac{2}{\pi }\approx 0.6366<1\Rightarrow \text{converges}.$

$\sqrt[n]{\frac{n^{10}}{3^n}}=\frac{(\sqrt[n]{n}{)}^{10}}{3}\to \frac{1^{10}}{3}=\frac{1}{3}<1.$ What/why: split the root over top and bottom; the top is $(n^{1/n}{)}^{10}\to 1$, the bottom is a clean $3$. $L=\frac{1}{3}$ $\Rightarrow$ converges.

The exponent is $n^2$, not $n$ — but the root test still shines because $\sqrt[n]{(\cdots {)}^{n^2}}=(\cdots {)}^n$: $\sqrt[n]{{\left(1-\frac{1}{n}\right)}^{n^2}}={\left(1-\frac{1}{n}\right)}^n.$ Why this step: $\sqrt[n]{x^{n^2}}=x^{n^2/n}=x^n$. Now use the famous limit ${\left(1-\frac{1}{n}\right)}^n\to e^{-1}.$ Why this limit: ${\left(1+\frac{c}{n}\right)}^n\to e^c$; here $c=-1$. So $L=e^{-1}=\frac{1}{e}\approx 0.3679<1\Rightarrow \text{converges}.$

$\sqrt[n]{\frac{2^n{n}^3}{5^n}}=\frac{2(\sqrt[n]{n}{)}^3}{5}\to \frac{2\cdot 1}{5}=\frac{2}{5}<1.$ What/why: the two exponentials root to their bases $2$ and $5$; the polynomial $(n^{1/n}{)}^3\to 1$. $L=\frac{2}{5}$ $\Rightarrow$ converges.

$\sqrt[n]{{\left(\frac{3n^2+1}{n^2+4}\right)}^n}=\frac{3n^2+1}{n^2+4}\to 3.$ Why this step: divide top and bottom by $n^2$: $\frac{3+1/n^2}{1+4/n^2}\to 3$. Since $L=3>1$ $\Rightarrow$ diverges.

$\sqrt[n]{\frac{1}{n(\ln n{)}^2}}=\frac{1}{n^{1/n}(\ln n{)}^{2/n}}.$ Why each piece: $n^{1/n}\to 1$ (standard limit), and $(\ln n{)}^{2/n}=e^{\frac{2\ln \ln n}{n}}\to e^0=1$ because $\frac{\ln \ln n}{n}\to 0$ (even $\ln \ln n$ grows far slower than $n$). Hence $L=\frac{1}{1\cdot 1}=1\Rightarrow \text{inconclusive}.$ What next: switch tools — the integral test shows this series converges, but the root test simply cannot see it.

$\sqrt[n]{\frac{n!}{n^n}}=\frac{(n!{)}^{1/n}}{n}.$ Why this is subtle: the factorial $n!$ is not a clean power. Use the known asymptotic $(n!{)}^{1/n}\sim \frac{n}{e}$ (from Stirling: $n!\sim \sqrt{2\pi n}(n/e{)}^n$, so $(n!{)}^{1/n}\sim n/e$). Then $L=\frac{(n!{)}^{1/n}}{n}\to \frac{n/e}{n}=\frac{1}{e}\approx 0.3679<1\Rightarrow \text{converges}.$ Comment: the Ratio test here is even cleaner — $\frac{a_{n+1}}{a_n}={\left(\frac{n}{n+1}\right)}^n\to e^{-1}$, same $L$. This is why the two tests are cousins: for many series they hand you the same number.

$\sqrt[n]{\frac{1}{n^{1+1/n}}}=n^{-(1+1/n)/n}=e^{-\frac{(1+1/n)\ln n}{n}}.$ Why the exponential form: to take a limit of a power with a moving exponent, rewrite $x^y=e^{y\ln x}$. The exponent is $-\frac{(1+1/n)\ln n}{n}\to -1\cdot 0=0,$ because $\frac{\ln n}{n}\to 0$. So $L=e^0=1\Rightarrow$ inconclusive. (Comparison shows it actually diverges, close to $\sum 1/n$.)

Here $a_n=\frac{x^n}{n{2}^n}$. Take the root: $\sqrt[n]{|a_n|}=\frac{|x|}{2(\sqrt[n]{n})}\to \frac{|x|}{2}.$ Why: $|x{|}^n$ roots to $|x|$, $2^n$ roots to $2$, and $\sqrt[n]{n}\to 1$. Convergence requires $L<1$: $\frac{|x|}{2}<1⟺|x|<2.$ So the radius of convergence is $R=2$ (compare Radius of convergence, Cauchy–Hadamard). Check endpoints separately (root test is silent there since $L=1$):

• $x=2$: $\sum \frac{2^n}{n{2}^n}=\sum \frac{1}{n}$ diverges.
• $x=-2$: $\sum \frac{(-2{)}^n}{n{2}^n}=\sum \frac{(-1{)}^n}{n}$ converges conditionally (alternating harmonic — the alternating series test gives convergence, but $\sum \frac{1}{n}$ shows it is not absolutely convergent, which is precisely why the root test, an absolute-convergence test, stays silent here).

Interval of convergence: $[-2,2)$.

The exponent $n^2$ makes the root test natural (it turns $n^2$ into $n$): $\sqrt[n]{{\left(\frac{n}{n+1}\right)}^{n^2}}={\left(\frac{n}{n+1}\right)}^n={\left(1-\frac{1}{n+1}\right)}^n.$ Why: $\sqrt[n]{x^{n^2}}=x^n$. As $n\to \infty$, ${\left(1-\frac{1}{n+1}\right)}^n\to e^{-1}$ (same ${\left(1+\frac{c}{n}\right)}^n\to e^c$ idea, $c=-1$). So $L=e^{-1}=\frac{1}{e}\approx 0.3679<1\Rightarrow \text{converges}.$ The ratio test would force you to simplify ${\left(\frac{n+1}{n+2}\right)}^{(n+1{)}^2}/{\left(\frac{n}{n+1}\right)}^{n^2}$ — algebra hell. Root wins.

$\sqrt[n]{{\left(\frac{pn}{n+1}\right)}^n}=\frac{pn}{n+1}\to p.$ Why: $\frac{pn}{n+1}=\frac{p}{1+1/n}\to p$. So $L=p$.

• $p<1$: $L<1$ $\Rightarrow$ converges.
• $p>1$: $L>1$ $\Rightarrow$ diverges.
• $p=1$: $L=1$, root test inconclusive. Then $a_n={\left(\frac{n}{n+1}\right)}^n\to e^{-1}\ne 0$, so by the n-th term divergence test it diverges.

Answer: converges iff $0<p<1$.

Compute the two subsequential values of $\sqrt[n]{a_n}$:

• even $n$: $\sqrt[n]{2^{-n}}=\frac{1}{2}$,
• odd $n$: $\sqrt[n]{3^{-n}}=\frac{1}{3}$.

Why limsup: the sequence $\sqrt[n]{a_n}$ oscillates between $\frac{1}{2}$ and $\frac{1}{3}$, so no single limit exists. The $\mathrm{limsup}$ is the largest subsequential limit: $L={\mathrm{limsup}}_{n\to \infty }\sqrt[n]{a_n}=\max \left(\frac{1}{2},\frac{1}{3}\right)=\frac{1}{2}<1.$ $L=\frac{1}{2}<1\Rightarrow$ converges absolutely. (This is exactly why the parent note defines $L$ with $\mathrm{limsup}$ — so the test always has a value even when the plain limit fails.)

For large $n$ the denominator is dominated by $3^n$ (the $n^3$ is negligible). Root it: $\sqrt[n]{\frac{2^n}{3^n+n^3}}=\frac{2}{\sqrt[n]{3^n+n^3}}.$ Why this step: factor the largest term out of the denominator: $3^n+n^3=3^n\left(1+\frac{n^3}{3^n}\right)$, so $\sqrt[n]{3^n+n^3}=3{\left(1+\frac{n^3}{3^n}\right)}^{1/n}.$ Why the bracket's root $\to 1$: write ${\epsilon }_n=\frac{n^3}{3^n}$, which $\to 0$ (exponential beats polynomial). Then ${\left(1+{\epsilon }_n\right)}^{1/n}=e^{\frac{\ln (1+{\epsilon }_n)}{n}},$ and the exponent $\frac{\ln (1+{\epsilon }_n)}{n}\to \frac{\ln (1+0)}{\infty }=\frac{0}{\infty }=0$, so the bracket's root $\to e^0=1$. Therefore $L=\frac{2}{3\cdot 1}=\frac{2}{3}<1\Rightarrow \text{converges}.$

Step 1 — the root of the square. For each $n$, $\sqrt[n]{|a_n{|}^2}={\left(\sqrt[n]{|a_n|}\right)}^2.$ Why: $|a_n{|}^2=(|a_n|{)}^2$, and taking the $n$-th root commutes with squaring: $(x{)}^{2/n}=(x^{1/n}{)}^2$. Step 2 — pass the limit through the square. Squaring is continuous, so ${\lim }_n{\left(\sqrt[n]{|a_n|}\right)}^2=({\lim }_n\sqrt[n]{|a_n|}{)}^2=L^2.$ Step 3 — conclusion. If $L<1$ then $L^2<1$ too (squaring a number in $[0,1)$ keeps it in $[0,1)$). By the root test applied to $\sum a_n^2$, that series also converges absolutely. So absolute convergence via the root test is inherited by the squared series. $■$

Root cancels the $n$-th power ::: yes — $\sqrt[n]{(\cdots {)}^n}=(\cdots )$ Polynomial factor $n^k$ roots to ::: $1$ $L=1$ means ::: inconclusive, switch tools Endpoints of a power-series interval need ::: separate manual testing When plain limit fails, use ::: $\mathrm{limsup}$ (largest subsequential limit)