Exercises — Limit comparison test

Let $a_n>0$ and $b_n>0$ for all large $n$, and set $L=\underset{n\to \infty }{\lim }\frac{a_n}{b_n}$.

• Main case $0<L<\infty$: $\sum a_n$ and $\sum b_n$ both converge or both diverge (same fate).
• Edge $L=0$: if $\sum b_n$ converges, then $\sum a_n$ converges. (Says nothing if $\sum b_n$ diverges.)
• Edge $L=\infty$: if $\sum b_n$ diverges, then $\sum a_n$ diverges. (Says nothing if $\sum b_n$ converges.)

Skeleton. Strongest power on top is $5n$ (behaves like $n$). Strongest on bottom is $2n^2$ (behaves like $n^2$). So $b_n=\frac{n}{n^2}=\frac{1}{n}.$ Why this $b_n$? For huge $n$ the $+3$ and $+n$ are dwarfed by the leading powers — only the biggest power in each slot decides the decay rate. Fate. $\sum 1/n$ is the Harmonic Series with $p=1$, which diverges. Since $a_n$ tracks $1/n$, the series diverges. ✔

Top $\sim n^2$, bottom $\sim n^5$, so $b_n=\frac{n^2}{n^5}=\frac{1}{n^3}.$ $\sum 1/n^3$ is a p-Series with $p=3>1$, which converges. So $\sum a_n$ converges. ✔

Pick. Top $\sim 4n^2$, bottom $\sim n^3$: $b_n=\frac{n^2}{n^3}=\frac{1}{n}$. Compute $L$. $L={\lim }_{n\to \infty }\frac{(4n^2-n)/(n^3+5)}{1/n}=\lim \frac{n(4n^2-n)}{n^3+5}=\lim \frac{4n^3-n^2}{n^3+5}.$ Divide top and bottom by $n^3$: $\frac{4-1/n}{1+5/n^3}\to 4.$ Conclude. $0<L=4<\infty$, and $\sum 1/n$ diverges (harmonic). So $\sum a_n$ diverges. ✔

Pick. Inside the root the strongest power is $4n^3$, so $\sqrt{4n^3+n}\approx \sqrt{4n^3}=2n^{3/2}$. The $2$ is a constant we can drop for the skeleton: $b_n=\frac{1}{n^{3/2}}.$ Compute $L$. $L=\lim \frac{1/\sqrt{4n^3+n}}{1/n^{3/2}}=\lim \frac{n^{3/2}}{\sqrt{4n^3+n}}.$ Factor $n^3$ out of the root: $\sqrt{4n^3+n}=n^{3/2}\sqrt{4+1/n^2}$, so $L=\lim \frac{n^{3/2}}{n^{3/2}\sqrt{4+1/n^2}}=\frac{1}{\sqrt{4}}=\frac{1}{2}.$ Conclude. $0<L=\frac{1}{2}<\infty$, and $\sum 1/n^{3/2}$ converges ($p=\frac{3}{2}>1$). So $\sum a_n$ converges. ✔

Pick. For large $n$, $4^n$ crushes $n^2$, so the denominator behaves like $4^n$, giving $b_n=\frac{3^n}{4^n}={\left(\frac{3}{4}\right)}^n,$ a Geometric Series with ratio $r=\frac{3}{4}$. Compute $L$. $L=\lim \frac{3^n/(4^n+n^2)}{3^n/4^n}=\lim \frac{4^n}{4^n+n^2}=\lim \frac{1}{1+n^2/4^n}=1,$ because $n^2/4^n\to 0$ (exponential beats polynomial). Conclude. $0<L=1<\infty$, and $\sum (3/4{)}^n$ converges ($|r|=\frac{3}{4}<1$). So $\sum a_n$ converges. ✔

Naive skeleton would say $b_n=1/n^3$. But $\ln n\to \infty$, so $\frac{(\ln n)/n^3}{1/n^3}=\ln n\to \infty$ — that's the $L=\infty$ case against a convergent $b_n$, which is inconclusive. We need a $b_n$ that is slightly bigger yet still convergent, so its extra power of $n$ can swallow the slow $\ln n$. Pick. $b_n=\frac{1}{n^{2.5}}$ ($p=2.5>1$, still convergent). Compute $L$. $L=\lim \frac{(\ln n)/n^3}{1/n^{2.5}}=\lim \frac{\ln n}{n^{0.5}}=0,$ since any positive power of $n$ eventually beats $\ln n$. Conclude. $L=0$ and $\sum 1/n^{2.5}$ converges, so by the $L=0$ rule $\sum a_n$ converges. ✔

Simplify the skeleton. Ignoring $\ln$, the fraction $\frac{n}{n^2+1}$ behaves like $\frac{1}{n}$, but there's a $\ln (n+1)$ dividing, which helps convergence a little. To get a clean verdict, compare against $b_n=1/n$ and expect $L=0$. $L=\lim \frac{n/[\ln (n+1)(n^2+1)]}{1/n}=\lim \frac{n^2}{\ln (n+1)(n^2+1)}=\lim \frac{1}{\ln (n+1)\cdot (1+1/n^2)}=0,$ because $\ln (n+1)\to \infty$. But $\sum 1/n$ diverges — so $L=0$ paired with a divergent $b_n$ is inconclusive. This is the wrong yardstick. Repair. The $\ln$ in the denominator is weak; the term is barely smaller than $1/n$. Compare against $b_n=\frac{1}{n\ln (n+1)}$ instead. A quick note on $\ln (n+1)$ vs $\ln n$: these are asymptotically equal, since $\frac{\ln (n+1)}{\ln n}=\frac{\ln n+\ln (1+1/n)}{\ln n}=1+\frac{\ln (1+1/n)}{\ln n}\to 1,$ so $\sum \frac{1}{n\ln (n+1)}$ and $\sum \frac{1}{n\ln n}$ share the same fate. The latter's fate comes from the Integral Test: $\int \frac{dx}{x\ln x}=\ln \ln x\to \infty$, so $\sum \frac{1}{n\ln n}$ diverges. $L=\lim \frac{n/[\ln (n+1)(n^2+1)]}{1/[n\ln (n+1)]}=\lim \frac{n^2}{n^2+1}=1.$ Conclude. $0<L=1<\infty$ against a divergent reference, so $\sum a_n$ diverges. ✔

Diagnosis. Factorials have no "leading power" to strip, so there's no natural $p$-series or geometric skeleton — LCT is a poor fit. This screams Ratio Test (ratios of consecutive terms kill factorials). Ratio Test. With $a_n=\frac{n!}{(2n)!}$, $\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(2n+2)!}\cdot \frac{(2n)!}{n!}=\frac{(n+1)}{(2n+2)(2n+1)}=\frac{n+1}{(2n+2)(2n+1)}.$ As $n\to \infty$ this $\to 0<1$. Converges. ✔ Lesson: LCT is for algebraic (power/root/exponential-base) terms; factorials belong to the Ratio Test.

Diagnosis. The whole thing is raised to the $n$; the base tends to $\frac{2}{5}$. This looks geometric, so build $b_n=(2/5{)}^n$ (a Geometric Series, $|r|=\frac{2}{5}<1$, converges). Compute $L$. $L=\lim \frac{{\left(\frac{2n+1}{5n+3}\right)}^n}{(2/5{)}^n}=\lim {\left(\frac{5(2n+1)}{2(5n+3)}\right)}^n=\lim {\left(\frac{10n+5}{10n+6}\right)}^n.$ Write the base as $1-\frac{1}{10n+6}$. Now use the classical exponential limit ${\left(1+\frac{u}{m}\right)}^m\to e^u$: here the "$-1$" in the numerator plays the role of $u$ and the denominator grows like $10n$, so ${\left(1-\frac{1}{10n+6}\right)}^n={\left[{\left(1-\frac{1}{10n+6}\right)}^{10n+6}\right]}^{\frac{n}{10n+6}}\to (e^{-1}{)}^{1/10}=e^{-1/10},$ a finite positive number (the inner bracket $\to e^{-1}$, and the exponent $\frac{n}{10n+6}\to \frac{1}{10}$). Conclude. $0<L=e^{-1/10}<\infty$, and $\sum (2/5{)}^n$ converges, so $\sum a_n$ converges. ✔ (The Root Test also nails this instantly — LCT works because we could name the geometric skeleton.)

Skeleton. Top $\sim n^p$, bottom $\sim n^3$, so $b_n=\frac{n^p}{n^3}=\frac{1}{n^{3-p}}.$ Compute $L$. $L=\lim \frac{n^p/(n^3+1)}{1/n^{3-p}}=\lim \frac{n^p\cdot n^{3-p}}{n^3+1}=\lim \frac{n^3}{n^3+1}=1.$ $0<L=1<\infty$ for every $p$, so $\sum a_n$ shares the fate of $\sum 1/n^{3-p}$. Boundary. This p-Series converges $⟺3-p>1⟺p<2$. Answer. $\sum a_n$ converges for $\boxed{p<2}$ and diverges for $p\ge 2$. ✔

Why LCT stalls. Every candidate $b_n=1/n^q$ gives $L=0$ (if $q<1$, divergent ref, inconclusive) or $L=\infty$ (if $q>1$, convergent ref, inconclusive) — the $(\ln n{)}^{\alpha }$ factor is sub-polynomial, so no power skeleton sits at the same rate. LCT can't resolve logarithmic borderlines. Right tool: Integral Test. With $f(x)=\frac{1}{x(\ln x{)}^{\alpha }}$, substitute $u=\ln x$, $du=dx/x$: ${\int }_2^{\infty }\frac{dx}{x(\ln x{)}^{\alpha }}={\int }_{\ln 2}^{\infty }\frac{du}{u^{\alpha }}.$ This is a $p$-integral in $u$: it converges $⟺\alpha >1$. Answer. $\sum a_n$ converges for $\boxed{\alpha >1}$, diverges for $0<\alpha \le 1$. ✔ Takeaway: LCT ranks series that decay like a power; logarithmic fine-structure needs the Integral Test.

When is LCT the wrong tool entirely? ::: When terms are factorials or $n$-th powers (use Ratio Test/Root Test), or when the decay differs from a power only by a logarithm (use the Integral Test).