Sigma (Σ) notation — evaluating, telescoping sums: Visual walkthrough

Before anything else, three plain words we will use constantly:

Step 1 — What "adding up" looks like as a row of blocks

WHAT. Forget formulas. Picture the sum $a_{1} + a_{2} + a_{3} + a_{4} + a_{5}$ as five separate blocks laid in a row, left to right.

WHY. Every claim on this page is really a claim about finite addition, and finite addition has one magic property we will lean on: you may reorder and regroup the blocks however you like and the total never changes (this is what "commutative and associative" mean). Seeing them as movable blocks makes that obvious.

PICTURE. Each block sits over its index $k$ on the axis. Nothing has cancelled yet — this is the raw material.

Figure — Sigma (Σ) notation — evaluating, telescoping sums

$block over k = 1 a_{1} + k = 2 a_{2} + k = 3 a_{3} + k = 4 a_{4} + k = 5 a_{5}$

Every symbol here: $a_{k}$ is one block, the little number $k$ underneath is which block. The $+$ signs just say "put them all on the same scale and read the total height."

Step 2 — Build a special term: a difference of the SAME rule at two spots

WHAT. Instead of letting each term be some random number, we choose to make it $a_{k} = f (k) - f (k + 1) .$

WHY. This is the whole game. If our building blocks all secretly come from one rule $f$ evaluated at neighbouring inputs, the blocks will overlap in a way we can exploit. Read the two pieces:

$f (k)$ — the rule at the spot we're standing on. Call it the head.
$- f (k + 1)$ — minus the rule at the next spot along. Call it the tail.

PICTURE. Draw each term as a head bar (pointing up, at position $k$ ) glued to a tail bar (pointing down, at position $k + 1$ ). The tail already leans one step to the right — that lean is what makes them line up later.

$a_{k} = head: rule here f (k) - tail: rule at NEXT spot f (k + 1)$

Why the same $f$ in both pieces, and why the input differs by exactly $1$ ? Because we want the tail of one term to be the exact negative of the head of the following term. Next step shows that collision.

Step 3 — Line the terms up and watch neighbours collide

WHAT. Write out the first few terms one under another, keeping each head over its own $k$ and each tail over $k + 1$ :

k=1:&\quad f(1) - f(2)\\ k=2:&\quad f(2) - f(3)\\ k=3:&\quad f(3) - f(4) \end{aligned}$$ **WHY.** Look down the columns. The tail of row $1$ is $-f(2)$; the head of row $2$ is $+f(2)$. They are the **same number with opposite signs**, so their sum is $0$. That is why we insisted on "same rule, next spot" back in Step 2 — it *manufactures* these matched $\pm f(2)$ pairs. **PICTURE.** The down-pointing tail bar of one term sits at the *exact* position and length of the up-pointing head bar of the next term. Overlaid, they erase — a $+f(2)$ block and a $-f(2)$ block stack to zero height. ![[deepdives/dd-maths-3.3.07-d2-s03.png]] $$f(1)\;\underbrace{-\,f(2)\;+\,f(2)}_{=\,0}\;\underbrace{-\,f(3)\;+\,f(3)}_{=\,0}\;-\,f(4)$$ Each brace is a domino knocking its neighbour flat. --- ## Step 4 — The full chain: everything in the middle annihilates **WHAT.** Now do it for all $n$ terms, from $k=1$ to $k=n$: $$\big[f(1)-f(2)\big]+\big[f(2)-f(3)\big]+\dots+\big[f(n)-f(n{+}1)\big].$$ **WHY.** By Step 1 we may drop the brackets and reorder freely. Every internal value $f(2), f(3), \dots, f(n)$ appears **twice**: once as a positive head, once as a negative tail. Twice-appearing with opposite signs $\Rightarrow$ each cancels itself. The only values that appear *only once* live at the two edges of the chain, where there is no neighbour to cancel against. **PICTURE.** A long strip of head/tail bars; a sweep of red crosses cancels every interior pair, leaving just the far-left head and the far-right tail standing tall. ![[deepdives/dd-maths-3.3.07-d2-s04.png]] $$\underbrace{f(1)}_{\text{no left neighbour → survives}}\;\underbrace{-f(2)+f(2)-\dots+f(n)}_{\text{all cancel in pairs}}\;\underbrace{-\,f(n{+}1)}_{\text{no right neighbour → survives}}$$ --- ## Step 5 — Read off the survivors: the telescoping formula **WHAT.** Two lonely bars remain: the very first head $f(1)$ and the very last tail $-f(n+1)$. Therefore $$\boxed{\;\sum_{k=1}^{n}\big[f(k)-f(k+1)\big] = f(1) - f(n+1)\;}$$ **WHY.** This is just Step 4 written cleanly. Term-by-term: - $f(1)$ — the head of the *first* term, which had no earlier tail to kill it. - $-f(n+1)$ — the tail of the *last* term, which had no later head to kill it. - Everything strictly between $1$ and $n+1$ cancelled. And if we start at $m$ instead of $1$, the same reasoning gives $$\sum_{k=m}^{n}\big[f(k)-f(k+1)\big] = f(m) - f(n+1).$$ **PICTURE.** The collapsing telescope: a many-jointed tube slides shut, and only the two end rings poke out — labelled $f(m)$ and $f(n+1)$. ![[deepdives/dd-maths-3.3.07-d2-s05.png]] > [!mnemonic] The one line to remember > **First head minus last tail.** Start at $m$, stop at $n$: the answer is $f(m) - f(n+1)$. --- ## Step 6 — A real term: manufacture the difference with partial fractions **WHAT.** In the wild you rarely *see* $f(k)-f(k+1)$; you see something like $\dfrac{1}{k(k+1)}$. We must *turn it into* a difference. Using [[Partial Fractions]]: $$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}.$$ **WHY.** We *want* a matched head/tail, so we look for a rule $f$ with $f(k)=\tfrac1k$. Then $f(k+1)=\tfrac1{k+1}$, and indeed $\tfrac1k-\tfrac1{k+1}$ equals our fraction — check by common denominator: $\tfrac{(k+1)-k}{k(k+1)}=\tfrac{1}{k(k+1)}$. Partial fractions is the *tool that reveals* the hidden telescoping shape; without it the collapse is invisible. **PICTURE.** The single tall bar $\tfrac{1}{k(k+1)}$ splits into a head $\tfrac1k$ and a tail $-\tfrac{1}{k+1}$ — the exact head/tail shape of Step 2. ![[deepdives/dd-maths-3.3.07-d2-s06.png]] Now apply Step 5 with $f(k)=\tfrac1k$, from $k=1$ to $n$: $$\sum_{k=1}^{n}\frac{1}{k(k+1)} = f(1)-f(n+1) = 1 - \frac{1}{n+1} = \frac{n}{n+1}.$$ **Check $n=1$:** left side $\tfrac{1}{1\cdot2}=\tfrac12$; right side $\tfrac{1}{2}$. ✓ --- ## Step 7 — The degenerate & edge cases (so nothing surprises you) **WHAT.** Four scenarios that break the "obvious" picture — each shown so you never hit an unshown one. **WHY / PICTURE.** See the four panels: ![[deepdives/dd-maths-3.3.07-d2-s07.png]] - **Single term, $n=m$.** The sum has *one* block: $f(m)-f(m+1)$. Formula: $f(m)-f((m)+1)=f(m)-f(m+1)$ — consistent. Nothing cancels because there is no neighbour; *both* head and tail survive as that single term. (Recall from the parent: number of terms $=n-m+1=1$.) - **Empty sum, $n=m-1$.** Zero terms. By convention an empty sum is $0$, and the formula gives $f(m)-f(m)=0$. ✓ Still consistent. - **Gap-2 terms, $\dfrac{1}{k(k+2)}$.** Here $f(k)=\tfrac1k$ but the tail leans **two** spots right: $\tfrac12\big(\tfrac1k-\tfrac1{k+2}\big)$. A tail at $k+2$ only meets a head *two rows down*, so the cancellation is *delayed*. That leaves **two** heads ($\tfrac11,\tfrac12$) and **two** tails ($-\tfrac1{n+1},-\tfrac1{n+2}$) uncancelled: $$\sum_{k=1}^{n}\frac{1}{k(k+2)} = \frac12\Big(1+\frac12-\frac{1}{n+1}-\frac{1}{n+2}\Big).$$ **Check $n=1$:** left $\tfrac{1}{1\cdot3}=\tfrac13$; right $\tfrac12\big(\tfrac32-\tfrac12-\tfrac13\big)=\tfrac12\cdot\tfrac23=\tfrac13$. ✓ - **Limiting behaviour, $n\to\infty$.** For $\tfrac1{k(k+1)}$ the answer $1-\tfrac1{n+1}$ marches toward $1$: the last tail $\tfrac1{n+1}$ shrinks to nothing. This is why telescoping is the friend of [[Convergence of Series]] — it hands you the *exact* partial sum, so the limit is a glance away. > [!mistake] The trap in the gap-2 case > **Feels right:** "gap of 2, so still one survivor each end." **Wrong:** the delay by one row means > **two** heads and **two** tails survive. Always *write three or four rows out* and see which values > never meet a partner before trusting the collapse. > [!mistake] Re-index without moving the limits > If you rename $j=k+1$ inside $\sum_{k=1}^{n}\tfrac{1}{k+1}$, the limits **must** move to > $j=2\ldots n+1$. The summand looks simpler so people freeze the bounds — but the bounds travel with > the substitution. (Same warning as the parent — it bites here constantly.) --- ## The one-picture summary ![[deepdives/dd-maths-3.3.07-d2-s08.png]] One diagram, the whole derivation: each term is a head bar plus a leaning tail bar; interior heads and tails annihilate in matched $\pm$ pairs; only $f(1)$ (far left) and $-f(n+1)$ (far right) survive — the telescope slid shut. > [!formula] Everything on this page, compressed > $$\sum_{k=m}^{n}\big[f(k)-f(k+1)\big] = f(m) - f(n+1)$$ > **Recipe:** (1) spot a fraction with factorable denominator → (2) split with [[Partial Fractions]] > into $f(k)-f(k+1)$ → (3) survivors are first head $f(m)$ minus last tail $f(n+1)$. > [!recall]- Feynman retelling — the whole walkthrough in plain words > Imagine a row of people, each holding a card that says "+ my number" in the right hand and > "– the next person's number" in the left hand. When everyone shows their cards, person 2's "+3" > card and person 1's "–3" card are pressed together and cancel to zero. This happens all the way down > the line: every inside number is *both* somebody's plus and somebody's minus, so it disappears. Only > the very first person's plus and the very last person's minus have no partner — they stay. So a huge > line of adding becomes just "first minus last." The trick to *use* it: when you're handed an ugly > fraction, split it (partial fractions) so each person is holding exactly a "+ my rule" and a > "– next rule." Then the row collapses like a folding telescope. > [!recall]- Quick self-test > Why does $f(2)$ vanish in a $\sum_{k=1}^n[f(k)-f(k+1)]$? ::: It appears twice — as $+f(2)$ (head of term $k=2$) and $-f(2)$ (tail of term $k=1$) — opposite signs, so they cancel. > What survives, and why only those? ::: $f(1)$ and $-f(n+1)$; the first head has no earlier tail and the last tail has no later head to cancel with. > Why does $\frac{1}{k(k+2)}$ leave two survivors each end? ::: The tail leans 2 spots right, so cancellation is delayed one row — two heads and two tails never meet a partner. > First step when you meet $\frac{1}{k(k+1)}$? ::: Partial fractions → $\frac1k-\frac1{k+1}$, revealing the $f(k)-f(k+1)$ shape. --- ## Connections - [[Partial Fractions]] — the tool of Step 6 that *creates* the head/tail shape. - [[Method of Differences]] — the general name for this collapse; telescoping is its finite case. - [[Convergence of Series]] — Step 7's limit: exact partial sums make limits trivial. - [[Mathematical Induction]] — an alternative rigorous proof of the same boxed formula. - [[Arithmetic Progressions]] — where linearity + $\sum k$ (Step 1's reordering) first appears.