Exercises — Divergence test (necessary but not sufficient)

What we do: compute $L=\underset{n\to \infty }{\lim }\frac{n}{n+1}$. Why: the test's entire input is this one limit. Divide top and bottom by $n$ (the highest power) — why: it isolates the leading behaviour and sends the small $\frac{1}{n}$ pieces to $0$: $\frac{n}{n+1}=\frac{1}{1+\frac{1}{n}}⟶\frac{1}{1+0}=1.$ Since $L=1\ne 0$, the test fires. Answer: the series diverges by the Divergence Test.

What we do: $L=\underset{n\to \infty }{\lim }\frac{5}{n^2}$. As $n\to \infty$, the denominator explodes, so $\frac{5}{n^2}\to 0$, giving $L=0$. Why this matters: $L=0$ is exactly the case where the test goes silent. Answer: inconclusive — the Divergence Test says nothing. (A different tool — e.g. the p-Series Test with $p=2>1$ — would show it converges, but that is not this test's job.)

What we do: look at the terms $\cos (\pi n)$ for $n=1,2,3,\dots$ $\cos \pi =-1,\cos 2\pi =1,\cos 3\pi =-1,\dots$ The terms bounce $-1,+1,-1,+1,\dots$ forever and never settle on one number. Why this decides it: if the terms don't approach a single value, then $\lim a_n$ Does Not Exist (DNE). For the test, DNE is treated exactly like "$\ne 0$." Answer: diverges (limit DNE $\ne 0$). See Limits of Sequences for why an oscillating sequence has no limit.

What we do: same-degree rational function → divide top and bottom by $n^3$ (highest power). Why $n^3$: it makes every lower-power term become a $\frac{1}{n},\frac{1}{n^2},\dots$ that dies at infinity, leaving only the leading coefficients. $\frac{4n^3-7n}{9n^3+2n^2+1}=\frac{4-\frac{7}{n^2}}{9+\frac{2}{n}+\frac{1}{n^3}}⟶\frac{4-0}{9+0+0}=\frac{4}{9}.$ $L=\frac{4}{9}\ne 0$. Answer: diverges.

What we do: recognise a standard exponential limit. The template is ${\left(1+\frac{x}{n}\right)}^n\to e^x$. Why this template: the base $\to 1$ while the power $\to \infty$ — a $1^{\infty }$ tug-of-war that resolves to an exponential, not to $1$. Here $x=3$: ${\lim }_{n\to \infty }{\left(1+\frac{3}{n}\right)}^n=e^3\approx \mathrm{20.09.}$ $L=e^3\ne 0$. Answer: diverges.

What we do: $L=\underset{n\to \infty }{\lim }\frac{\ln n}{n}$. Both parts grow, but $n$ grows much faster than $\ln n$, so the ratio $\to 0$. $\frac{\ln n}{n}⟶0.$ $L=0$. Why this is the point: the terms genuinely vanish, so the test is silent. Answer: inconclusive (the Integral Test later shows this one actually diverges — a beautiful reminder that "$L=0$" hides both fates).

What we do: take both limits. $\frac{1}{\sqrt{n}}\to 0,\frac{1}{n^2}\to 0.$ Both give $L=0$, so for both the Divergence Test is inconclusive. Why this pairing is instructive: these two have opposite true fates — by the p-Series Test, (a) $\sum n^{-1/2}$ diverges ($p=\frac{1}{2}\le 1$) while (b) $\sum n^{-2}$ converges ($p=2>1$). Same "$L=0$", opposite outcomes. That single fact is why the test can never be a convergence detector. Answer: inconclusive for both.

What we do: examine ${\lim }_{n\to \infty }{r}^n$ by cases on $|r|$.

• $|r|>1$: $|r{|}^n\to \infty$, so $r^n\nrightarrow 0$. Test fires → diverges.
• $r=1$: terms are $1,1,1,\cdots \to 1\ne 0$. Test fires → diverges.
• $r=-1$: terms $1,-1,1,\dots$ → limit DNE $\ne 0$. Test fires → diverges.
• $|r|<1$: $r^n\to 0$, so $L=0$. Test is silent → inconclusive. Answer: the Divergence Test proves divergence for all $\boxed{|r|\ge 1}$. For $|r|<1$ it is inconclusive (the geometric-series formula, a stronger tool, then shows convergence to $\frac{1}{1-r}$). Why the boundary lands at $1$: $|r|<1$ is precisely the regime where repeated multiplication shrinks a number to nothing; at $|r|=1$ shrinking stops.

(i) Convergence by definition (from Partial Sums and Series Convergence): a series is the limit of its partial sums. ${\lim }_{N\to \infty }{S}_N={\lim }_{N\to \infty }\frac{3N}{N+2}=\frac{3}{1+0}=3.$ The partial sums settle to $3$, so the series converges to $3$. (ii) Recover the term using the Telescoping Series trick $a_N=S_N-S_{N-1}$: $a_N=\frac{3N}{N+2}-\frac{3(N-1)}{N+1}=\frac{3N(N+1)-3(N-1)(N+2)}{(N+2)(N+1)}.$ Expand the numerator: $3(N^2+N)-3(N^2+N-2)=6$. So $a_N=\frac{6}{(N+1)(N+2)}⟶0.$ $L=0$, so the Divergence Test is inconclusive. The lesson: even though we know it converges (from the partial sums), the Divergence Test alone would only say "I don't know." Consistent with the theory — $L=0$ is necessary here and indeed holds, but it never certifies convergence.

What we do: show the terms don't go to $0$, in fact blow up. Write the term as a product of $n$ fractions: $\frac{n^n}{n!}=\underset{=1}{\underbrace{\frac{n}{n}}}\cdot \underset{\ge 1}{\underbrace{\frac{n}{n-1}}}\cdot \underset{\ge 1}{\underbrace{\frac{n}{n-2}}}\cdots \underset{=n}{\underbrace{\frac{n}{1}}}.$ Every factor is $\ge 1$ and the last is $n$, so $\frac{n^n}{n!}\ge n\to \infty$. Why this bound: we only need to know the terms fail to vanish; a clean lower bound that itself $\to \infty$ is enough. Hence $L=\infty \ne 0$. Answer: diverges by the Divergence Test.

First series, $\sin (1/n)$: as $n\to \infty$, $\frac{1}{n}\to 0$, and $\sin (\text{small})\to \sin 0=0$. So $L=0$ → inconclusive. (Because $\sin (1/n)\approx \frac{1}{n}$ for large $n$, a comparison later shows it actually diverges like the harmonic series — but the Divergence Test cannot see that.) Second series, $\sin (n)$: here $n$ is in radians and marches around the circle without ever settling. The values $\sin (n)$ are dense in $[-1,1]$ and do not approach any single number → $\lim$ DNE $\ne 0$. Answer: first is inconclusive; second diverges (limit DNE).

Proof (mirrors the parent derivation): Let $S_N={\sum }_{n=1}^N{a}_n$. "Converges to $S$" means ${\lim }_{N\to \infty }{S}_N=S$ (finite). Use the identity $a_N=S_N-S_{N-1}$. Shifting the index by one does not change a limit, so $S_{N-1}\to S$ too. Since both limits exist and are finite, the limit of the difference is the difference of the limits: ${\lim }_{N\to \infty }{a}_N={\lim }_{N\to \infty }(S_N-S_{N-1})=S-S=0.■$ Why the reverse fails (one sentence): the argument runs one direction only — the Harmonic Series $\sum \frac{1}{n}$ has $a_n\to 0$ yet its partial sums $S_N\to \infty$, so "$a_n\to 0$" cannot force convergence.

What we know: since $\sum a_n$ converges, by L5·Q2 we must have $a_n\to 0$. Careful — the test only fires on a nonzero/DNE limit of $(a_n+b_n)$. We cannot always conclude via the Divergence Test alone, because $b_n$ might itself go to $0$ (a divergent series can have vanishing terms, like the harmonic one). Then $a_n+b_n\to 0$ and the test is silent.

• If $b_n\nrightarrow 0$: then $a_n+b_n\to 0+(\text{nonzero or DNE})\ne 0$, so the Divergence Test proves $\sum (a_n+b_n)$ diverges.
• If $b_n\to 0$ (still divergent, e.g. harmonic): $a_n+b_n\to 0$, Divergence Test inconclusive — yet the sum still diverges, provable by linearity (convergent + divergent = divergent), a separate argument. Answer: the sum always diverges, but the Divergence Test can prove it only when $b_n\nrightarrow 0$. When $b_n\to 0$ you need the linearity theorem instead.