Exercises — Ratio test — proof, limitations

$\frac{a_{n+1}}{a_n}=\frac{3^{n+1}/(n+1)!}{3^n/n!}=\frac{3^{n+1}}{3^n}\cdot \frac{n!}{(n+1)!}=\frac{3}{n+1}.$ WHAT: we divided term $n+1$ by term $n$. WHY these cancel: $3^{n+1}/3^n=3$ and $n!/(n+1)!=1/(n+1)$ — factorials telescope. $L={\lim }_{n\to \infty }\frac{3}{n+1}=0<1\Rightarrow \text{converges absolutely}.$

$\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{5^{n+1}}\cdot \frac{5^n}{n!}=\frac{(n+1)!}{n!}\cdot \frac{5^n}{5^{n+1}}=\frac{n+1}{5}.$ $L={\lim }_{n\to \infty }\frac{n+1}{5}=\infty >1\Rightarrow \text{diverges}.$ Factorials always eventually crush a fixed exponential.

$\frac{a_{n+1}}{a_n}=\frac{1/(n+1{)}^2}{1/n^2}=\frac{n^2}{(n+1{)}^2}\to 1.$ $L=1\Rightarrow \text{inconclusive}.$ The series does converge (it is a $p$-series with $p=2$), but the ratio test cannot see it — power-law terms always give $L=1$.

$\frac{a_{n+1}}{a_n}=\frac{(n+1{)}^2}{2^{n+1}}\cdot \frac{2^n}{n^2}={\left(\frac{n+1}{n}\right)}^2\cdot \frac{1}{2}.$ WHY: the $2^n$ gives $1/2$; the polynomial part is a ratio of squares. $L={\lim }_{n\to \infty }\frac{1}{2}{\left(1+\frac{1}{n}\right)}^2=\frac{1}{2}<1\Rightarrow \text{converges absolutely}.$

$\frac{a_{n+1}}{a_n}=\frac{(2n+2)!}{((n+1)!{)}^2}\cdot \frac{(n!{)}^2}{(2n)!}.$ Group the two factorial families. For the top: $\frac{(2n+2)!}{(2n)!}=(2n+2)(2n+1)$. For the bottom: $\frac{((n+1)!{)}^2}{(n!{)}^2}=(n+1{)}^2$. So $\frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)}{(n+1{)}^2}=\frac{2(n+1)(2n+1)}{(n+1{)}^2}=\frac{2(2n+1)}{n+1}.$ $L={\lim }_{n\to \infty }\frac{4n+2}{n+1}=4>1\Rightarrow \text{diverges}.$

$\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1{)}^{n+1}}\cdot \frac{n^n}{n!}=(n+1)\cdot \frac{n^n}{(n+1{)}^{n+1}}=\frac{n^n}{(n+1{)}^n}={\left(\frac{n}{n+1}\right)}^n.$ WHY this limit: rewrite as ${\left(1+\frac{1}{n}\right)}^{-n}$. The classic limit ${\left(1+\frac{1}{n}\right)}^n\to e$ gives $L=\frac{1}{e}\approx 0.3679<1\Rightarrow \text{converges absolutely}.$

$\left|\frac{a_{n+1}}{a_n}\right|=\frac{|x{|}^{n+1}}{(n+1{)}^2}\cdot \frac{n^2}{|x{|}^n}=|x|{\left(\frac{n}{n+1}\right)}^2\to |x|.$ So $L=|x|$. Ratio test: converges for $|x|<1$, diverges for $|x|>1$ — the radius of convergence is $R=1$. Endpoints ($L=1$, ratio test blind):

• $x=1$: $\sum \frac{1}{n^2}$ — a convergent $p$-series ($p=2$). Converges.
• $x=-1$: $\sum \frac{(-1{)}^n}{n^2}$ — converges absolutely (since $\sum 1/n^2$ does). Converges. Answer: converges for all $x\in [-1,1]$.

Using Exercise 2.3's ratio for the $\frac{n!}{n^n}$ part and attaching $|x|$: $\left|\frac{a_{n+1}}{a_n}\right|=|x|{\left(\frac{n}{n+1}\right)}^n\to \frac{|x|}{e}.$ $L=\frac{|x|}{e}<1⟺|x|<e$. So radius $R=e$: converges for $|x|<e$, diverges for $|x|>e$. Endpoint $|x|=e$ gives $L=1$ (inconclusive); a finer tool (Raabe's test or Stirling) is needed — the ratio test has done its job of pinning the radius.

The whole term is raised to the $n$-th power, so the root test cuts through it: $\sqrt[n]{a_n}=\frac{n}{2n+1}\to \frac{1}{2}<1\Rightarrow \text{converges}.$ The ratio test works too but the ratio ${\left(\frac{n+1}{2n+3}\right)}^{n+1}/{\left(\frac{n}{2n+1}\right)}^n$ is painful — recognising an "everything-to-the-$n$" form and reaching for the root test is the analysis skill here.

$\frac{a_{n+1}}{a_n}=\frac{2^{n+1}+(n+1)}{3^{n+1}+(n+1{)}^2}\cdot \frac{3^n+n^2}{2^n+n}.$ Dominant-term thinking: for large $n$, $2^n+n\approx 2^n$ and $3^n+n^2\approx 3^n$. So $\frac{a_{n+1}}{a_n}\approx \frac{2^{n+1}}{3^{n+1}}\cdot \frac{3^n}{2^n}=\frac{2}{3}.$ Rigorously, divide top/bottom by the leading power: the limit is $L=\frac{2}{3}<1\Rightarrow$ converges absolutely. (Alternatively compare with $\sum (2/3{)}^n$ — same conclusion, and it confirms the ratio limit.)

$\frac{a_{n+1}}{a_n}=\frac{n\ln n}{(n+1)\ln (n+1)}.$ Both $\frac{n}{n+1}\to 1$ and $\frac{\ln n}{\ln (n+1)}\to 1$, so $L=1$ — inconclusive (as expected for a power/log term). Switch to the integral test: ${\int }_2^{\infty }\frac{dx}{x\ln x}$. Substitute $u=\ln x$, $du=\frac{dx}{x}$: $\int \frac{du}{u}=\ln u=\ln (\ln x){|}_2^{\infty }=\infty .$ The integral diverges $\Rightarrow$ the series diverges. The ratio test could never have found this — logs make $L=1$.

Pick $r=\frac{3}{4}$ (any number in $(\frac{1}{2},1)$). Since the ratio $\to \frac{1}{2}<r$, there is an $N$ with $|a_{n+1}|<r|a_n|$ for all $n\ge N$. Chaining (as in the parent proof): $|a_{N+k}|<r^k|a_N|.$ As $k\to \infty$, $r^k\to 0$ because $0<r<1$, so $|a_{N+k}|\to 0$, hence $a_n\to 0$. $■$ This is exactly why $L<1$ passes the term test automatically — the geometric bound forces the terms to zero.

Any two $p$-series with $p\le 1$ vs $p>1$ work. Take:

• Divergent: $\sum \frac{1}{\sqrt{n}}$ ($p=\frac{1}{2}\le 1$). Ratio $=\sqrt{\frac{n}{n+1}}\to 1$, so $L=1$; diverges by the $p$-test.
• Convergent: $\sum \frac{1}{n^3}$ ($p=3>1$). Ratio $={\left(\frac{n}{n+1}\right)}^3\to 1$, so $L=1$; converges by the $p$-test. Both sit at $L=1$ yet split — the boundary is genuinely undecidable by ratio alone. See the figure below for why every power law flattens to $1$.

From Exercise 2.2, $\frac{(2n)!}{(n!{)}^2}$ has ratio $\to 4$, so its reciprocal $\frac{(n!{)}^2}{(2n)!}$ has ratio $\to \frac{1}{4}$. Attaching $x$: $\left|\frac{a_{n+1}}{a_n}\right|=x\cdot \frac{(n+1{)}^2}{(2n+2)(2n+1)}=x\cdot \frac{n+1}{2(2n+1)}\to \frac{x}{4}.$ $L=\frac{x}{4}<1⟺x<4$. Radius $R=4$: converges for $0<x<4$, diverges for $x>4$. Endpoint $x=4$: $L=1$, inconclusive — Stirling shows the term behaves like $\frac{1}{\sqrt{\pi n}}$, so $\sum$ diverges there. Ratio test correctly locates $R=4$.

FALSE. The condition "$<1$ for every $n$" is weaker than "$L<1$": the ratio can approach $1$ from below without any fixed $r<1$ trapping it. Counterexample: the harmonic series $\sum \frac{1}{n}$. Here $\frac{a_{n+1}}{a_n}=\frac{n}{n+1}<1$ for every $n$, yet the series diverges ($L=1$). The moral: the ratio test needs the limit $L$ to be strictly below $1$, giving a fixed gap $[L,1)$ for geometric domination. A ratio merely creeping up toward $1$ leaves no such gap.