Limit comparison test

Tags: #calculus #series #convergence Subject: Maths — Calculus III: Sequences & Series

The big picture

We want to decide whether $\sum a_n$ converges or diverges, where $a_n > 0$ (positive-term series). Comparing directly with a known series (the Direct Comparison Test) is fragile: you need an exact inequality $a_n \le b_n$ that may be a pain to prove. The Limit Comparison Test (LCT) is the smoother tool — it only cares whether $a_n$ and $b_n$ behave the same way for large $n$ .

WHAT it says

The "main" case is $0 < L < \infty$ . The other two are edge cases that still give one-directional information.

WHY it's true (derivation from scratch)

Assume the main case: $L = \lim a_n/b_n$ with $0 < L < \infty$ .

Step 1 — Translate the limit into a sandwich. Why this step? "Limit equals $L$ " means the ratio eventually lives in any small window around $L$ . We pick a window that keeps everything positive.

Choose $\varepsilon = L/2$ . By the definition of the limit, there exists $N$ such that for all $n > N$ : $\left| \frac{a_n}{b_n} - L \right| < \frac{L}{2}.$ Unpacking the absolute value: $L - \frac{L}{2} < \frac{a_n}{b_n} < L + \frac{L}{2} \quad\Longrightarrow\quad \frac{L}{2} < \frac{a_n}{b_n} < \frac{3L}{2}.$

Step 2 — Clear the fraction (legal because $b_n>0$ ). Why this step? Multiplying by the positive $b_n$ turns a ratio statement into a direct comparison statement. $\frac{L}{2}\, b_n \;<\; a_n \;<\; \frac{3L}{2}\, b_n \qquad (n > N).$

Step 3 — Feed each inequality into the Direct Comparison Test.

If $\sum b_n$ converges: then $\sum \frac{3L}{2} b_n$ converges (constant multiple). Since $a_n < \frac{3L}{2} b_n$ , the right inequality gives $\sum a_n$ converges.
If $\sum b_n$ diverges: then $\sum \frac{L}{2} b_n$ diverges. Since $a_n > \frac{L}{2} b_n$ , the left inequality forces $\sum a_n$ to diverge.

So $\sum a_n$ and $\sum b_n$ live or die together. $\blacksquare$

HOW to use it (the recipe)

Look at $a_n$ for large $n$ . Keep only the dominant powers in numerator and denominator.
Build $b_n$ from those dominant terms — usually a $p$ -series $\sum 1/n^p$ or a geometric series, whose fate you already know.
Compute $L = \lim a_n / b_n$ .
Read off the conclusion. (If you chose $b_n$ well, $L$ is a nice finite positive number.)

Reference facts you compare against: $\sum \frac{1}{n^p}\ \text{converges} \iff p>1, \qquad \sum r^n\ \text{converges} \iff |r|<1.$

Worked examples

Worked example Example 1 —

$\displaystyle\sum_{n=1}^\infty \frac{3n^2+5}{n^4 - n + 2}$ Pick $b_n$ . Strongest power on top is $3n^2$ , on bottom is $n^4$ . $b_n = \frac{n^2}{n^4} = \frac{1}{n^2}.$ Why this step? For huge $n$ , the $+5$ and $-n+2$ are negligible; the ratio of leading powers controls behaviour. Compute $L$ . $L = \lim_{n\to\infty} \frac{(3n^2+5)/(n^4-n+2)}{1/n^2} = \lim \frac{n^2(3n^2+5)}{n^4 - n + 2} = \lim \frac{3n^4 + 5n^2}{n^4 - n + 2} = 3.$ Why this step? Divide top & bottom by $n^4$ : $(3 + 5/n^2)/(1 - 1/n^3 + 2/n^4) \to 3$ . Conclude. $0 < L = 3 < \infty$ , and $\sum 1/n^2$ converges ( $p=2>1$ ). Therefore the series converges. ✔

Worked example Example 2 —

$\displaystyle\sum_{n=1}^\infty \frac{1}{\sqrt{n^2+1}}$ Pick $b_n$ . $\sqrt{n^2+1}\approx n$ for large $n$ , so $b_n = 1/n$ . Compute $L$ . $L = \lim \frac{1/\sqrt{n^2+1}}{1/n} = \lim \frac{n}{\sqrt{n^2+1}} = \lim \frac{1}{\sqrt{1 + 1/n^2}} = 1.$ Why this step? Factor $n$ out of the root: $\sqrt{n^2+1}=n\sqrt{1+1/n^2}$ . Conclude. $0 < L = 1 < \infty$ , and $\sum 1/n$ (harmonic) diverges. So the series diverges. ✔

Note direct comparison is annoying here ( $1/\sqrt{n^2+1} < 1/n$ goes the wrong way for proving divergence) — LCT saves us.

Worked example Example 3 —

$\displaystyle\sum_{n=1}^\infty \frac{\ln n}{n^2}$ using $L=0$ case Pick $b_n = 1/n^{1.5}$ . Why? Since $\ln n$ grows slower than any positive power $n^\varepsilon$ , the term $\frac{\ln n}{n^2}$ actually decays slightly slower than $1/n^2$ (indeed $\frac{\ln n}{n^2} > \frac{1}{n^2}$ for large $n$ ). To get a clean $L=0$ verdict we deliberately compare against a slightly larger, still convergent $b_n=1/n^{1.5}$ — its extra power of $n^{0.5}$ in the denominator overwhelms the slow $\ln n$ , driving the ratio to $0$ . Compute $L$ . $L = \lim \frac{(\ln n)/n^2}{1/n^{1.5}} = \lim \frac{\ln n}{n^{0.5}} = 0$ (logarithm beats by power-of- $n$ in the denominator). Conclude. $L=0$ and $\sum 1/n^{1.5}$ converges ( $p=1.5>1$ ). By the $L=0$ rule, the series converges. ✔

Common mistakes

Forecast-then-Verify drill

Before computing, predict convergence, then confirm with LCT.

$\displaystyle\sum \frac{2n}{n^3+1}$ → forecast: behaves like $1/n^2$ → converges. Verify: $L=2$ . ✔
$\displaystyle\sum \frac{\sqrt{n}+1}{n+4}$ → behaves like $\sqrt n / n = 1/n^{1/2}$ , $p=\tfrac12\le1$ → diverges. Verify: $L=1$ . ✔
$\displaystyle\sum \frac{1}{2^n + n}$ → behaves like $1/2^n$ geometric → converges. Verify: $L=1$ . ✔

Flashcards

What does the Limit Comparison Test require about the terms

a_n,b_n

They must be positive (for all large

n

In the LCT, what is computed?

L=\lim_{n\to\infty} a_n/b_n

0<L<\infty

in the LCT, what can you conclude?

\sum a_n

and

\sum b_n

both converge or both diverge.

L=0

, what extra condition gives convergence of

\sum a_n

\sum b_n

must converge.

L=\infty

, what extra condition gives divergence of

\sum a_n

\sum b_n

must diverge.

Why does multiplying the ratio inequality by

b_n

stay valid?

Because

b_n>0

, so inequality direction is preserved.

Which

\varepsilon

is chosen in the proof and why?

\varepsilon=L/2

, so the lower bound

L/2

stays strictly positive.

For

\sum \frac{3n^2+5}{n^4-n+2}

, what is the right

b_n

1/n^2

(ratio of leading powers).

Why is

a_n\to 0

not enough for convergence?

It's necessary not sufficient; e.g.

\sum 1/n

diverges though

1/n\to0

When is LCT inconclusive?

When

L=0

with

\sum b_n

divergent, or

L=\infty

with

\sum b_n

convergent.

p

-series

\sum 1/n^p

converges exactly when?

p>1

Recall Feynman: explain to a 12-year-old

Imagine two runners. We can't easily measure how far each goes alone, but we know one of them perfectly. If, after a long time, the unknown runner is always going at roughly the same speed as the runner we understand (their speed ratio settles to a steady positive number), then they'll either both reach the finish line or both run forever together. Series convergence is just "does the running total stop somewhere?" — and matching speeds means matching destinies.

Connections

Direct Comparison Test — LCT is built on top of it via the sandwich step.
p-Series — the favourite $b_n$ benchmark ( $\sum 1/n^p$ ).
Geometric Series — benchmark when $a_n$ has $r^n$ .
Ratio Test — better when factorials/exponentials dominate.
Integral Test — alternative for monotone positive terms.
Harmonic Series — the classic " $a_n\to0$ but diverges" counterexample.
Squeeze Theorem — same "trap the quantity in a window" technique as the $\varepsilon=L/2$ step.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Limit Comparison Test ka core idea simple hai: agar do positive series ke terms bade $n$ ke liye "same speed" se chhote hote hain, to dono ka fate ek jaisa hota hai — ya dono converge karenge ya dono diverge. Hum yeh "same speed" measure karte hain ratio ka limit lekar: $L = \lim a_n/b_n$ . Agar $L$ ek finite aur positive number nikla (matlab $0 < L < \infty$ ), to bas $\sum b_n$ ka behaviour dekh lo aur wahi answer $\sum a_n$ ka bhi hoga.

Trick yeh hai ki $b_n$ chunna kaise hai. Rule: "sabse strong power rakho, baaki phenk do" — numerator aur denominator dono mein leading term le lo. Jaise $\frac{3n^2+5}{n^4-n+2}$ mein top ka strong term $n^2$ , bottom ka $n^4$ , to $b_n = 1/n^2$ . Ab $1/n^2$ to $p$ -series hai with $p=2>1$ , yeh converge karta hai, isliye original series bhi converge. Aur agar $b_n = 1/n$ (harmonic) bana to woh diverge karta hai.

Proof ka magic step: limit ki definition se ratio eventually $L/2$ aur $3L/2$ ke beech aa jaata hai. $b_n$ positive hai isliye usse multiply karke direct comparison ban jaata hai — $\frac{L}{2}b_n < a_n < \frac{3L}{2}b_n$ . Phir Direct Comparison Test laga do, kaam khatam.

Ek common galti se bacho: sirf $a_n \to 0$ dekh ke "converge" mat bol dena — harmonic series $\sum 1/n$ mein bhi terms $0$ pe jaate hain par series diverge karti hai. LCT decay ki rate compare karta hai, sirf "0 pe jaata hai ya nahi" nahi. Yeh test exam mein rational aur root-wale terms ke liye sabse fast tarika hai.

The big picture

WHAT it says

WHY it's true (derivation from scratch)

HOW to use it (the recipe)

Worked examples

Common mistakes

Forecast-then-Verify drill

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus III — Sequences & Series

Connections