Exercises — Absolute vs conditional convergence

Step — throw away the signs. $|a_n|=\frac{1}{n^3}$, so $\sum |a_n|=\sum \frac{1}{n^3}$. Why: absolute convergence is decided entirely by the positive series $\sum |a_n|$.

Step — recognise the type. This is a p-series with $p=3$. A $p$-series converges exactly when $p>1$. Here $p=3>1$, so $\sum \frac{1}{n^3}$ converges.

Verdict: since $\sum |a_n|$ converges, the series is absolutely convergent. The alternating signs are irrelevant.

Step — absolute test. $|a_n|=\frac{1}{\sqrt{n}}=\frac{1}{n^{1/2}}$, a $p$-series with $p=\frac{1}{2}\le 1$ → diverges. So it is not absolutely convergent.

Step — test $\sum a_n$ itself with the Alternating Series Test (AST). The magnitudes $b_n=\frac{1}{\sqrt{n}}$ are (i) positive, (ii) decreasing, (iii) $\to 0$. All three hold, so $\sum a_n$ converges.

Verdict: converges but not absolutely → conditionally convergent.

Step — the cheapest test first: do the terms go to zero? ${\lim }_{n\to \infty }\left|(-1{)}^n\frac{n}{n+1}\right|={\lim }_{n\to \infty }\frac{n}{n+1}=1\ne 0.$ Why this test first: if the terms don't shrink to $0$, no series can converge — this is the Term Test for Divergence, the fastest kill switch.

Verdict: $a_n\nrightarrow 0$, so $\sum a_n$ diverges. (Neither absolute nor conditional applies — it's simply not convergent.)

Step — pick the tool. Factorials or powers like $2^n$ in the denominator scream Ratio Test, because the ratio of consecutive terms collapses the powers cleanly. We apply it to $|a_n|=\frac{n}{2^n}$.

Step — compute the ratio. $L={\lim }_{n\to \infty }\frac{|a_{n+1}|}{|a_n|}={\lim }_{n\to \infty }\frac{(n+1)/2^{n+1}}{n/2^n}={\lim }_{n\to \infty }\frac{n+1}{2n}=\frac{1}{2}.$ Why: the $2^n$ cancels to leave a single factor $\frac{1}{2}$; the polynomial ratio $\frac{n+1}{n}\to 1$.

Step — read the verdict. $L=\frac{1}{2}<1$, and the ratio test applied to $|a_n|$ decides absolute convergence directly.

Verdict: absolutely convergent.

Step — absolute test. $|a_n|=\frac{1}{n\ln n}$. This is the textbook integral-test series: ${\int }_2^{\infty }\frac{dx}{x\ln x}=[\ln (\ln x){]}_2^{\infty }=\infty$, so $\sum \frac{1}{n\ln n}$ diverges. Not absolutely convergent. Why the integral test: $\frac{1}{x\ln x}$ is positive and decreasing, and its antiderivative $\ln (\ln x)$ is known — the integral converts the sum into a growth we can read off.

Step — test $\sum a_n$ with the AST. Magnitudes $\frac{1}{n\ln n}$ are positive, decreasing (both $n$ and $\ln n$ grow), and $\to 0$. AST → converges.

Verdict: conditionally convergent.

Step — absolute test by comparison. $|a_n|=\frac{1}{n^2+1}\le \frac{1}{n^2}$ for all $n\ge 1$. Why Comparison Test: $\frac{1}{n^2+1}$ has no closed antiderivative worth using, but it is dominated by a known convergent $p$-series $\sum \frac{1}{n^2}$ ($p=2>1$).

Step — conclude. Since $0\le |a_n|\le \frac{1}{n^2}$ and $\sum \frac{1}{n^2}$ converges, $\sum |a_n|$ converges.

Verdict: absolutely convergent.

Step — identify the convergence type. $\sum \frac{1}{n}$ (the Harmonic series) diverges, while the alternating version converges. So this is conditionally convergent.

Step — invoke the theorem. The Riemann Rearrangement Theorem says: for a conditionally convergent series, rearranging can produce any real value. So there is no order-independent "true value" — both students computed a legitimate sum of their ordering. Why this happens: the positive-only part $\sum \frac{1}{2k-1}$ and negative-only part $\sum \frac{-1}{2k}$ each diverge (to $+\infty$ and $-\infty$). Rearranging redistributes how much of each infinite reservoir you draw before crossing to the other, steering the running total anywhere.

Step — Student B's value. The "one positive, two negatives" pattern is the classic rearrangement whose sum is $\frac{1}{2}\ln 2$. Sketch: group as $\left(1-\frac{1}{2}\right)-\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{6}\right)-\frac{1}{8}+\cdots =\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\cdots =\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{3}-\cdots \right)=\frac{1}{2}\ln 2.$

Verdict: Student A gets $\ln 2\approx 0.693$; Student B gets $\frac{1}{2}\ln 2\approx 0.347$. Both are correct for their ordering — that's the whole lesson of conditional convergence.

Step — decide the general claim: it is TRUE. Suppose $\sum a_n$ converges and $\sum a_n^2$ diverges. We show $\sum a_n$ cannot be absolutely convergent, which forces it to be conditional.

Step — the key comparison. Since $\sum a_n$ converges, its terms $\to 0$, so eventually $|a_n|<1$. For those $n$, $a_n^2=|a_n{|}^2\le |a_n|.$ Why this inequality: squaring a number smaller than $1$ makes it smaller, so $a_n^2$ sits below $|a_n|$.

Step — contrapositive via Comparison Test. If $\sum |a_n|$ converged, then by comparison $0\le a_n^2\le |a_n|$ would force $\sum a_n^2$ to converge too. But we are told $\sum a_n^2$ diverges — contradiction. Hence $\sum |a_n|$ diverges.

Step — conclude. $\sum a_n$ converges but $\sum |a_n|$ diverges → conditionally convergent. The implication is TRUE.

Step — check the concrete case. For $a_n=\frac{(-1{)}^n}{\sqrt{n}}$: $\sum a_n$ converges by AST (Problem 1.2), and $a_n^2=\frac{1}{n}$ gives $\sum a_n^2=\sum \frac{1}{n}$ which diverges. Both premises hold, and indeed this series is conditionally convergent (Problem 1.2 verdict) — exactly as the theorem predicts. ✓

Takeaway: here $\sum a_n^2$ diverging is a valid shortcut certifying non-absolute convergence — but only after you already know $\sum a_n$ itself converges.

Step — absolute behaviour. $\sum |a_n|=\sum \frac{1}{n^p}$ is a $p$-series: converges iff $p>1$.

• So for $p>1$: absolutely convergent.

Step — when absolute fails ($p\le 1$), test $\sum a_n$ via AST. Magnitudes $\frac{1}{n^p}$ decrease to $0$ iff $p>0$.

• For $0<p\le 1$: AST applies (decreasing $\to 0$), so $\sum a_n$ converges, but $\sum |a_n|$ diverges → conditionally convergent.

Step — the $p\le 0$ tail. If $p\le 0$ then $\frac{1}{n^p}=n^{-p}\nrightarrow 0$ (it stays $\ge 1$ or grows). Term Test → divergent.

Full answer:

• $p>1$: absolutely convergent.
• $0<p\le 1$: conditionally convergent.
• $p\le 0$: divergent.

Figure 2 shows this as a number line so the three regions and their boundaries are visible at a glance: the divergent stretch $p\le 0$ (white, terms don't vanish), the conditional band $0<p\le 1$ (amber), and the absolute ray $p>1$ (cyan). Note the endpoint markers — $p=1$ belongs to conditional, while $p=0$ is excluded from it (divergent).

Step — radius via Ratio Test on $|a_n|$. With $a_n=\frac{(x-3{)}^n}{n}$, $L={\lim }_{n\to \infty }\left|\frac{(x-3{)}^{n+1}/(n+1)}{(x-3{)}^n/n}\right|=|x-3|{\lim }_{n\to \infty }\frac{n}{n+1}=|x-3|.$ Absolute convergence needs $L<1$: $|x-3|<1$, i.e. $2<x<4$. Radius $R=1$, centre $3$.

Step — right endpoint $x=4$. Series becomes $\sum \frac{1^n}{n}=\sum \frac{1}{n}$ = harmonic → diverges.

Step — left endpoint $x=2$. Series becomes $\sum \frac{(-1{)}^n}{n}$ → converges by AST, but $\sum \frac{1}{n}$ diverges → conditionally convergent.

Full answer:

• $2<x<4$: absolutely convergent.
• $x=2$: conditionally convergent.
• $x=4$: divergent.
• Otherwise ($x<2$ or $x>4$): divergent. Interval of convergence: $[2,4)$.

(a) — the mechanism. Split $S$ into its positive terms $P=1+\frac{1}{3}+\frac{1}{5}+\cdots$ and (magnitudes of) negatives $N=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots$. Each is a scaled Harmonic series: $P\ge \frac{1}{2}\sum \frac{1}{n}=\infty ,N=\frac{1}{2}\sum \frac{1}{n}=\infty .$ Why this matters: you have two infinite reservoirs, one of $+$'s and one of $-$'s. To reach a target $T$: add positive terms until the running total first exceeds $T$, then add negatives until it drops below $T$, then positives again, and so on. Because each reservoir is infinite you never run out, and because individual terms $\to 0$ the overshoot each time shrinks to $0$ — so the running total converges to exactly $T$. That is the constructive proof behind the Riemann Rearrangement Theorem.

(b) — AST error bound. For an alternating series with decreasing magnitudes $b_n\to 0$, the error after $N$ terms is at most the first omitted term: $|S-S_N|\le b_{N+1}=\frac{1}{N+1}$. Why: consecutive partial sums straddle the limit, so the gap is smaller than the next step.

Require $\frac{1}{N+1}\le 0.01\Rightarrow N+1\ge 100\Rightarrow N\ge 99$.

Answer: $N=99$ terms guarantee the partial sum is within $0.01$ of $\ln 2$. (Contrast the absolutely convergent $\sum \frac{1}{n^2}$-type series, whose tails shrink far faster — a vivid measure of how slowly conditional convergence settles.)

First question you always ask ::: Does $\sum |a_n|$ converge? If yes ::: Absolutely convergent — stop, no second test. If no ::: Test $\sum a_n$ itself (usually AST): converges → conditional; diverges → divergent. Endpoints of a power series ::: Always substituted and tested by hand (Ratio Test is silent at $L=1$).