Noesis
4.3.4Calculus III — Sequences & Series

Geometric series — convergence condition, proof

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🇮🇳 Yeh note Hinglish mein padho →

A geometric series is a sum where each term is a fixed multiple of the one before it. The whole theory of "infinite sums that make sense" starts here — it's the simplest infinite series we can fully understand.

The Setup — WHAT are we summing?

Deriving the Partial Sum from Scratch

We never sum infinitely many things directly. We sum the first NN terms, get a formula, then take a limit. The first NN terms (indices 00 to N−1N-1): SN=a+ar+ar2+⋯+arN−1S_N = a + ar + ar^2 + \cdots + ar^{N-1}

Why exclude r=1r=1? Then every term is just aa, so SN=aNS_N = aN — a separate, trivial case (it diverges unless a=0a=0).

Taking the Limit — the Convergence Condition

The infinite sum is defined as ∑n=0∞arn=lim⁡N→∞SN\displaystyle\sum_{n=0}^\infty ar^n = \lim_{N\to\infty} S_N.

lim⁡N→∞SN=a1−r lim⁡N→∞(1−rN)\lim_{N\to\infty} S_N = \frac{a}{1-r}\,\lim_{N\to\infty}\big(1 - r^N\big)

Everything hinges on lim⁡N→∞rN\displaystyle\lim_{N\to\infty} r^N. HOW does rNr^N behave?

  • If ∣r∣<1|r|<1: repeatedly multiplying by something smaller than 11 in size drives rN→0r^N \to 0.
  • If ∣r∣>1|r|>1: ∣rN∣=∣r∣N→∞|r^N|=|r|^N \to \infty, no limit.
  • If r=1r=1: rN=1r^N=1 always, but this is the excluded case (SN=aN→±∞S_N=aN\to\pm\infty).
  • If r=−1r=-1: rNr^N flips 1,−1,1,−1,…1,-1,1,-1,\dots — never settles.
Figure — Geometric series — convergence condition, proof

Worked Examples

Common Mistakes (Steel-manned)

Recall Feynman: explain to a 12-year-old

Imagine you're walking to a wall. First you walk half the way, then half of what's left, then half of that, forever. Each step is smaller than the last. Even though there are infinitely many steps, you cover a total distance of exactly the whole way to the wall — a finite number! That's a geometric series with r=12r=\frac12. But if instead every step got bigger (like doubling), you'd walk forever and never stop — that series "diverges". The magic rule: the steps must keep shrinking (size of rr under 1) for the total to settle down.

Active Recall — Flashcards

What is the convergence condition for ∑arn\sum a r^n?
∣r∣<1|r|<1 (with a≠0a\neq0); otherwise it diverges.
What does a convergent geometric series sum to?
a1−r\dfrac{a}{1-r} where aa is the first term, rr the common ratio.
Derive the partial sum SNS_N: what trick is used?
Compute SN−rSNS_N - rS_N; middle terms cancel (telescope), giving SN=a(1−rN)1−rS_N=\frac{a(1-r^N)}{1-r}.
Why does ∣r∣<1|r|<1 guarantee convergence?
Because then rN→0r^N\to 0, so SN=a(1−rN)1−r→a1−rS_N=\frac{a(1-r^N)}{1-r}\to\frac{a}{1-r}.
What happens at r=1r=1?
Formula invalid; every term is aa so SN=aN→∞S_N=aN\to\infty (diverges unless a=0a=0).
What happens at r=−1r=-1?
rNr^N oscillates 1,−1,1,…1,-1,1,\dots; partial sums don't settle — diverges.
Sum of 12+14+18+⋯\frac12+\frac14+\frac18+\cdots?
11 (here a=r=12a=r=\tfrac12, S=1/21/2S=\frac{1/2}{1/2}).
Write 0.7‾0.\overline{7} as a fraction via geometric series.
7/101−1/10=79\frac{7/10}{1-1/10}=\frac79.
Why is plugging a1−r\frac{a}{1-r} wrong when ∣r∣>1|r|>1?
The derivation assumed rN→0r^N\to0, which fails; the limit doesn't exist so the value is meaningless.

Connections

  • Sequences — limits and convergence (defines lim⁡rN\lim r^N we relied on)
  • Partial sums and series convergence (series = limit of partial sums)
  • Ratio Test (generalises "constant ratio" idea to varying ratios)
  • Power series and radius of convergence (geometric series is the model: ∑xn=11−x\sum x^n=\frac{1}{1-x})
  • Repeating decimals as fractions
  • Telescoping series (same cancellation idea used in the proof)

Concept Map

has

sum first N terms

multiply by r and subtract

yields

requires

take limit N to infinity

governs

if abs r less than 1

if abs r at least 1

gives

valid iff

sanity check

Geometric series a plus ar plus ar2

Common ratio r

Partial sum S_N

Telescoping cancellation

S_N equals a times 1 minus r^N over 1 minus r

Exclude r equals 1

Behaviour of r^N

r^N goes to 0

Series diverges

Sum equals a over 1 minus r

abs r less than 1

Limits match a and infinity

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Geometric series matlab aisa sum jisme har agla term, pichhle term ka ek fixed multiple hota hai — woh multiple ko hum common ratio rr kehte hain. Series likhi jaati hai a+ar+ar2+⋯a + ar + ar^2 + \cdots. Sabse bada sawaal yeh hai: kya infinite terms jod ke ek finite number milega? Iska poora jawaab depend karta hai ki rr ki size kitni hai.

Proof ka asli jaadu ek chhota sa trick hai. Pehle sirf NN terms ka sum SNS_N lo. Phir poore sum ko rr se multiply karo, aur rSNrS_N ko SNS_N se subtract kar do. Beech ke saare terms cancel ho jaate hain (telescope), aur bachta hai SN(1−r)=a(1−rN)S_N(1-r) = a(1-r^N), yaani SN=a(1−rN)1−rS_N = \frac{a(1-r^N)}{1-r}. Ab infinite sum ke liye N→∞N\to\infty lo. Yahaan sab kuch rNr^N par tika hai — agar ∣r∣<1|r|<1 hai to rN→0r^N \to 0, aur sum settle ho jaata hai a1−r\frac{a}{1-r} par. Agar ∣r∣≥1|r|\ge 1 hai to rNr^N ya to badhta jaata hai ya oscillate karta hai, sum diverge ho jaata hai.

To yaad rakhne wali baat: "ratio under one, the sum is done" — yaani ∣r∣<1|r|<1 hone par hi formula a1−r\frac{a}{1-r} valid hai. Sabse common galti yeh hai ki students ∣r∣≥1|r|\ge1 par bhi formula thok dete hain aur ek bekaar number nikaal lete hain (jaise −1-1). Pehle hamesha ∣r∣<1|r|<1 check karo, tabhi formula lagao. Aur aa matlab series ka pehla actual term, coefficient nahi — index start (n=0n=0 ya n=1n=1) dekh ke pehla term nikaalo.

Yeh topic kyun important hai? Kyunki yeh sabse simple infinite series hai jise hum poori tarah samajh sakte hain, aur isi se aage ke bade ideas aate hain — power series (11−x=∑xn\frac{1}{1-x}=\sum x^n), ratio test, aur repeating decimals ko fraction banana. Foundation strong, to aage sab easy.

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