Series — partial sums, convergence definition

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WHAT is a series?

WHY define it through partial sums? Because "$a_1+a_2+\cdots$ forever" is not a defined arithmetic operation. Addition is a binary operation; we only ever add two things at a time. By forming $s_1, s_2, s_3,\dots$ we replace the undefined infinite process with a perfectly legal infinite sequence, and a sequence either has a limit or it doesn't.

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WHAT does convergence mean?

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HOW: deriving the geometric series sum from scratch

This is the model example — derive, don't memorise.

Take $a_n = a r^{\,n-1}$, so $\sum_{n=1}^{\infty} ar^{n-1} = a + ar + ar^2 + \cdots$.

Step 1 — write the partial sum. $s_N = a + ar + ar^2 + \cdots + ar^{N-1}.$ Why this step? Convergence is defined via $s_N$, so we must get a closed form for $s_N$.

Step 2 — multiply by $r$ and subtract. $r\,s_N = ar + ar^2 + \cdots + ar^{N}.$ $s_N - r\,s_N = a - ar^{N}.$ Why this step? Multiplying by $r$ shifts every term one slot; subtracting cancels the entire middle, leaving only the first and last pieces. This is the classic telescoping trick.

Step 3 — solve for $s_N$ (assume $r\ne 1$). $s_N(1-r) = a(1 - r^{N}) \implies s_N = \frac{a(1-r^{N})}{1-r}.$ Why this step? Now $s_N$ is an explicit formula in $N$, so we can take a limit.

Step 4 — take the limit. The only $N$-dependent part is $r^{N}$.

• If $|r|<1$: $r^{N}\to 0$, so $s_N \to \dfrac{a}{1-r}$. Converges.
• If $|r|>1$: $|r^{N}|\to\infty$, $s_N$ blows up. Diverges.
• If $r=1$: $s_N = aN \to \pm\infty$. Diverges.
• If $r=-1$: $s_N$ alternates $a,0,a,0,\dots$ — no limit. Diverges.

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HOW: a telescoping series from scratch

Consider $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$.

Step 1 — partial fractions. $\dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}$. Why? It rewrites each term as a difference so consecutive terms cancel.

Step 2 — partial sum.

= \left(1-\tfrac12\right)+\left(\tfrac12-\tfrac13\right)+\cdots+\left(\tfrac1N-\tfrac1{N+1}\right).$$ Everything cancels except the ends: $s_N = 1 - \dfrac{1}{N+1}$. *Why?* That's the whole point of telescoping — the inner terms annihilate. **Step 3 — limit.** $s_N = 1 - \frac{1}{N+1} \to 1$. So the series **converges to $1$**. --- ## A divergent example you must know > [!example] The harmonic series diverges > $\sum_{n=1}^{\infty}\frac1n$ has terms $\to 0$, yet **diverges**. > Group terms: $1 + \tfrac12 + \underbrace{(\tfrac13+\tfrac14)}_{>\,1/2} + \underbrace{(\tfrac15+\cdots+\tfrac18)}_{>\,1/2} + \cdots$ > Each block exceeds $\tfrac12$, so $s_N$ grows without bound. > **Why it matters:** terms going to $0$ is *not enough* for convergence. > [!formula] The $n$-th Term Test (a necessary condition) > If $\sum a_n$ converges then $\displaystyle\lim_{n\to\infty} a_n = 0$. > **Derivation:** $a_N = s_N - s_{N-1}$. If $s_N \to S$ and $s_{N-1}\to S$, then > $a_N \to S - S = 0$. **Contrapositive (the usable test):** if $a_n \not\to 0$, the series diverges. > ⚠️ The converse is FALSE — harmonic series is the counterexample. --- > [!mistake] Steel-manned common errors > **Mistake 1: "$a_n \to 0$, so the series converges."** > *Why it feels right:* if you're adding tinier and tinier crumbs, surely the total settles. > *The fix:* the harmonic series adds tiny crumbs forever and still reaches $\infty$. Smallness > of terms doesn't guarantee the *accumulation* stays bounded. $a_n\to0$ is **necessary, not sufficient**. > > **Mistake 2: confusing the sequence $(a_n)$ with the series $\sum a_n$.** > *Why it feels right:* both come from the same numbers. *The fix:* $(a_n)\to 0$ asks about the > *terms*; $\sum a_n$ converging asks about the *running totals* $(s_N)$. Always go through $s_N$. > > **Mistake 3: using $\frac{a}{1-r}$ when $|r|\ge 1$.** > *Why it feels right:* the formula is so handy you forget its passport. *The fix:* the formula > only exists because $r^N\to 0$, which needs $|r|<1$. Outside that, there is no finite sum. > > **Mistake 4: thinking divergence means "goes to infinity".** > *Why it feels right:* most divergent examples blow up. *The fix:* $\sum(-1)^n$ has $s_N$ bouncing > $-1,0,-1,0,\dots$ — bounded but **no limit**, hence divergent. --- > [!recall]- Feynman: explain to a 12-year-old > Imagine you owe yourself candy and each day you get a piece: 1 candy, then half a candy, then a > quarter, then an eighth... Each day your *pile* grows a little. A "series" is the total pile. > But you can't wait forever, so instead you peek at the pile each night and ask: "Is my pile > getting closer and closer to some fixed amount, like 2 candies, and basically stopping there?" > If yes — the series **converges** to 2. If your pile keeps growing forever with no ceiling > (like adding $1, \tfrac12, \tfrac13, \tfrac14\dots$), or it keeps flip-flopping and never > settles, then there's **no final pile** — it **diverges**. The trick is: don't try to add > forever. Just watch the running total and see where it's headed. > [!mnemonic] Remember the logic chain > **"Partial sums Point the Path."** A series = sequence of **P**artial sums; its limit (if it > exists) is the sum. And for the term test: **"Terms not zero ⇒ no go."** --- ## Active recall > [!recall] Quick self-test > 1. Define $s_N$ and state exactly when $\sum a_n$ converges. > 2. Derive $s_N$ for a geometric series and find when it converges. > 3. Why does $a_n\to0$ NOT imply convergence? Give the example. > 4. Prove the $n$-th term test from $a_N = s_N - s_{N-1}$. #flashcards/maths What is a series defined in terms of? ::: The limit of its sequence of partial sums $s_N=\sum_{n=1}^N a_n$. $\sum a_n$ converges to $S$ means exactly what? ::: $\lim_{N\to\infty} s_N = S$, a finite number. $N$-th partial sum $s_N$ formula? ::: $s_N = a_1+a_2+\cdots+a_N = \sum_{n=1}^N a_n$. Geometric series $\sum_{n=1}^\infty ar^{n-1}$ converges when, and to what? ::: When $|r|<1$, to $a/(1-r)$. Closed form of geometric partial sum ($r\ne1$)? ::: $s_N = a(1-r^N)/(1-r)$. Why does $|r|<1$ matter for geometric convergence? ::: Because $r^N\to0$ only when $|r|<1$; otherwise $s_N$ has no finite limit. Sum of $\sum 1/(n(n+1))$ and why? ::: Equals $1$; telescopes to $s_N=1-1/(N+1)\to1$. State the $n$-th term test. ::: If $\sum a_n$ converges then $a_n\to0$; contrapositive: $a_n\not\to0\Rightarrow$ diverges. Is $a_n\to0$ sufficient for convergence? ::: No — harmonic series $\sum 1/n$ has $a_n\to0$ but diverges. Derive $a_N$ from partial sums. ::: $a_N = s_N - s_{N-1}$. Does divergence always mean $\to\infty$? ::: No — e.g. $\sum(-1)^n$ has partial sums bouncing, bounded but no limit. $\varepsilon$-definition of $s_N\to S$? ::: $\forall\varepsilon>0\,\exists M: N\ge M\Rightarrow |s_N-S|<\varepsilon$. --- ## Connections - [[Sequences — limits and convergence]] (a series IS a sequence in disguise — its partial sums) - [[Geometric series]] (the prototype convergent series, derived above) - [[n-th Term Test for Divergence]] - [[Harmonic series]] (key divergent counterexample) - [[Telescoping series]] - [[p-series and the Integral Test]] (deciding convergence beyond geometric) - [[Comparison Test]] · [[Ratio & Root Tests]] - [[Power series and radius of convergence]] (where geometric reasoning generalises) ## 🖼️ Concept Map ```mermaid flowchart TD A[Sequence a_n] -->|add terms| B[Series sum a_n] B -->|not defined as| C[Infinite addition] C -->|replaced by| D[Partial sum s_N] A -->|finite sum| D D -->|form new sequence| E[Sequence of partial sums] E -->|take limit N to inf| F[lim s_N] F -->|equals finite S| G[Series converges to S] F -->|DNE or infinite| H[Series diverges] G -->|formalised by| I[Epsilon-M definition] B -->|model example| J[Geometric series a r^n-1] J -->|telescoping trick| K[s_N = a 1-r^N over 1-r] K -->|if abs r less than 1| G ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, ek **series** ka matlab hai infinite numbers ko add karna — par hum infinite baar > addition kar hi nahi sakte. Toh trick yeh hai: hum **partial sums** banate hain. Pehle 1 term > jodo ($s_1$), phir 2 terms ($s_2$), phir 3 ($s_3$)... yeh running totals ka ek naya sequence > ban jaata hai. Ab sawaal simple ho gaya: kya yeh running totals ka sequence kisi ek fixed number > ke paas settle ho raha hai? Agar haan — series **converge** karti hai aur wahi number uska sum > hai. Agar nahi (ya infinity chala gaya, ya bounce karta rahe) — toh series **diverge** karti hai. > > Sabse important formula hai geometric series ka, aur hum usse rato mat — derive karo. $s_N$ likho, > $r$ se multiply karke subtract karo (beech ke saare terms cancel ho jaate hain), aur milta hai > $s_N = a(1-r^N)/(1-r)$. Ab limit lo: $|r|<1$ ho toh $r^N \to 0$, aur sum ban jaata hai $a/(1-r)$. > Agar $|r|\ge 1$ ho toh sum exist hi nahi karta — formula tab galat hai, yaad rakhna. > > Sabse bada **trap**: students sochte hain "terms zero ki taraf ja rahe hain, toh sum finite hoga". > Galat! Harmonic series $1+\tfrac12+\tfrac13+\cdots$ ke terms zero jaate hain phir bhi sum infinity > ho jaata hai. Toh $a_n\to0$ sirf *necessary* condition hai, *sufficient* nahi. Iska ulta — agar > $a_n$ zero nahi ja raha, toh series pakka diverge karti hai (yeh hai $n$-th term test). Exam mein > yeh sabse pehle check karo. > > Yaad rakhne ka tareeka: **"Partial sums Point the Path"** — hamesha $s_N$ ke through socho, kabhi > direct infinite sum ke through nahi. Tabhi convergence ka concept clear rahega. ![[audio/4.3.03-Series-—-partial-sums,-convergence-definition.mp3]]