Series of real terms: convergence, standard tests

At a Glance

• Frequency: 2 sub-parts across 2 of 13 years (2022, 2023)
• Priority tier: T3
• Marks (count): 10 (1), 15 (1)
• Average solve time: ~7 min
• Difficulty mix: medium 2
• Section: A | Dominant type: proof

Why This Chapter Matters

Both questions involve the ratio test applied to series with parameters ($x > 0$), where the ratio test is conclusive everywhere except at a single boundary value where Raabe’s test is needed. Mastering the two-layer decision tree (ratio test $\to$ if $L=1$, apply Raabe) is the key skill. The 2022 question reaches its boundary via Stirling’s approximation; the 2023 question uses Raabe’s test directly. Together they cover every exam scenario for this atom.

Minimum Theory

Ratio test. For a positive-term series $\sum a_n$, let $L = \lim_{n\to\infty} |a_{n+1}/a_n|$. If $L < 1$: convergent. If $L > 1$: divergent. If $L = 1$: inconclusive — use a finer test.

Raabe’s test. When the ratio test gives $L = 1$, compute $R = \lim_{n\to\infty} n\!\left(\dfrac{a_n}{a_{n+1}} - 1\right)$. If $R > 1$: convergent. If $R < 1$: divergent. If $R = 1$: still inconclusive.

Stirling’s approximation. $n! \sim \sqrt{2\pi n}\, n^n e^{-n}$ as $n \to \infty$. Useful when the ratio test gives $L = 1$ and the general term $a_n$ involves $n^n/n!$ — Stirling shows $a_n \sim 1/\sqrt{2\pi n}$, then compare to $\sum 1/\sqrt{n}$ (divergent by $p$-series with $p = 1/2 < 1$).

Standard limit. $\lim_{n\to\infty}(1+1/n)^n = e$. This appears whenever the ratio $a_{n+1}/a_n$ contains a factor $(1+1/n)^n$ — the limit delivers $e$ and determines the radius of convergence.

Question Archetypes

ratio-test (1 question; 2022)

Recognition Cues — The series has general term involving $n^n x^n / n!$ or similar combinations where $(1+1/n)^n$ emerges in the ratio. The boundary $x = 1/e$ (radius of convergence) is a Stirling borderline — the general term at the boundary is $\sim 1/\sqrt{n}$, signalling divergence.

Solution Template

1. Identify $a_n$ and compute $a_{n+1}/a_n$ algebraically.
2. Simplify to isolate the factor $(1+1/n)^n \to e$.
3. Take $\lim a_{n+1}/a_n = xe$ (or similar); apply ratio test: converges if $xe < 1$, diverges if $xe > 1$.
4. At the boundary $x = 1/e$ (ratio $\to 1$): substitute $x = 1/e$ into $a_n$, apply Stirling to get $a_n \sim 1/\sqrt{2\pi n}$. Compare to $\sum 1/\sqrt{n}$ (divergent); conclude divergent at $x = 1/e$.
5. State the full conclusion.

Worked Example

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

Test for convergence or divergence of $x + \dfrac{2^2 x^2}{2!} + \dfrac{3^3 x^3}{3!} + \cdots + \dfrac{n^n x^n}{n!} + \cdots$ ($x > 0$).

Set $a_n = \dfrac{n^n x^n}{n!}$.

Step 1 — Ratio test.

$\frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}x^{n+1}/(n+1)!}{n^n x^n/n!} = x \cdot \frac{(n+1)^{n+1}}{(n+1) \cdot n^n} = x \cdot \frac{(n+1)^n}{n^n} = x\!\left(1 + \frac{1}{n}\right)^n.$

Step 2 — Take the limit.

$\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = x \cdot e.$

Step 3 — Apply ratio test.

• $xe < 1$, i.e., $x < 1/e$: series converges.
• $xe > 1$, i.e., $x > 1/e$: series diverges.
• $x = 1/e$: inconclusive — need another test.

Step 4 — Boundary $x = 1/e$.

At $x = 1/e$: $a_n = \dfrac{n^n}{n! \cdot e^n}$. By Stirling, $n! \sim \sqrt{2\pi n} \cdot n^n e^{-n}$, so

$a_n \sim \frac{n^n}{\sqrt{2\pi n} \cdot n^n e^{-n} \cdot e^n} = \frac{1}{\sqrt{2\pi n}}.$

The series $\sum 1/\sqrt{n}$ diverges ($p$-series, $p = 1/2 < 1$). By limit comparison, the series diverges at $x = 1/e$.

$\boxed{\text{Series converges for } x < 1/e; \text{ diverges for } x \ge 1/e.}$

Common Traps

• In simplifying $a_{n+1}/a_n$: the factor $(n+1)^{n+1}$ splits as $(n+1) \cdot (n+1)^n$; the extra $(n+1)$ cancels with the $(n+1)!$ denominator, leaving $(n+1)^n / n^n = (1 + 1/n)^n \to e$.
• The radius of convergence is $1/e$, not $1/e^2$ or $1/n$. State the limit precisely.
• At the boundary $x = 1/e$, Stirling is the cleanest tool; the series diverges here too.

ratio-raabe-test (1 question; 2023)

Recognition Cues — The series involves double-factorial products $(2n-1)!! / (2n)!!$ and a parameter $x > 0$. The ratio test gives $L = x^2$, which equals $1$ at $x = 1$. Raabe’s test is then applied directly using the expansion of $a_n / a_{n+1}$.

Solution Template

1. Write out $a_n$ and compute $a_{n+1}/a_n$ by pairing the double-factorial ratios carefully.
2. Take $L = \lim a_{n+1}/a_n$. Apply the ratio test for $L \ne 1$.
3. At the boundary ($L = 1$): compute $a_n / a_{n+1}$ and expand as $1 + c/n + O(1/n^2)$.
4. Apply Raabe’s test: $R = \lim n(a_n/a_{n+1} - 1)$. If $R > 1$: converges.
5. State the full conclusion.

Worked Example

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

Test the convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \cdot \frac{x^{2n+1}}{2n+1}$, $x > 0$.

Let $a_n = \dfrac{(2n-1)!!}{(2n)!!} \cdot \dfrac{x^{2n+1}}{2n+1}$.

Step 1 — Ratio test.

$\frac{a_{n+1}}{a_n} = \frac{(2n+1)!!}{(2n+2)!!} \cdot \frac{(2n)!!}{(2n-1)!!} \cdot \frac{2n+1}{2n+3} \cdot x^2 = \frac{2n+1}{2n+2} \cdot \frac{2n+1}{2n+3} \cdot x^2.$

As $n \to \infty$, both fractions $\to 1$, so

$L = \lim_{n\to\infty}\frac{a_{n+1}}{a_n} = x^2.$

• $x^2 < 1$ (i.e., $0 < x < 1$): series converges.
• $x^2 > 1$ (i.e., $x > 1$): series diverges.
• $x^2 = 1$ (i.e., $x = 1$): ratio test inconclusive.

Step 2 — Boundary $x = 1$ via Raabe’s test.

At $x = 1$, compute $a_n/a_{n+1}$:

$\frac{a_n}{a_{n+1}} = \frac{(2n+2)(2n+3)}{(2n+1)^2} = 1 + \frac{6n+5}{(2n+1)^2}.$

Apply Raabe’s test:

$R = \lim_{n\to\infty} n\!\left(\frac{a_n}{a_{n+1}} - 1\right) = \lim_{n\to\infty} \frac{n(6n+5)}{(2n+1)^2} = \frac{6}{4} = \frac{3}{2} > 1.$

Since $R > 1$, the series converges at $x = 1$.

$\boxed{\text{Series converges for } 0 < x \le 1;\text{ diverges for } x > 1.}$

Common Traps

• The ratio $a_{n+1}/a_n$ contains two factors involving $(2n+1)$: one from $(2n+1)!!/(2n-1)!!$ and one from the explicit $1/(2n+1)$. Combine them carefully.
• Skipping Raabe at $x = 1$ is the most common error — the ratio test gives $L = 1$ exactly at $x = 1$ and is inconclusive; the series converges there, but this requires Raabe.
• Raabe is applied to $a_n/a_{n+1}$ (not $a_{n+1}/a_n$); the formula is $R = \lim n(a_n/a_{n+1} - 1)$.

Marks-Aware Writing

15-mark answer (ratio-test): Four steps — compute ratio, take the limit (invoking $(1+1/n)^n \to e$), apply ratio test with all three cases, handle boundary via Stirling. The boundary analysis (Step 4) is essential and carries marks.

10-mark answer (ratio-raabe-test): Two steps — ratio test with explicit limit $L = x^2$, Raabe computation at $x = 1$ with $R = 3/2 > 1$. Show the algebra for the ratio $a_n/a_{n+1}$ in Raabe clearly.

Practice Set

• 2022-P2-Q4b (15 m) — — Hint: after the ratio test gives $xe$, apply Stirling at $x = 1/e$ to see $a_n \sim 1/\sqrt{n}$.
• 2023-P2-Q1c (10 m) — — Hint: ratio test gives $L = x^2$; at $x=1$, apply Raabe to $a_n/a_{n+1} = (2n+2)(2n+3)/(2n+1)^2$.