Absolute and conditional convergence

At a Glance

• Frequency: 4 sub-parts across 4 of 13 years (2015, 2016, 2018, 2025)
• Priority tier: T3
• Marks (count): 10 (3), 15 (1)
• Average solve time: ~7 min
• Difficulty mix: easy 3, medium 1
• Section: A | Dominant type: computation

Why This Chapter Matters

Alternating-series questions appear in Section A compulsory (Q1) almost every year, always for 10 marks. The question always asks: does the series converge? Is the convergence absolute or conditional? The answer is always: yes (Leibniz); conditional (absolute series diverges by comparison with $\sum1/n^p$ for $p\le1$). The 2016 question requires proving Leibniz and the harmonic divergence from scratch — that proof is worth knowing in full. The 2022 and 2023 questions add the ratio test and Raabe’s test for the absolute part.

Minimum Theory

Absolute convergence. $\sum a_n$ is absolutely convergent if $\sum|a_n|$ converges. Absolute convergence implies convergence.

Conditional convergence. $\sum a_n$ converges but $\sum|a_n|$ diverges.

Leibniz alternating series test. If $a_n>0$, $a_n$ is decreasing, and $a_n\to0$, then $\sum(-1)^{n+1}a_n$ converges.

Proof sketch: Even partial sums $S_{2m}$ are non-decreasing and bounded above by $a_1$; odd partial sums satisfy $S_{2m+1}=S_{2m}+a_{2m+1}$; both converge to the same limit.

Convergence tests for $\sum|a_n|$:

Raabe’s test. If $\lim_{n\to\infty}n\!\left(\dfrac{a_n}{a_{n+1}}-1\right)=L$, then the series converges absolutely if $L>1$ and diverges if $L<1$.

Question Archetypes

conditional-convergence (3 question(s); 2015, 2016, 2018)

Recognition Cues — Alternating series $\sum(-1)^{n+1}a_n$ where $a_n$ involves $1/n$ or $1/(n+a)^p$; asked for “absolute and conditional” convergence or “range of $p$.”

Solution Template

1. Verify Leibniz conditions: $a_n>0$, $a_n$ decreasing (compute $f'(x)<0$ for large $x$), $a_n\to0$.
2. Conclude convergence by Leibniz.
3. Test $\sum|a_n|$ by limit comparison with $\sum1/n^p$.
4. Classify: converges absolutely iff $p>1$; conditionally iff $0<p\le1$.

Worked Example 1

2015 Paper 2, 2015-P2-Q1c (10 marks)

Test convergence and absolute convergence of $\displaystyle\sum_{n=1}^\infty(-1)^{n+1}\frac{n}{n^2+1}$.

Set $a_n=n/(n^2+1)$. Leibniz: $a_n>0$; $f(x)=x/(x^2+1)$ has $f'(x)=(1-x^2)/(x^2+1)^2<0$ for $x>1$, so $a_n$ is decreasing; $a_n\sim1/n\to0$. All conditions hold — series converges.

For absolute convergence: $a_n\sim1/n$. Limit comparison with $\sum1/n$: $\lim na_n=\lim n^2/(n^2+1)=1>0$. Since $\sum1/n$ diverges ($p=1$), so does $\sum a_n$. Not absolutely convergent — conditionally convergent. $\blacksquare$

Worked Example 2

2018 Paper 2, 2018-P2-Q1d (10 marks)

$\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n+1}}{(n+a)^p}$, $a>0$. Find: (i) range of $p$ for absolute convergence; (ii) range of $p$ for conditional convergence.

Leibniz: $b_n=1/(n+a)^p>0$, decreasing ($(n+a)^p$ increasing), $b_n\to0$ for all $p>0$. Series converges for all $p>0$.

Absolute series: $\sum1/(n+a)^p$. Limit comparison with $\sum1/n^p$ gives ratio $\to1$, same convergence behaviour. Converges iff $p>1$.

$\boxed{(i)\;p>1;\quad(ii)\;0<p\le1.}$

Worked Example 3

2016 Paper 2, 2016-P2-Q2a (15 marks)

Show $\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n+1}$ is conditionally convergent. Prove any theorems used.

Convergence (Leibniz test + proof): $a_n=1/(n+1)>0$, strictly decreasing, $a_n\to0$. Leibniz proof: even partial sums $S_{2m}=\sum_{k=1}^m(a_{2k-1}-a_{2k})$ are non-decreasing (each bracket $\ge0$) and bounded above by $a_1$; odd sums $S_{2m+1}=S_{2m}+a_{2m+1}\to S$ since $a_{2m+1}\to0$. Both subsequences converge to the same limit. ✓

Divergence of $\sum|a_n|$: $\sum_{n=1}^\infty1/(n+1)=\sum_{k=2}^\infty1/k$ is the harmonic series minus its first term. Proof of harmonic divergence: group in blocks $(\frac12)+(\frac13+\frac14)+(\frac15+\cdots+\frac18)+\cdots$; the $j$-th block ($2^{j-1}$ terms starting from $2^{j-1}+1$) sums to $\ge2^{j-1}\cdot1/2^j=1/2$. Infinitely many blocks each contribute $\ge1/2$, so partial sums $\to\infty$. ✓

Hence conditionally convergent. $\blacksquare$

Common Traps

• Verify all three Leibniz conditions. Monotonicity is the one most often skipped. Use $f'(x)<0$ or explicit comparison $a_{n+1}<a_n$.
• $a_n\to0$ doesn’t imply convergence. The harmonic series $\sum1/n$ has $a_n\to0$ but diverges. Leibniz additionally requires the alternating sign and monotonicity.
• Limit comparison: the ratio must be in $(0,\infty)$. If $\lim a_n/b_n=0$ or $\infty$, the standard limit comparison fails; use the direct comparison instead.

ratio-test (1 question(s); 2022)

Worked Example

2022 Paper 2, 2022-P2-Q4b (15 marks)

Test $\displaystyle\sum_{n=1}^\infty\frac{n^n x^n}{n!}$ for $x>0$.

$a_{n+1}/a_n = x(1+1/n)^n\to xe$ as $n\to\infty$. Ratio test: converges if $xe<1$, diverges if $xe>1$.

At $x=1/e$: $a_n\sim1/\sqrt{2\pi n}$ (Stirling: $n!\sim\sqrt{2\pi n}(n/e)^n$). Since $\sum1/\sqrt{n}$ diverges (p-series, $p=1/2<1$), diverges also at $x=1/e$.

$\boxed{\text{Converges for }x<1/e;\;\text{diverges for }x\ge1/e.}$

ratio-raabe-test (1 question(s); 2023)

Worked Example

2023 Paper 2, 2023-P2-Q1c (10 marks)

Test $\displaystyle\sum_{n=1}^\infty\frac{(2n-1)!!}{(2n)!!}\cdot\frac{x^{2n+1}}{2n+1}$ for $x>0$.

Ratio $a_{n+1}/a_n=(2n+1)^2/((2n+2)(2n+3))\cdot x^2\to x^2$. Converges for $x^2<1$, diverges for $x^2>1$, inconclusive at $x=1$.

Raabe at $x=1$: $\lim n(a_n/a_{n+1}-1)=\lim n\cdot(2n+2)(2n+3)/(2n+1)^2-n\to3/2>1$. Converges at $x=1$.

$\boxed{\text{Converges for }0<x\le1;\;\text{diverges for }x>1.}$

Marks-Aware Writing

For 10-mark convergence questions: three explicit steps — Leibniz conditions stated and verified; conclude convergence; limit-comparison with $\sum1/n^p$ for the absolute series. Box the final classification.

For the 15-mark question (2016) requiring proofs: write the full Leibniz proof (even partial sums non-decreasing and bounded; odd sums sandwich) AND the dyadic-blocking proof of harmonic divergence. These proofs together are 8–10 marks; missing either means half-marks at most.

Practice Set

• 2025-P2-Q1c (10 m) — — Hint: Leibniz; absolute series is the harmonic series, $p=1$, diverges.