Exercises — Alternating series test — Leibniz test, proof

(a) Signs run $+,-,+,-,\dots$ so yes, alternating. Here $a_n=\frac{(-1{)}^{n-1}}{n^2}$ and $b_n=|a_n|=\frac{1}{n^2}>0$. ✓

(b) The trick: $\cos (n\pi )$ equals $-1$ when $n$ is odd and $+1$ when $n$ is even, i.e. $\cos (n\pi )=(-1{)}^n$. So the term is $a_n=\frac{(-1{)}^n}{\sqrt{n}}$, which is alternating, with $b_n=|a_n|=\frac{1}{\sqrt{n}}$ (the leading sign is negative, which is fine — the test only cares that signs alternate). ✓

(c) All terms are positive — the sign never flips. Not alternating. It converges (a $p$-series with $p=2$), but not by Leibniz.

$b_n=\frac{1}{n^2}$.

• Decreasing? Bigger $n$ ⟹ bigger $n^2$ ⟹ smaller $\frac{1}{n^2}$. Looks decreasing. ✓
• Limit 0? As $n\to \infty$, $\frac{1}{n^2}\to 0$. ✓ Both hold, so Leibniz will apply (we verify rigorously at L2).

$b_n=\frac{1}{n^2}$.

• Condition 1 (decreasing): compare $b_{n+1}$ and $b_n$. $\frac{1}{(n+1{)}^2}<\frac{1}{n^2}$ because $(n+1{)}^2>n^2$ for $n\ge 1$. So $b_{n+1}\le b_n$. ✓
• Condition 2 (limit 0): $\underset{n\to \infty }{\lim }\frac{1}{n^2}=0$. ✓

Both hold ⟹ the series converges by the Leibniz test. $■$

$b_n=\frac{1}{\sqrt{n}}>0$ and the signs alternate.

• Decreasing: $\sqrt{n+1}>\sqrt{n}$ ⟹ $\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}$. ✓
• Limit 0: $\frac{1}{\sqrt{n}}\to 0$. ✓

Converges by Leibniz. (Aside: $\sum 1/\sqrt{n}$ is a $p$-series with $p=\frac{1}{2}\le 1$, so it diverges — hence this is only conditionally convergent, see Absolute vs Conditional Convergence.)

$b_n=\frac{n}{2n+1}$. Check condition 2 first (it's the cheap kill-switch): ${\lim }_{n\to \infty }\frac{n}{2n+1}={\lim }_{n\to \infty }\frac{1}{2+\frac{1}{n}}=\frac{1}{2}\ne 0.$ Terms do not vanish, so by the nth-Term Test for Divergence the series diverges. Leibniz never even gets to apply — condition 2 is a prerequisite, not an option.

$b_n=\frac{\ln n}{n}$. Direct comparison of $b_{n+1}$ vs $b_n$ is messy, so we use the tool built for "is this shape going down?": treat $b_n$ as a continuous function $f(x)=\frac{\ln x}{x}$ and check its derivative (the sign of $f^{\prime }$ tells us if the graph rises or falls — that is exactly what "decreasing" means). $f^{\prime }(x)=\frac{\frac{1}{x}\cdot x-\ln x\cdot 1}{x^2}=\frac{1-\ln x}{x^2}.$ The denominator $x^2>0$, so the sign is the sign of $1-\ln x$. This is negative when $\ln x>1$, i.e. $x>e\approx 2.718$. So $f$ is decreasing for $x\ge 3$ — the sequence $b_n$ is eventually decreasing (from $n=3$ on). Leibniz only needs "for all large $n$," so ✓.

• Limit 0: $\frac{\ln n}{n}\to 0$ (logarithm crawls; $n$ sprints). ✓

Both conditions hold eventually ⟹ converges.

No. The sequence goes to $0$ but is not monotone decreasing: e.g. $b_2=\frac{1}{3}<b_3=\frac{1}{2}$, a step up — and this happens infinitely often, not just at the start, so it is not "eventually decreasing" either. The proof's bracket argument requires $b_{n+1}\le b_n$ so that each $(b_{2k-1}-b_{2k})\ge 0$; without it the brackets can be negative and the "increasing, bounded" squeeze collapses.

So the Leibniz test does not apply — it can neither confirm nor deny convergence here. (You'd need a different tool.) The lesson: monotonicity is not optional.

The conditions hold ($b_n=1/n^3$ decreasing, $\to 0$), so the remainder bound applies: $|s-s_N|\le b_{N+1}=\frac{1}{(N+1{)}^3}.$ We want this $<0.001=\frac{1}{1000}$: $\frac{1}{(N+1{)}^3}<\frac{1}{1000}⟺(N+1{)}^3>1000⟺N+1>10⟺N>9.$ So $N=10$ terms suffice. (Compare with the alternating harmonic series needing $N=200$ for $0.005$ — the extra powers of $n$ make $1/n^3$ shrink fast.)

Two-step routine (see Absolute vs Conditional Convergence):

1. Does it converge at all? By Leibniz (L2.2), $b_n=1/\sqrt{n}$ is decreasing $\to 0$ ⟹ converges.
2. Does it converge absolutely? Look at $\sum |a_n|=\sum \frac{1}{\sqrt{n}}$. This is a $p$-series with $p=\frac{1}{2}\le 1$ ⟹ diverges.

Converges but not absolutely ⟹ conditionally convergent. The alternation is what rescued it.

$b_n=\frac{1}{n^p}$ with $p>0$.

• Convergence (Leibniz): for any $p>0$, $n^p$ increases ⟹ $b_n$ decreases, and $b_n=n^{-p}\to 0$. Both conditions hold ⟹ converges for all $p>0$.
• Absolute convergence: $\sum \frac{1}{n^p}$ is a $p$-series, which converges iff $p>1$.

Summary:

• $0<p\le 1$: converges conditionally (e.g. $p=1$ is the alternating harmonic series).
• $p>1$: converges absolutely.

Every $p>0$ converges; the line $p=1$ separates conditional from absolute.

Substitute each endpoint and read off the resulting number series.

• At $x=1$: $\sum \frac{1^n}{n}=\sum \frac{1}{n}$ — the harmonic series, which diverges.
• At $x=-1$: $\sum \frac{(-1{)}^n}{n}$ — the alternating harmonic series (up to overall sign). Here $b_n=1/n$ decreases $\to 0$, so by Leibniz it converges (conditionally).

So the interval of convergence is $[-1,1)$: closed at $-1$ (Leibniz saves it), open at $+1$ (harmonic diverges). This is the classic reason Leibniz appears in power-series endpoint analysis.

The flaw: $b_n$ must be a positive, decreasing sequence — but here the proposed $b_n=\frac{1}{n}+\frac{(-1{)}^{n-1}}{n^2}$ itself contains a hidden sign flip $(-1{)}^{n-1}$, so it is not monotone (it wiggles). Leibniz's hypotheses are violated; you cannot cite the test.

But does it converge? Split the series (legal because we'll show both pieces converge): $\sum (-1{)}^{n-1}\frac{1}{n}+\sum (-1{)}^{n-1}\cdot \frac{(-1{)}^{n-1}}{n^2}=\sum \frac{(-1{)}^{n-1}}{n}+\sum \frac{1}{n^2}.$ The first is the alternating harmonic series (converges, to $\ln 2$); the second is $\sum \frac{1}{n^2}$ (converges, a $p$-series with $p=2$). Sum of two convergent series converges. So the series does converge — just not by a naive Leibniz argument. Right answer, wrong justification is still wrong.

• L1 — spot alternation, name $a_n$ and $b_n$.
• L2 — run both conditions; use $\to 0$ as the fast kill-switch.
• L3 — prove monotonicity with derivatives; recognise non-monotone failure.
• L4 — remainder bound $|s-s_N|\le b_{N+1}$ for accuracy; conditional vs absolute.
• L5 — parameters, power-series endpoints, and disguised non-monotone terms.