Sigma (Σ) notation — evaluating, telescoping sums: Worked examples

You already met the parent topic: what $\sum$ means, the three algebra rules, Gauss's trick, and the "dominoes that cancel" idea of telescoping. This page does one thing: it throws every kind of sum at you and works each to the end, so you never meet a case in an exam you haven't already seen tamed.

The scenario matrix

Here is the full menu of cases this topic can throw at you. Every cell is covered by a worked example below.

#	Case class	What makes it tricky	Example
A	Linear term $ak + b$	Just linearity + Gauss	Ex 1
B	Quadratic / cubic term	Needs $\sum k^{2}$ , $\sum k^{3}$	Ex 2
C	Gap-1 telescope $\frac{1}{k ( k + 1 )}$	Neighbours cancel, 1 survivor each end	Ex 3
D	Gap-2 telescope $\frac{1}{k ( k + 2 )}$	Delayed cancel, 2 survivors each end	Ex 4
E	Shifted lower limit (starts at $m \neq = 1$ )	Must re-index or subtract a head	Ex 5
F	Limiting value $n \to \infty$	Telescoping → convergence	Ex 6
G	Degenerate / empty & single-term	$n < m$ gives $0$ ; $n = m$ gives one term	Ex 7
H	Real-world word problem	Translate story → Σ	Ex 8
I	Exam twist: -difference telescope	Rationalise to expose $f (k) - f (k + 1)$	Ex 9
J	Sign / alternating trap	$(- 1)^{k}$ is NOT a constant to pull out	Ex 10

The worked examples

Ex 1 — Case A: a linear term

Worked example Evaluate

$k = 1 \sum 30 (5 k - 4)$ . Forecast: Guess the ballpark first. The terms run from $5 (1) - 4 = 1$ up to $5 (30) - 4 = 146$ , so ~30 terms averaging ~73 → around $2000$ . Hold that thought.

Step 1. Split by linearity: $\sum_{k = 1}^{30} (5 k - 4) = 5 \sum_{k = 1}^{30} k - \sum_{k = 1}^{30} 4$ . Why this step? $\sum$ distributes over $+$ and $-$ , and constants factor out — this turns one unfamiliar sum into two known pieces.

Step 2. $\sum_{k = 1}^{30} k = \frac{30 \cdot 31}{2} = 465$ . Why this step? Gauss's formula $\sum_{k = 1}^{n} k = \frac{n ( n + 1 )}{2}$ .

Step 3. $\sum_{k = 1}^{30} 4 = 30 \cdot 4 = 120$ (a constant added $30$ times). Why this step? The rule $\sum_{k = 1}^{n} c = n c$ .

Step 4. Combine: $5 (465) - 120 = 2325 - 120 = 2205$ .

Verify: In the ballpark ( $\approx 2000$ ) ✓. Cross-check via Arithmetic Progressions: an AP with first term $1$ , last term $146$ , $30$ terms sums to $\frac{30}{2} (1 + 146) = 15 \cdot 147 = 2205$ ✓.

Ex 2 — Case B: quadratic + cubic in one term

Ex 3 — Case C: the gap-1 telescope

Worked example Evaluate

$k = 1 \sum n \frac{1}{k ( k + 1 )}$ and give the value for $n = 4$ . Forecast: Denominators grow fast, so later terms are tiny. The sum should creep toward but never reach $1$ .

Step 1. Partial Fractions: write $\frac{1}{k ( k + 1 )} = \frac{A}{k} + \frac{B}{k + 1}$ . Clearing gives $1 = A (k + 1) + B k$ ; setting $k = 0 \Rightarrow A = 1$ , $k = - 1 \Rightarrow B = - 1$ . So $\frac{1}{k ( k + 1 )} = \frac{1}{k} - \frac{1}{k + 1}$ . Why this step? This is exactly the shape $f (k) - f (k + 1)$ with $f (k) = \frac{1}{k}$ — the fuel for telescoping.

Step 2. Look at the figure below: each red tail cancels the next blue head.

Why this step? Seeing the cancellation prevents the classic error of thinking "nothing simplifies."

Step 3. By the telescoping formula $\sum_{k = 1}^{n} [f (k) - f (k + 1)] = f (1) - f (n + 1) = 1 - \frac{1}{n + 1} = \frac{n}{n + 1}$ .

Verify: For $n = 4$ : formula gives $\frac{4}{5} = 0.8$ . Add directly: $\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} = 0.5 + 0.1 \overline{6} + 0.083 \overline{3} + 0.05 = 0.8$ ✓. And it stays below $1$ as forecast ✓.

Ex 4 — Case D: the gap-2 telescope (two survivors each end)

Worked example Evaluate

$k = 1 \sum n \frac{1}{k ( k + 2 )}$ and give the value for $n = 3$ . Forecast: The gap between factors is $2$ , so a term won't cancel its immediate neighbour — it cancels the one two rows later. Expect two leftover heads and two leftover tails.

Step 1. Partial fractions: $\frac{1}{k ( k + 2 )} = \frac{1}{2} (\frac{1}{k} - \frac{1}{k + 2})$ . Why this step? Now $f (k) = \frac{1}{k}$ but the shift is by $2$ , so cancellation is delayed by one term.

Step 2. The figure shows which terms cancel and which survive.

Why this step? A gap-2 diagram makes "why two survivors" visible instead of memorised.

Step 3. Surviving heads: $\frac{1}{1} + \frac{1}{2}$ . Surviving tails: $- \frac{1}{n + 1} - \frac{1}{n + 2}$ . So $\sum_{k = 1}^{n} \frac{1}{k ( k + 2 )} = \frac{1}{2} (1 + \frac{1}{2} - \frac{1}{n + 1} - \frac{1}{n + 2}) = \frac{1}{2} (\frac{3}{2} - \frac{1}{n + 1} - \frac{1}{n + 2}) .$

Verify: For $n = 3$ : formula gives $\frac{1}{2} (\frac{3}{2} - \frac{1}{4} - \frac{1}{5}) = \frac{1}{2} \cdot \frac{30 - 5 - 4}{20} = \frac{1}{2} \cdot \frac{21}{20} = \frac{21}{40} = 0.525$ . Direct: $\frac{1}{3} + \frac{1}{8} + \frac{1}{15} = 0.3 \overline{3} + 0.125 + 0.0 \overline{6} = 0.525$ ✓.

Ex 5 — Case E: a shifted lower limit

Worked example Evaluate

$k = 5 \sum 20 \frac{1}{k ( k + 1 )}$ . Forecast: Same telescoping structure as Ex 3, but starting at $k = 5$ . Only the survivors change — heads move to $f (5)$ .

Step 1. Use the general telescoping formula $\sum_{k = m}^{n} [f (k) - f (k + 1)] = f (m) - f (n + 1)$ with $f (k) = \frac{1}{k}$ , $m = 5$ , $n = 20$ . Why this step? Never assume the lower limit is $1$ . Here every interior term from $k = 5$ to $20$ still cancels; only the first head $f (5)$ and last tail $f (21)$ remain.

Step 2. Value $= f (5) - f (21) = \frac{1}{5} - \frac{1}{21}$ .

Step 3. Common denominator: $\frac{21 - 5}{105} = \frac{16}{105}$ .

Verify (the re-index safety check): Compute instead as [whole from 1 to 20] minus [head from 1 to 4]: $(1 - \frac{1}{21}) - (1 - \frac{1}{5}) = \frac{1}{5} - \frac{1}{21} = \frac{16}{105}$ ✓. Two routes agree — this guards against the [!mistake] of forgetting to move both limits when shifting the index.

Ex 6 — Case F: the limiting value ( $n \to \infty$ )

Ex 7 — Case G: degenerate and single-term sums

Worked example Evaluate (a)

$k = 1 \sum 0 k^{2}$ , (b) $k = 7 \sum 7 (3 k - 1)$ , (c) $k = 1 \sum n \frac{1}{k ( k + 1 )}$ at $n = 1$ . Forecast: (a) upper limit below lower ⇒ nothing to add ⇒ $0$ . (b) one lonely term. (c) should match a single fraction.

Step 1 (a). When $n < m$ the sum is empty (the convention: an empty sum is $0$ ). Why this step? "Add all integers from $1$ up to $0$ " describes no integers, and the identity element of addition is $0$ . Also the formula would give $\frac{0 \cdot 1 \cdot 1}{6} = 0$ , consistent.

Step 2 (b). $\sum_{k = 7}^{7} (3 k - 1)$ has exactly $7 - 7 + 1 = 1$ term: $3 (7) - 1 = 20$ . Why this step? The count formula $n - m + 1$ still works and equals $1$ ; only $k = 7$ is included.

Step 3 (c). $\sum_{k = 1}^{1} \frac{1}{k ( k + 1 )} = \frac{1}{1 \cdot 2} = \frac{1}{2}$ ; the closed form $\frac{n}{n + 1} = \frac{1}{2}$ agrees. Why this step? Always sanity-test a closed form at the smallest legal $n$ .

Verify: (a) $= 0$ ✓; (b) $= 20$ ✓; (c) $= 0.5$ matches $\frac{1}{2}$ ✓.

Ex 8 — Case H: a word problem

Worked example A staircase-lighting art piece has lamp

$k$ shining brightness $\frac{6}{k ( k + 3 )}$ units, for lamps $k = 1$ to $k = n$ . What is the total brightness, and its ceiling as more lamps are added forever? Forecast: Gap between factors is $3$ , so expect three survivors at each end. As $n \to \infty$ the tails die, leaving a finite ceiling.

Step 1. Factor out and split: $\frac{6}{k ( k + 3 )} = 6 \cdot \frac{1}{3} (\frac{1}{k} - \frac{1}{k + 3}) = 2 (\frac{1}{k} - \frac{1}{k + 3})$ . Why this step? $\frac{1}{k ( k + d )} = \frac{1}{d} (\frac{1}{k} - \frac{1}{k + d})$ ; here $d = 3$ .

Step 2. Gap-3 telescope: heads $\frac{1}{1} + \frac{1}{2} + \frac{1}{3}$ survive; tails $- \frac{1}{n + 1} - \frac{1}{n + 2} - \frac{1}{n + 3}$ survive. So $Total = 2 (\frac{11}{6} - \frac{1}{n + 1} - \frac{1}{n + 2} - \frac{1}{n + 3}),$ since $1 + \frac{1}{2} + \frac{1}{3} = \frac{6 + 3 + 2}{6} = \frac{11}{6}$ . Why this step? Three-step gap means the first three heads and last three tails never find a partner.

Step 3. As $n \to \infty$ all three tails $\to 0$ , giving ceiling $2 \cdot \frac{11}{6} = \frac{11}{3} \approx 3.667$ units.

Verify: For a modest $n = 3$ : total $= 2 (\frac{11}{6} - \frac{1}{4} - \frac{1}{5} - \frac{1}{6}) = 2 \cdot \frac{110 - 15 - 12 - 10}{60} = 2 \cdot \frac{73}{60} = \frac{73}{30} \approx 2.433$ ; direct $\frac{6}{4} + \frac{6}{10} + \frac{6}{18} = 1.5 + 0.6 + 0.3 \overline{3} = 2.4 \overline{3}$ ✓. Below the ceiling $3.667$ as expected ✓.

Ex 9 — Case I: the exam twist (a square-root telescope)

Worked example Evaluate

$k = 1 \sum n \frac{1}{k + k + 1}$ and give the value for $n = 8$ . Forecast: No obvious $f (k) - f (k + 1)$ ... until we make one. This is the classic "rationalise to reveal the telescope" twist.

Step 1. Multiply top and bottom by the conjugate $k + 1 - k$ : $\frac{1}{k + k + 1} \cdot \frac{k + 1 - k}{k + 1 - k} = \frac{k + 1 - k}{( k + 1 ) - k} = k + 1 - k .$ Why this step? Rationalising turns a stubborn fraction into a clean difference $g (k + 1) - g (k)$ with $g (k) = k$ . That is a telescope (run in the "+ then −" order).

Step 2. Sum: $\sum_{k = 1}^{n} (k + 1 - k) = n + 1 - 1 = n + 1 - 1$ . Why this step? This is $\sum [g (k + 1) - g (k)] = g (n + 1) - g (1)$ — the telescoping formula with the sign flipped, so the last head and first tail survive.

Step 3. So the value is $n + 1 - 1$ .

Verify: For $n = 8$ : $9 - 1 = 3 - 1 = 2$ . Direct sum of $\frac{1}{k + k + 1}$ for $k = 1..8$ equals $2$ numerically ✓.

Ex 10 — Case J: the sign / alternating trap

Recall One-line summary of the whole matrix

Every sum is "rewrite the term into something addable." Polynomials → split into $\sum k, \sum k^{2}, \sum k^{3}$ . Fractions with factored denominators → partial fractions → telescope (gap $d$ ⇒ $d$ survivors each end). Roots → rationalise → telescope. Signs like $(- 1)^{k}$ stay inside. Empty sums are $0$ ; check closed forms at the smallest $n$ ; take $n \to \infty$ by killing the tail.

Quick self-test

Gap-

d

telescope

\frac{1}{k ( k + d )}

leaves how many survivors each end?

Exactly

d

What is an empty sum

\sum_{k = 1}^{0} a_{k}

equal to?

0

(the identity of addition).

How do you telescope

\frac{1}{k + k + 1}

Rationalise with the conjugate →

k + 1 - k

Can

(- 1)^{k}

be pulled out of a Σ?

No — it varies with

k

, so it is not a constant.

Limit of

\sum_{k = 1}^{n} \frac{1}{k ( k + 1 )}

n \to \infty

1

(since

\frac{n}{n + 1} \to 1

Connections

Partial Fractions — the engine that creates every telescoping split here.
Method of Differences — Ex 3–5, 8, 9 are all instances of it.
Arithmetic Progressions — cross-check for the linear Ex 1.
Convergence of Series — Ex 6 and Ex 8's ceiling come straight from exact partial sums.
Mathematical Induction — how to prove the closed forms rigorously.
Binomial Theorem — where $(- 1)^{k}$ sums like Ex 10 reappear.