Noesis
4.3.8Calculus III — Sequences & Series

Direct comparison test

1,844 words8 min readdifficulty · medium6 backlinks
🇮🇳 Yeh note Hinglish mein padho →

WHAT is it?

WHY does it work? (Derivation from scratch)

We build it from the Monotone Convergence Theorem: a sequence that is increasing and bounded above converges.

Let an≥0a_n \ge 0. Define the partial sums Sn=a1+a2+⋯+an.S_n = a_1 + a_2 + \cdots + a_n.

Step 1 — Partial sums are increasing. Sn+1−Sn=an+1≥0S_{n+1} - S_n = a_{n+1} \ge 0, so SnS_n is monotone increasing. Why this step? Non-negativity of terms is what guarantees the partial sums never go down — this is why DCT demands an≥0a_n \ge 0.

Step 2 — Bound the partial sums. Since an≤bna_n \le b_n, Sn=∑k=1nak  ≤  ∑k=1nbk  ≤  ∑k=1∞bk=:B.S_n = \sum_{k=1}^n a_k \;\le\; \sum_{k=1}^n b_k \;\le\; \sum_{k=1}^\infty b_k =: B. Why this step? If ∑bn\sum b_n converges, its partial sums are bounded by the finite total BB. Term-by-term domination passes that bound to SnS_n.

Step 3 — Apply Monotone Convergence. SnS_n is increasing and bounded above by BB. Therefore SnS_n converges to a finite limit. Hence ∑an\sum a_n converges. ∎

The contrapositive gives the divergence half. If ∑an\sum a_n diverges, then Sn→∞S_n \to \infty. Since ∑k=1nbk≥Sn\sum_{k=1}^n b_k \ge S_n, the bigger partial sums also →∞\to \infty, so ∑bn\sum b_n diverges. ∎

Figure — Direct comparison test

HOW to use it (a recipe)

  1. Make sure terms are eventually ≥0\ge 0.
  2. Guess convergence or divergence (look at the dominant power → think pp-series / geometric).
  3. Build the inequality in the correct direction:
    • To prove convergence, find a bigger convergent bnb_n.
    • To prove divergence, find a smaller divergent bnb_n.
  4. Justify the inequality algebraically.
  5. State the known benchmark result and conclude.

Worked Examples

Common Mistakes (Steel-man them)

Recall Feynman: explain to a 12-year-old

Imagine you're adding up smaller and smaller piles of candy forever. You can't add them all by hand, so you cheat: you find a different candy pile you already understand.

  • If your pile is always smaller than one that adds up to a finite jar, your pile also fits in a jar — it stops growing. (converges)
  • If your pile is always bigger than one that grows to infinity, your pile grows to infinity too. (diverges) You never count your own candies — you just compare pile-to-pile with a friend you trust!

Flashcards

What two conditions does the Direct Comparison Test require on the terms?
They must be non-negative (0≤an≤bn0 \le a_n \le b_n), at least eventually (for n≥Nn \ge N).
If 0≤an≤bn0\le a_n\le b_n and ∑bn\sum b_n converges, what about ∑an\sum a_n?
It converges (trapped under a finite roof).
If 0≤an≤bn0\le a_n\le b_n and ∑an\sum a_n diverges, what about ∑bn\sum b_n?
It diverges (pushed above an infinite floor).
Which theorem underlies the proof of DCT?
The Monotone Convergence Theorem (increasing + bounded above ⇒ converges).
Why must terms be non-negative for DCT?
So partial sums are monotone increasing, letting Monotone Convergence apply.
To prove a series CONVERGES with DCT, you compare it to a ___ series that is ___.
bigger; convergent.
To prove a series DIVERGES with DCT, you compare it to a ___ series that is ___.
smaller; divergent.
Does showing an≤1n2a_n \le \frac{1}{n^2} (smaller, convergent) prove ∑an\sum a_n converges?
No — that's the useless direction; ana_n could still diverge.
For ∑2+sin⁡nn2\sum \frac{2+\sin n}{n^2}, what bound gives convergence?
2+sin⁡nn2≤3n2\frac{2+\sin n}{n^2} \le \frac{3}{n^2}, and ∑3/n2\sum 3/n^2 converges.
What test replaces DCT when terms change sign?
Compare absolute values (absolute convergence) or use the Alternating Series Test.

Connections

  • p-series test — the usual benchmark (p>1p>1 converge, p≤1p\le1 diverge).
  • Geometric series — another common comparison partner.
  • Limit comparison test — gentler cousin when the inequality is awkward.
  • Monotone convergence theorem — the engine behind the proof.
  • Harmonic series — the canonical divergent benchmark.
  • Absolute convergence — what to do when terms aren't all positive.
  • Integral test — alternate way to classify the benchmark series.

Concept Map

guarantees

increasing + bounded converges

passes bound

with bound gives

from

convergence half

contrapositive

required by

supplies bn

applies

Monotone Convergence Theorem

Non-negative terms an,bn >= 0

Term-by-term 0 <= an <= bn

Partial sums Sn increasing

Sn bounded above by B

Direct Comparison Test

Sum bn conv => Sum an conv

Sum an div => Sum bn div

Benchmarks p-series / geometric

Recipe: guess then bound

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Direct Comparison Test ka idea ekdum simple hai: jab tumhe ek series ka sum nikalna mushkil ho, to use sum mat karo — kisi jaani-pehchaani series se compare kar lo. Condition sirf itni hai ki saare terms non-negative hone chahiye (an≥0a_n \ge 0). Agar tumhari series har term mein kisi convergent series se chhoti hai, to tumhari bhi convergent hai (finite chhat ke neeche trap ho gayi). Aur agar tumhari series kisi divergent series se badi hai, to tumhari bhi diverge karegi (infinite floor ke upar push ho gayi).

Yaad rakhne wali sabse important baat: direction. Convergence prove karni hai to upar se trap karo (bigger convergent dhundo), aur divergence prove karni hai to neeche se push karo (smaller divergent dhundo). Ulta karoge — jaise an≤1/n2a_n \le 1/n^2 dikha ke convergence claim — to galat ho jayega, kyunki chhoti convergent series se kuch prove nahi hota.

WHY kaam karta hai? Kyunki terms positive hain, to partial sums SnS_n hamesha badhte hain. Aur agar bigger series ka total finite BB hai, to Sn≤BS_n \le B bounded ho jaata hai. "Increasing + bounded above" matlab Monotone Convergence Theorem se limit exist karti hai. Bas yahi proof ka dimaag hai.

Practical tip (80/20): pehle dominant power dekho. 2+sin⁡nn2\frac{2+\sin n}{n^2} jaise mein sin⁡\sin ko uske maximum (yahan 3) se replace karke clean bound bana lo. Benchmark ke liye pp-series (p>1p>1 converge), harmonic (1/n1/n diverge), aur geometric series yaad rakho — 90% problems inhi se ho jaati hain.

Go deeper — visual, from zero

D1FoundationsEvery symbol from zero · 2,690 wordsD2Visual walkthroughThe derivation in pictures · 2,658 wordsD3Worked examplesEvery scenario covered · 2,836 wordsD4ExercisesGraded, full solutions · 2,207 wordsD5Question bankConcept traps, quick reveals · 1,839 words

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Connections

Alternating series test — Leibniz test, proofMaths · 4.3.10Geometric series — convergence condition, proofMaths · 4.3.4Limit comparison testMaths · 4.3.9Absolute vs conditional convergenceMaths · 4.3.11Integral test — proof, p-seriesMaths · 4.3.7Improper integrals — Type I (infinite limits), Type II (discontinuous integrand)Maths · 4.2.11Series — partial sums, convergence definitionMaths · 4.3.3Divergence test (necessary but not sufficient)Maths · 4.3.6Integral test — proof, p-seriesMaths · 4.3.7Limit comparison testMaths · 4.3.9Absolute vs conditional convergenceMaths · 4.3.11