Telescoping series: Worked examples — Maths deep dive

Before we start, one word we lean on constantly: a partial sum $S_{N}$ just means "add only the first $N$ terms and stop." Watching $S_{N}$ as $N$ grows is how we decide what the infinite sum is — see Sequence of Partial Sums.

The scenario matrix

Every telescoping problem is one (or a blend) of these cells:

Cell	What changes	Danger	Example
C1 Gap 1	$a_{n} = b_{n} - b_{n + 1}$ , clean	none — the textbook case	Ex 1
C2 Gap 2	$a_{n} = b_{n} - b_{n + 2}$	2 survivors each end	Ex 2
C3 Shifted start	series begins at $n = 3$ , not $1$	lead term is $b_{3}$ , not $b_{1}$	Ex 3
C4 Non-vanishing tail	$b_{n} \to L \neq = 0$	answer is $b_{1} - L$ , not $b_{1}$	Ex 4
C5 Root / conjugate form	difference hidden by	must rationalise to see it	Ex 5
C6 Product/log form	$\prod$ or $ln$ , second difference	telescopes as ratios	Ex 6
C7 Word problem	real quantities (medicine, tiles)	translate to $b_{n} - b_{n + 1}$	Ex 7
C8 Exam twist	diverges / degenerate	recognise it does not collapse to a finite number	Ex 8

We now hit each cell.

C1 — the clean gap-1 case

Worked example Ex 1 (cell C1):

$n = 1 \sum \infty \frac{1}{( 2 n - 1 ) ( 2 n + 1 )}$ Forecast: guess — is the sum bigger or smaller than $1$ ? (It's $\frac{1}{2}$ ; hold that thought.)

Step 1. Split with Partial Fraction Decomposition: $\frac{1}{( 2 n - 1 ) ( 2 n + 1 )} = \frac{A}{2 n - 1} + \frac{B}{2 n + 1}$ . Why this step? We want the shape $b_{n} - b_{n + 1}$ ; partial fractions is the only tool that manufactures that difference from a product in the denominator.

Step 2. Clear denominators: $1 = A (2 n + 1) + B (2 n - 1)$ . Put $n = \frac{1}{2} \Rightarrow 1 = 2 A \Rightarrow A = \frac{1}{2}$ . Put $n = - \frac{1}{2} \Rightarrow 1 = - 2 B \Rightarrow B = - \frac{1}{2}$ . $\frac{1}{( 2 n - 1 ) ( 2 n + 1 )} = \frac{1}{2} (\frac{1}{2 n - 1} - \frac{1}{2 n + 1}) .$ Why this step? Now $b_{n} = \frac{1}{2 n - 1}$ and $b_{n + 1} = \frac{1}{2 n + 1}$ — the term is exactly $\frac{1}{2} (b_{n} - b_{n + 1})$ , a gap-1 telescope.

Step 3. Partial sum: $S_{N} = \frac{1}{2} (b_{1} - b_{N + 1}) = \frac{1}{2} (1 - \frac{1}{2 N + 1})$ . Why this step? "First minus last" — the master result from the parent note.

Step 4. Limit: $\frac{1}{2 N + 1} \to 0$ , so $\sum = \frac{1}{2} (1 - 0) = \frac{1}{2}$ . See Limits of Sequences.

Verify: add first 3 terms: $\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} = 0.333 + 0.0667 + 0.0286 = 0.4286$ , and $\frac{1}{2} (1 - \frac{1}{7}) = 0.4286$ ✓. Matches, and it's below $\frac{1}{2}$ as it should be while $N$ is finite.

C2 — gap 2 (two survivors each end)

Worked example Ex 2 (cell C2):

$n = 1 \sum \infty \frac{2}{n ( n + 2 )}$ Forecast: because two terms survive at each end, expect the answer to be a "sum of two boundary pieces," not one.

Step 1. $\frac{2}{n ( n + 2 )} = \frac{A}{n} + \frac{B}{n + 2}$ . Clear: $2 = A (n + 2) + B n$ . $n = 0 \Rightarrow A = 1$ ; $n = - 2 \Rightarrow B = - 1$ . So term $= \frac{1}{n} - \frac{1}{n + 2}$ . Why this step? Same partial-fraction engine, but note the denominators differ by 2, not 1.

Step 2. This is $b_{n} - b_{n + 2}$ with $b_{n} = \frac{1}{n}$ — a gap-2 telescope. Write 5 terms: $S_{N} = (1 - \frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5}) + (\frac{1}{4} - \frac{1}{6}) + \dots$ Why this step? The parent note's Mistake 2: with a gap of 2 you must physically see which pieces meet. The $- \frac{1}{3}$ waits for the third term's $+ \frac{1}{3}$ , not the next one.

Step 3. After cancellation the survivors are the first two " $+$ " pieces ( $1, \frac{1}{2}$ ) and the last two " $-$ " pieces ( $- \frac{1}{N + 1}, - \frac{1}{N + 2}$ ): $S_{N} = 1 + \frac{1}{2} - \frac{1}{N + 1} - \frac{1}{N + 2} .$ Why this step? Gap $k$ ⟹ $k$ survivors each side. Here $k = 2$ .

Step 4. Limit: both tails $\to 0$ , so $\sum = 1 + \frac{1}{2} = \frac{3}{2}$ .

Verify: first 4 terms $= 0.6667 + 0.25 + 0.1333 + 0.0833 = 1.1333$ ; formula $S_{4} = 1.5 - \frac{1}{5} - \frac{1}{6} = 1.1333$ ✓.

C3 — shifted starting index

C4 — the tail does NOT vanish

Worked example Ex 4 (cell C4):

$n = 1 \sum \infty (\frac{n}{n + 1} - \frac{n + 1}{n + 2})$ Forecast: it's tempting to say the answer is the first " $+$ " piece $\frac{1}{2}$ . Resist — the tail here does not go to zero!

Step 1. It's already a difference: $b_{n} = \frac{n}{n + 1}$ , and $a_{n} = b_{n} - b_{n + 1}$ . Gap-1. Why this step? No partial fractions needed — the telescoping form is handed to us; just identify $b_{n}$ .

Step 2. Partial sum: $S_{N} = b_{1} - b_{N + 1} = \frac{1}{2} - \frac{N + 1}{N + 2}$ . Why this step? Master result again — but now watch the second piece.

Step 3. Limit of the tail: $N \to \infty lim \frac{N + 1}{N + 2} = 1$ , not $0$ . Why this step? $\frac{N + 1}{N + 2} = \frac{1 + 1/ N}{1 + 2/ N} \to \frac{1}{1} = 1$ . This is the whole point of cell C4.

Step 4. So $\sum = \frac{1}{2} - 1 = - \frac{1}{2}$ — a negative sum, and not $b_{1} = \frac{1}{2}$ .

Verify: $S_{3} = \frac{1}{2} - \frac{4}{5} = - 0.3$ ; and $\frac{1}{2} - \frac{N + 1}{N + 2}$ at $N = 3$ is $\frac{1}{2} - 0.8 = - 0.3$ ✓. As $N \to \infty$ it heads to $- 0.5$ . ✓

C5 — hidden difference behind a root

Worked example Ex 5 (cell C5):

$n = 1 \sum \infty \frac{1}{n + n + 1}$ Forecast: there's no obvious difference here. Guess whether this even converges — the terms shrink like $\frac{1}{2 n}$ , which... look closely.

Step 1. Rationalise: multiply top and bottom by $n + 1 - n$ : $\frac{1}{n + n + 1} \cdot \frac{n + 1 - n}{n + 1 - n} = \frac{n + 1 - n}{( n + 1 ) - n} = n + 1 - n .$ Why this step? The conjugate trick is our tool: $(n + 1 + n) (n + 1 - n) = (n + 1) - n = 1$ turns the ugly fraction into a clean difference $b_{n + 1} - b_{n}$ with $b_{n} = n$ .

Step 2. So $a_{n} = n + 1 - n = b_{n + 1} - b_{n}$ (a rising telescope). Partial sum: $S_{N} = N + 1 - 1 = N + 1 - 1.$ Why this step? Write it out: $(2 - 1) + (3 - 2) + \dots + (N + 1 - N)$ — the middle cancels, leaving last minus first.

Step 3. Limit: $N + 1 - 1 \to \infty$ . The series diverges. Why this step? Here $b_{n} = n \to \infty$ , so no finite sum exists — the telescope keeps extending, it never folds shut. Telescoping tells us the exact behaviour either way (see Convergence Tests for Series).

Verify: $S_{3} = (2 - 1) + (3 - 2) + (4 - 3) = 4 - 1 = 1$ ; formula $4 - 1 = 1$ ✓. Growing without bound. ✓ $diverges$

C6 — product / log form (second difference)

Worked example Ex 6 (cell C6):

$n = 2 \prod \infty (1 - \frac{1}{n ^{2}})$ Forecast: this is a product, but products of " $1 - tiny$ " tend to a finite limit. Guess: $\frac{1}{2}$ ? $1$ ? $0$ ?

Step 1. Factor each term: $1 - \frac{1}{n ^{2}} = \frac{( n - 1 ) ( n + 1 )}{n ^{2}}$ . Why this step? Difference of squares exposes the ratio structure that will telescope in a product.

Step 2. Partial product to $N$ : $P_{N} = \prod_{n = 2}^{N} \frac{( n - 1 ) ( n + 1 )}{n ^{2}} = shifts down (n = 2 \prod N \frac{n - 1}{n}) \cdot shifts up (n = 2 \prod N \frac{n + 1}{n}) .$ Why this step? Splitting into two telescoping products is the multiplicative version of $b_{n} - b_{n + 1}$ : consecutive ratios cancel.

Step 3. Each product telescopes: $\prod_{n = 2}^{N} \frac{n - 1}{n} = \frac{1}{2} \cdot \frac{2}{3} \dots \frac{N - 1}{N} = \frac{1}{N}, \prod_{n = 2}^{N} \frac{n + 1}{n} = \frac{3}{2} \cdot \frac{4}{3} \dots \frac{N + 1}{N} = \frac{N + 1}{2} .$ Why this step? In a telescoping product, numerator of one factor cancels the denominator of the next — the middle collapses exactly as in a sum, just multiplied.

Step 4. Multiply: $P_{N} = \frac{1}{N} \cdot \frac{N + 1}{2} = \frac{N + 1}{2 N} \to \frac{1}{2}$ . Why this step? $\frac{N + 1}{2 N} = \frac{1 + 1/ N}{2} \to \frac{1}{2}$ . (Take $ln$ and you recover the parent's $\sum ln (1 - \frac{1}{n ^{2}}) = - ln 2$ , since $ln \frac{1}{2} = - ln 2$ ✓.)

Verify: $P_{4} = (1 - \frac{1}{4}) (1 - \frac{1}{9}) (1 - \frac{1}{16}) = 0.75 \cdot 0.8889 \cdot 0.9375 = 0.625$ ; and $\frac{5}{8} = 0.625$ ✓. Heads to $\frac{1}{2}$ .

C7 — real-world word problem

Worked example Ex 7 (cell C7): drug clearance

A patient's kidneys process a drug so that on day $n$ the amount of drug removed equals $R_{n} = \frac{100}{n} - \frac{100}{n + 1}$ milligrams (a manufactured telescoping model). How many mg are removed in total over all time? Forecast: since day-1 removal already includes the biggest piece, guess the total is near $50$ mg.

Step 1. Recognise $R_{n} = b_{n} - b_{n + 1}$ with $b_{n} = \frac{100}{n}$ mg. Gap-1 telescope. Why this step? A word problem is solved the instant you spot the $b_{n} - b_{n + 1}$ shape hiding in the units.

Step 2. Total over $N$ days: $S_{N} = b_{1} - b_{N + 1} = 100 - \frac{100}{N + 1}$ mg. Why this step? First minus last — the "middle" days' contributions cancel in the running total.

Step 3. As $N \to \infty$ : total removed $= 100 - 0 = 100 mg$ . Why this step? $\frac{100}{N + 1} \to 0$ , so the body eventually clears $100$ mg total.

Verify (units + numbers): each $R_{n}$ is mg $-$ mg $=$ mg ✓. After 4 days: $100 - \frac{100}{5} = 80$ mg removed; direct sum $R_{1} + R_{2} + R_{3} + R_{4} = 50 + 16. \overline{6} + 8. \overline{3} + 5 = 80$ mg ✓. My forecast of " $\approx 50$ " was wrong — the tail contributes a lot; total is $100$ mg.

C8 — exam twist: the trap that diverges

Worked example Ex 8 (cell C8):

$n = 1 \sum \infty (\frac{1}{n} - \frac{1}{n + 1} + \frac{1}{n})$ — spot the trick Forecast: it looks telescoping. Does it converge?

Step 1. Simplify the summand first: $\frac{1}{n} - \frac{1}{n + 1} + \frac{1}{n} = \frac{2}{n} - \frac{1}{n + 1}$ . Why this step? Never telescope blindly — combine like terms so you see the true term. The extra $+ \frac{1}{n}$ breaks the clean difference.

Step 2. Rewrite as $(\frac{1}{n} - \frac{1}{n + 1}) + \frac{1}{n}$ . The first bracket telescopes to $1$ ; the leftover $\sum \frac{1}{n}$ is the harmonic series. Why this step? Separating a genuinely-telescoping part from a non-telescoping part reveals the trap.

Step 3. $n = 1 \sum \infty \frac{1}{n}$ diverges (grows like $ln N$ ). A finite number plus $\infty$ is $\infty$ . Why this step? By Convergence Tests for Series, the harmonic tail dominates; the telescoping part cannot rescue it.

Step 4. Conclusion: the series diverges. $no finite sum$

Verify: partial sums grow: $S_{1} = 1.5$ , $S_{2} \approx 2.17$ , $S_{3} \approx 2.68$ , and they keep climbing past every bound (harmonic growth) ✓. The exam wanted you to assume it telescopes — the fix is always simplify first.

Which cell did each example hit?

The green cells (Ex 1,2,3,6,7) converge to a clean value; the red cells (Ex 5, 8) diverge; the yellow cell (Ex 4) converges but to $b_{1} - L$ with $L \neq = 0$ . Cover all three colours and you've seen every scenario.

Recall Rapid self-test

Sum of $\sum_{n = 1}^{\infty} \frac{1}{( 2 n - 1 ) ( 2 n + 1 )}$ ? ::: $\frac{1}{2}$ How many terms survive each end for a gap-2 telescope? ::: two $\sum_{n = 3}^{\infty} \frac{1}{n ( n + 1 )}$ ? ::: $\frac{1}{3}$ (lead term is $b_{3}$ ) Why is $\sum (\frac{n}{n + 1} - \frac{n + 1}{n + 2}) = - \frac{1}{2}$ , not $\frac{1}{2}$ ? ::: because $lim b_{n} = 1 \neq = 0$ , so sum $= b_{1} - lim b_{n} = \frac{1}{2} - 1$ Does $\sum \frac{1}{n + n + 1}$ converge? ::: no — it equals $N + 1 - 1 \to \infty$ Value of $\prod_{n = 2}^{\infty} (1 - \frac{1}{n ^{2}})$ ? ::: $\frac{1}{2}$

Connections

Telescoping series — the parent; this page is its full example bank.
Partial Fraction Decomposition — engine for cells C1, C2.
Method of Differences — the same collapse in finite sums.
Sequence of Partial Sums — every "Verify" watches $S_{N}$ .
Limits of Sequences — decides C4 (tail $\to 1$ ) and C5 (tail $\to \infty$ ).
Convergence Tests for Series — needed to call C8 divergent.
Geometric Series — contrast: another exact-sum family.