Exercises — Direct comparison test

Goal = convergence, so we must trap our series under a finite roof. That means $b_n$ must be bigger (an upper bound) and must converge. Concretely: $\frac{1}{n^3+1}<\frac{1}{n^3}$, and $\sum 1/n^3$ is a $p$-series with $p=3>1$ → converges. Answer: bigger and convergent.

Goal = divergence, so we push up from an infinite floor: $b_n$ must be smaller (a lower bound) and divergent. Concretely $\frac{1}{\sqrt{n}}=\frac{1}{n^{1/2}}$ is itself a $p$-series with $p=\frac{1}{2}\le 1$, so it diverges outright — but as a comparison, note $\frac{1}{\sqrt{n}}\ge \frac{1}{n}$, and the smaller harmonic $\sum \frac{1}{n}$ diverges. Answer: smaller and divergent.

Yes. $\sum \frac{1}{2^n}$ is geometric with $r=\frac{1}{2}$, $|r|<1$, so it converges. The inequality provides a bigger convergent partner — the correct arrow for a convergence claim. Valid. ✅

Guess: dominant power $n^2$ → behaves like $\frac{1}{n^2}$ → converges. Since $3n^2+2>n^2$ for all $n\ge 1$, we get $0\le \frac{1}{3n^2+2}<\frac{1}{n^2}$. $\sum 1/n^2$ is a $p$-series, $p=2>1$ → converges. Bigger convergent roof ⇒ converges. ✅

Guess: $\frac{n}{n^2}=\frac{1}{n}$ → diverges. We need a smaller divergent partner. Since $n^2+1\le n^2+n^2=2n^2$ for $n\ge 1$, we have $\frac{n}{n^2+1}\ge \frac{n}{2n^2}=\frac{1}{2n}\ge 0$. $\sum \frac{1}{2n}=\frac{1}{2}\sum \frac{1}{n}$ diverges (harmonic). Smaller divergent floor ⇒ diverges. ✅ Why $2n^2$? We deliberately inflate the denominator to a clean multiple of $n^2$ so the term collapses to a harmonic-type lower bound.

$\cos n$ wiggles in $[-1,1]$, so $3\le 4+\cos n\le 5$; in particular the term is $\ge 0$. Bound the numerator by its maximum: $\frac{4+\cos n}{n^3}\le \frac{5}{n^3}$. $\sum \frac{5}{n^3}=5\sum \frac{1}{n^3}$, $p=3>1$ → converges. Bigger convergent roof ⇒ converges. ✅

The denominator is dominated by $3^n$, and $2^n+3^n>3^n$, so $0\le \frac{1}{2^n+3^n}<\frac{1}{3^n}$. $\sum {\left(\frac{1}{3}\right)}^n$ is geometric with $r=\frac{1}{3}$, $|r|<1$ → converges. Bigger convergent roof ⇒ converges. ✅

Guess: $\sim \frac{1}{n^2}$ → converges. Need a bigger convergent partner. For $n\ge 2$, $n^2-n=n^2\left(1-\frac{1}{n}\right)\ge n^2\cdot \frac{1}{2}=\frac{n^2}{2}$ (since $1-\frac{1}{n}\ge \frac{1}{2}$). Therefore $\frac{1}{n^2-n}\le \frac{2}{n^2}$. $\sum \frac{2}{n^2}$ converges ($p=2$). ⇒ converges. ✅ Why the $\frac{1}{2}$ move? Directly $\frac{1}{n^2-n}>\frac{1}{n^2}$ points the wrong way for convergence, so we manufacture a valid upper bound by controlling how much $n^2-n$ dips below $n^2$.

Guess: $\sqrt{n}\cdot (n+1)\sim n^{3/2}$ → like $\frac{1}{n^{3/2}}$ → converges ($p=\frac{3}{2}>1$). Since $n+1>n$, $\sqrt{n}(n+1)>\sqrt{n}\cdot n=n^{3/2}$, so $0\le \frac{1}{\sqrt{n}(n+1)}<\frac{1}{n^{3/2}}$. $\sum n^{-3/2}$ converges. Bigger convergent roof ⇒ converges. ✅

Guess: ratio $\to (2/3{)}^n$ → geometric-like, converges. Need a bigger convergent partner. For $n\ge 1$, $3^n-1\ge 3^n-3^{n-1}=3^{n-1}\cdot 2=\frac{2}{3}{3}^n$. Actually simpler: $3^n-1\ge \frac{1}{2}{3}^n$ for $n\ge 1$ (since $3^n\ge 2$ gives $3^n-1\ge \frac{3^n}{2}$). So $\frac{2^n}{3^n-1}\le \frac{2^n}{\frac{1}{2}{3}^n}=2{\left(\frac{2}{3}\right)}^n$. $\sum 2(2/3{)}^n$ is geometric with $r=\frac{2}{3}<1$ → converges. Bigger convergent roof ⇒ converges. ✅

Key fact: $\ln n<n$ for all $n\ge 1$ (the log grows slower than the line). So $\frac{1}{\ln n}>\frac{1}{n}\ge 0$ for $n\ge 3$. $\sum \frac{1}{n}$ diverges (harmonic). Smaller divergent floor ⇒ diverges. ✅ Why compare to $1/n$? We suspect divergence because $\ln n$ is smaller than $n$, so $\frac{1}{\ln n}$ is larger than $\frac{1}{n}$ — we sit above a known-infinite floor.

Both pieces are non-negative, so a valid single upper bound is a sum of upper bounds. Each piece converges: $\sum \frac{1}{n^2}$ ($p=2$) and $\sum \frac{1}{2^n}$ (geometric $r=\frac{1}{2}$). A sum of two convergent series converges. With DCT directly: $\frac{1}{n^2}+\frac{1}{2^n}\le \frac{1}{n^2}+\frac{1}{n^2}=\frac{2}{n^2}$ for $n\ge 1$ (since $2^n\ge n^2$ for $n\ge 4$; check small $n$ separately — finitely many terms don't matter). $\sum \frac{2}{n^2}$ converges. ⇒ converges. ✅

Since $-1\le \sin n\le 1$, the numerator $n+\sin n\ge n-1\ge 0$ for $n\ge 1$ (equality only shrinks it). For $n\ge 2$, $n+\sin n\ge n-1\ge \frac{n}{2}$. So $\frac{n+\sin n}{n^2}\ge \frac{n/2}{n^2}=\frac{1}{2n}\ge 0$. $\sum \frac{1}{2n}$ diverges (harmonic). Smaller divergent floor ⇒ diverges. ✅ Insight: the dominant term is $\frac{n}{n^2}=\frac{1}{n}$; the $\sin n$ is a bounded nuisance we absorb with a lower bound.

Guess: top $\sim 2^n$, bottom $\sim 3^n$ → like $(2/3{)}^n$ → converges. Upper bound: $1+2^n\le 2\cdot 2^n$ (since $1\le 2^n$) and $1+3^n\ge 3^n$. So $\frac{1+2^n}{1+3^n}\le \frac{2\cdot 2^n}{3^n}=2{\left(\frac{2}{3}\right)}^n.$ Geometric $r=\frac{2}{3}<1$ → converges. Bigger convergent roof ⇒ converges. ✅

Try DCT: $\frac{1}{n\ln n}<\frac{1}{n}$ — but $\frac{1}{n}$ diverges, so a smaller-than-divergent comparison is useless. And $\frac{1}{n\ln n}>\frac{1}{n^2}$ for large $n$ — but $\frac{1}{n^2}$ converges, so a bigger-than-convergent comparison is also useless. DCT is squeezed out from both sides. Correct tool — Integral test: ${\int }_2^{\infty }\frac{dx}{x\ln x}=[\ln (\ln x){]}_2^{\infty }=\infty$. The integral diverges, so the series diverges. ✅ Lesson: DCT needs a benchmark on the correct side; this series lives between the two benchmark families, so DCT can't reach it — you escalate to the integral test.

Terms change sign, so DCT does not apply directly (it needs $a_n\ge 0$). Instead compare absolute values: $\left|\frac{(-1{)}^n}{n^2}\right|=\frac{1}{n^2}$, and $\sum \frac{1}{n^2}$ converges. By DCT the series of absolute values converges, hence the original is absolutely convergent (Absolute convergence), therefore convergent. ✅ Why the detour? DCT's proof rests on monotone partial sums, which requires non-negative terms — so we apply it to $|a_n|$ instead.

Claim: converges iff $p>1$ (same threshold as the pure $p$-series).

• If $p>1$: for $n\ge 2$, $n^p-1\ge \frac{1}{2}{n}^p$ (since $n^p\ge 2$ gives $n^p-1\ge \frac{n^p}{2}$), so $\frac{1}{n^p-1}\le \frac{2}{n^p}$. $\sum \frac{2}{n^p}$ converges ($p>1$). Bigger convergent roof ⇒ converges.
• If $p\le 1$: $n^p-1<n^p$, so $\frac{1}{n^p-1}>\frac{1}{n^p}\ge 0$. $\sum \frac{1}{n^p}$ diverges ($p\le 1$). Smaller divergent floor ⇒ diverges. Both directions pinned by DCT. Answer: converges $⟺p>1$. ✅

For $x\ge 0$ we have $0\le \sin x\le x$ (the sine never overshoots its input on $[0,\infty )$ where it's non-negative; here $\frac{1}{n^2}\in (0,1]$, so $\sin (\frac{1}{n^2})\ge 0$). Thus $0\le \sin \left(\frac{1}{n^2}\right)\le \frac{1}{n^2}$. $\sum \frac{1}{n^2}$ converges. Bigger convergent roof ⇒ converges. ✅ Why $\sin x\le x$? It's the clean upper bound that converts a transcendental term into a $p$-series comparison — the whole reason DCT is powerful.

Every problem reduced to one question: which trusted friend, on which side? Convergence ⇒ trap under a bigger convergent roof (numerator up, denominator down). Divergence ⇒ push above a smaller divergent floor (numerator down, denominator up). When the target squeezes between benchmarks (logs!) or changes sign, DCT alone can't do it — reach for the Integral test, Limit comparison test, or Absolute convergence.