Arithmetic-geometric progression — finding sum: Worked examples

Before anything else, three symbols you must own (the parent defined these; we re-anchor them so line one is self-contained):

Intuition WHY that finite formula looks the way it does (a 30-second re-derivation)

Do not swallow the boxed $S_{n}$ as a black box — the parent earned it, so let us re-walk the skeleton. Write $S_{n}$ , then $r S_{n}$ shifted one place right, and subtract. Everything in the bulk cancels except three leavings:

the very first term $a$ (nothing sat above it) → gives $\frac{a}{1 - r}$ after dividing by $(1 - r)$ ;
the middle strip, which becomes a pure GP $d r + d r^{2} + \dots + d r^{n - 1}$ of $n - 1$ terms → gives $\frac{d r ( 1 - r ^{n - 1} )}{( 1 - r ) ^{2}}$ (one $(1 - r)$ from summing the GP, one more from the $(1 - r)$ we factored out — hence the square);
the last shifted term $(a + (n - 1) d) r^{n}$ (nothing sat below it) → gives the $- \frac{( a + ( n - 1 ) d ) r ^{n}}{1 - r}$ correction. So the three fractions are literally "head, middle GP, tail." Reading them this way, you can reconstruct the formula from scratch even if you forget it.

The scenario matrix

Every AGP you meet lands in exactly one of these boxes. The "Example" column tells you which worked example nails that box. Read the ratio $r$ on a number line — each interval and each point where behaviour changes is its own case.

Cell	Ratio $r$	What is being asked	The danger	Example
C0a	$r = 0$ (boundary)	trivial sum	only the first term survives	Ex 0
C1	$0 < r < 1$ (positive, shrinking)	infinite sum	none — the friendly case	Ex 1
C2	$- 1 < r < 0$ (negative, shrinking)	infinite sum	alternating signs; sign of the tilt	Ex 2
C0b	$r = - 1$ (boundary)	does it converge?	$	r
C3	$r > 1$ (growing)	finite sum only	infinite sum does NOT exist	Ex 3
C4	$r < - 1$ (growing, alternating)	finite sum only	signs AND divergence both trap you	Ex 4
C5	$r = 1$ (degenerate)	sum with no geometric part	boxed formula divides by zero	Ex 5
C6	$	r	<1 $, ser i ess t a r t s a t$ k=m\neq1$	infinite tail
C7	word problem (real staircase / bouncing)	model then sum	translating physics into $a, d, r$	Ex 7
C8	exam twist: differentiate a GP	prove the AGP identity itself	knowing why $\sum n x^{n - 1} = \frac{1}{( 1 - x ) ^{2}}$	Ex 8

On the number line the behaviour changes at exactly three points — $r = - 1, 0, 1$ . The convergence rule " $∣ r ∣ < 1$ " is not a fourth point; it is the condition that packages the two endpoints $r = \pm 1$ together as the outer edge of the shrinking zone. Between these three points lie the four open intervals C1–C4. We now walk every cell, endpoints included.

C0a · The boundary $r = 0$

Worked example Example 0 — the collapse to a single term

Evaluate $S = 5 + 8 \cdot 0 + 11 \cdot 0^{2} + 14 \cdot 0^{3} + \dots$ , i.e. $a = 5, d = 3, r = 0$ .

Forecast: every geometric factor $0^{k - 1}$ for $k \geq 2$ is zero. So only the very first term can survive. Guess: $S = 5$ ?

Kill the tail. For $k \geq 2$ , the factor $r^{k - 1} = 0^{k - 1} = 0$ , so every term after the first is $0$ . Why this step? Multiplying by zero annihilates a term regardless of how big its arithmetic coefficient is — the geometric part dominates completely.
Only $k = 1$ lives. The $k = 1$ term is $a \cdot r^{0} = a \cdot 1 = 5$ (recall $0^{0} = 1$ by convention for the leading term). Why this step? The zeroth power is $1$ , so the first term is untouched by the $r = 0$ collapse.
Read the sum. $S = 5$ . Why this step? Nothing else contributes, so the whole "series" is just its head.

Verify: using the infinite formula (legal since $∣0∣ < 1$ ): $\frac{a}{1 - 0} + \frac{d \cdot 0}{( 1 - 0 ) ^{2}} = \frac{5}{1} + 0 = 5$ . ✓ The tilt term vanishes because it carries a factor $r$ .

C1 · Positive shrinking ratio

Worked example Example 1 — the friendly infinite sum

Evaluate $S = 3 + 5 \cdot \frac{1}{3} + 7 \cdot \frac{1}{9} + 9 \cdot \frac{1}{27} + \dots$

Forecast: the arithmetic coefficients $3, 5, 7, 9$ climb by $2$ ; the geometric part shrinks by $\frac{1}{3}$ each time. Guess: does the answer land near $3 + 1.67 + 0.78 + \dots$ , i.e. somewhere around $6$ ?

Identify $a, d, r$ . First coefficient $a = 3$ ; jumps of $d = 2$ ; ratio $r = \frac{1}{3}$ . Why this step? The infinite formula is written in terms of exactly these three numbers, so we must read them off before touching it.
Check the licence. $∣ r ∣ = \frac{1}{3} < 1$ , so $S_{\infty}$ exists. Why this step? The infinite formula is legal only when $∣ r ∣ < 1$ (see the matrix). Skipping this check is the #1 exam trap.
Plug in. $S = \frac{a}{1 - r} + \frac{d r}{( 1 - r ) ^{2}} = \frac{3}{1 - \frac{1}{3}} + \frac{2 \cdot \frac{1}{3}}{( 1 - \frac{1}{3} ) ^{2}} = \frac{3}{\frac{2}{3}} + \frac{\frac{2}{3}}{\frac{4}{9}} .$ Why this step? Direct substitution — the GP heart $\frac{a}{1 - r}$ plus the arithmetic tilt.
Simplify. $\frac{3}{2/3} = \frac{9}{2}$ and $\frac{2/3}{4/9} = \frac{2}{3} \cdot \frac{9}{4} = \frac{3}{2}$ . So $S = \frac{9}{2} + \frac{3}{2} = 6.$ Why this step? We turn the "divide by a fraction" into "multiply by its reciprocal" so the two clean fractions can be added — turning the formula's output into a single number we can compare with the forecast.

Verify: partial sums $3, 4.667, 5.444, 5.778, 5.914, \dots$ march up toward $6$ . ✓ And it matches our forecast.

Reading the figure below. The horizontal axis is the term index $k = 1, 2, \dots, 8$ . For each $k$ I plot two bars: the tall violet bar is the bare arithmetic coefficient $a + (k - 1) d = 3, 5, 7, \dots$ (a straight-line climb), and the orange bar next to it is the actual AGP term $(a + (k - 1) d) r^{k - 1}$ after the geometric shrink is applied. Notice how the orange bars, labelled with their values in magenta, collapse toward zero even while the violet bars keep rising — that visual gap IS the statement "geometric decay beats linear growth," which is exactly why the infinite sum is finite.

Figure — Arithmetic-geometric progression — finding sum

C2 · Negative shrinking ratio

Worked example Example 2 — alternating signs, still converges

Evaluate $S = 1 - 2 \cdot \frac{1}{3} + 3 \cdot \frac{1}{9} - 4 \cdot \frac{1}{27} + \dots$ , i.e. $a = 1, d = 1, r = - \frac{1}{3}$ .

Forecast: signs flip $+, -, +, -$ . Will the tilt term $\frac{d r}{( 1 - r ) ^{2}}$ be negative (since $r < 0$ )? Guess the sum is a bit below $\frac{a}{1 - r}$ .

Read $a, d, r$ . $a = 1, d = 1, r = - \frac{1}{3}$ . The alternating signs come entirely from $r$ being negative. Why this step? You never write " $-$ " by hand into an AGP; a negative $r$ manufactures the alternation automatically.
Licence check. $∣ r ∣ = \frac{1}{3} < 1$ . Converges. The magnitude is what matters, not the sign. Why this step? Convergence depends on $∣ r ∣$ , so a negative ratio is just as valid as a positive one of the same size.
GP heart. $\frac{a}{1 - r} = \frac{1}{1 - ( - \frac{1}{3} )} = \frac{1}{\frac{4}{3}} = \frac{3}{4} .$ Why this step? $1 - r$ becomes $1 + \frac{1}{3}$ — a common slip is to write $1 - \frac{1}{3}$ . Track the double negative.
Tilt term. $\frac{d r}{( 1 - r ) ^{2}} = \frac{1 \cdot ( - \frac{1}{3} )}{( \frac{4}{3} ) ^{2}} = \frac{- \frac{1}{3}}{\frac{16}{9}} = - \frac{1}{3} \cdot \frac{9}{16} = - \frac{3}{16} .$ Why this step? Here $r < 0$ makes the tilt negative — confirming our forecast that the sum sits below $\frac{3}{4}$ .
Add. $S = \frac{3}{4} - \frac{3}{16} = \frac{12 - 3}{16} = \frac{9}{16} .$

Verify: partial sums $1, 0.333, 0.667, 0.519, 0.580, \dots$ oscillate and squeeze toward $0.5625 = \frac{9}{16}$ . ✓

C0b · The boundary $r = - 1$

Worked example Example 0b — alternating but NOT shrinking

Does $S = 1 - 2 + 3 - 4 + 5 - \dots$ (so $a = 1, d = 1, r = - 1$ ) have an infinite sum?

Forecast: the geometric factor $(- 1)^{k - 1}$ never shrinks — its size stays $1$ forever. So the terms $1, - 2, 3, - 4, \dots$ get bigger, not smaller. Guess: no finite sum.

Test the licence. $∣ r ∣ = ∣ - 1∣ = 1$ , which is not $< 1$ . The infinite formula is forbidden. Why this step? The convergence licence is strictly $∣ r ∣ < 1$ ; the endpoint $∣ r ∣ = 1$ fails it, so we may not write $\frac{a}{1 - r} + \frac{d r}{( 1 - r ) ^{2}}$ (indeed $1 - r = 2$ would give a tempting-but-wrong number).
Look at the terms directly. $∣ T_{k} ∣ = k \cdot 1 = k \to \infty$ : the terms do not even approach $0$ . Why this step? A necessary condition for any series to converge is that its terms shrink to $0$ . Here they blow up, so convergence is impossible — no cleverness rescues it.
Watch the partial sums. $1, - 1, 2, - 2, 3, - 3, \dots$ swing further and further apart; they settle on no single value. Why this step? Divergence by oscillation: the partial sums have no limit, confirming the series has no sum.
Conclusion. For $n$ terms use the finite $S_{n}$ ; there is no infinite sum at $r = - 1$ . Why this step? This seals the endpoint — $∣ r ∣ = 1$ (both $r = 1$ and $r = - 1$ ) always excludes the infinite formula.

Verify: finite $S_{4} = 1 - 2 + 3 - 4 = - 2$ from the boxed $S_{n}$ with $a = 1, d = 1, r = - 1, n = 4$ (checked below); but the partial sums $1, - 1, 2, - 2, 3, - 3, \dots$ have no limit — no $S_{\infty}$ . ✓

C3 · Growing ratio — finite only

Worked example Example 3 —

$r > 1$ , so infinity is forbidden Sum the first four terms of $1 \cdot 1 + 2 \cdot 2 + 3 \cdot 4 + 4 \cdot 8 + \dots$ where $a = 1, d = 1, r = 2, n = 4$ .

Forecast: direct addition is $1 + 4 + 12 + 32 = 49$ . We will confirm the boxed finite formula reproduces exactly $49$ — and note the infinite sum is meaningless here.

Refuse the infinite formula. $∣ r ∣ = 2 \geq 1$ , so $S_{\infty}$ does not exist; the terms $1, 4, 12, 32, \dots$ blow up. Why this step? Reaching for $\frac{a}{1 - r} + \frac{d r}{( 1 - r ) ^{2}}$ here would give a finite-looking number that is a lie — the series diverges.
Use the full $S_{n}$ with $n = 4$ . $S_{4} = \frac{a}{1 - r} + \frac{d r ( 1 - r ^{n - 1} )}{( 1 - r ) ^{2}} - \frac{( a + ( n - 1 ) d ) r ^{n}}{1 - r} .$ Why this step? The finite formula is valid for any $r \neq = 1$ — including $r > 1$ — because no limit is taken.
Substitute $a = 1, d = 1, r = 2, n = 4$ . $S_{4} = \frac{1}{1 - 2} + \frac{1 \cdot 2 ( 1 - 2 ^{3} )}{( 1 - 2 ) ^{2}} - \frac{( 1 + 3 ) 2 ^{4}}{1 - 2} .$ Why this step? Careful substitution of all four numbers keeps the three fractions honest before we crunch signs.
Crunch each piece. $\frac{1}{- 1} = - 1$ ; $\frac{2 ( 1 - 8 )}{1} = 2 (- 7) = - 14$ ; $- \frac{4 \cdot 16}{- 1} = + 64.$ Why this step? We evaluate the three "head / middle GP / tail" fractions separately so the minus-over-minus in the tail (giving $+ 64$ ) cannot slip through unnoticed — sign discipline is everything when $r > 1$ .
Add. $S_{4} = - 1 - 14 + 64 = 49.$

Verify: direct sum $1 + 4 + 12 + 32 = 49$ . ✓ Matches exactly.

C4 · Growing AND alternating — finite only

Worked example Example 4 —

$r < - 1$ , both traps at once Sum the first three terms of $1 + 2 (- 2) + 3 (- 2)^{2}$ , so $a = 1, d = 1, r = - 2, n = 3$ .

Forecast: direct: $1 + 2 (- 2) + 3 (4) = 1 - 4 + 12 = 9$ . This case has BOTH a negative sign (from $r < 0$ ) and divergence (from $∣ r ∣ > 1$ ). Only the finite formula survives.

Direct sum first (our target). $1 \cdot 1 + 2 \cdot (- 2) + 3 \cdot (4) = 1 - 4 + 12 = 9.$ Why this step? With just three terms, a direct sum gives a rock-solid answer to check the formula against.
No infinite sum. $∣ r ∣ = 2 \geq 1$ : divergent, and the signs would swing wildly. Finite formula only. Why this step? Reinforces that $∣ r ∣ \geq 1$ blocks the infinite formula regardless of sign.
Apply $S_{n}$ with $n = 3$ . $S_{3} = \frac{1}{1 - ( - 2 )} + \frac{1 \cdot ( - 2 ) ( 1 - ( - 2 ) ^{2} )}{( 1 - ( - 2 ) ) ^{2}} - \frac{( 1 + 2 ) ( - 2 ) ^{3}}{1 - ( - 2 )} .$ Why this step? Substituting $r = - 2$ inside brackets first prevents sign errors on the powers.
Simplify each. $1 - (- 2) = 3$ . First: $\frac{1}{3}$ . Second: $\frac{- 2 ( 1 - 4 )}{9} = \frac{- 2 ( - 3 )}{9} = \frac{6}{9} = \frac{2}{3}$ . Third: $- \frac{3 ( - 8 )}{3} = - \frac{- 24}{3} = + 8.$ Why this step? We expand each hidden double negative ( $(- 2)^{2} = 4$ , $(- 2)^{3} = - 8$ ) one fraction at a time, because with $r < - 1$ a single mis-tracked sign wrecks the whole answer.
Add. $S_{3} = \frac{1}{3} + \frac{2}{3} + 8 = 1 + 8 = 9.$

Verify: equals the direct sum $9$ . ✓

C5 · The degenerate case $r = 1$

Worked example Example 5 — when the boxed formula breaks

Sum $S = 1 + 2 + 3 + \dots + n$ viewed as an AGP with $a = 1, d = 1, r = 1$ . Take $n = 5$ .

Forecast: with $r = 1$ the geometric part is all $1$ 's, so the "AGP" is just an ordinary AP. The famous answer $\frac{n ( n + 1 )}{2}$ should appear — and plugging $r = 1$ into the boxed formula must be forbidden (division by $1 - r = 0$ ).

Spot the division by zero. The finite formula has $(1 - r)$ and $(1 - r)^{2}$ in denominators. At $r = 1$ those are $0$ — undefined. The formula simply does not apply. Why this step? Blindly substituting $r = 1$ produces $\frac{1}{0}$ ; you must switch tools.
Recognise what remains. With $r = 1$ every $r^{k - 1} = 1$ , so $T_{k} = a + (k - 1) d$ — a pure AP. Why this step? The geometric machinery was only needed because the terms scaled; with no scaling, ordinary AP summation is exact and simpler.
Use the AP sum. $S_{n} = \frac{n}{2} (2 a + (n - 1) d)$ . With $a = 1, d = 1$ : $S_{n} = \frac{n}{2} (2 + (n - 1)) = \frac{n ( n + 1 )}{2} .$ Why this step? This is the correct engine for the degenerate cell — see Arithmetic Progression.
Evaluate $n = 5$ . $S_{5} = \frac{5 \cdot 6}{2} = 15.$ Why this step? Substituting the specific $n = 5$ turns the general AP formula into a concrete number we can check by hand.

Verify: $1 + 2 + 3 + 4 + 5 = 15$ . ✓

C6 · Series that starts in the middle

Worked example Example 6 — re-indexing before you sum

Evaluate $S = k = 3 \sum \infty \frac{k}{2 ^{k - 1}} = \frac{3}{4} + \frac{4}{8} + \frac{5}{16} + \dots$

Forecast: the parent proved $\sum_{k = 1}^{\infty} \frac{k}{2 ^{k - 1}} = 4$ . Our series just drops the first two terms $\frac{1}{2 ^{0}} = 1$ and $\frac{2}{2 ^{1}} = 1$ . Guess: $S = 4 - 1 - 1 = 2$ ?

Subtraction route (safest). $S = (\sum_{k = 1}^{\infty} \frac{k}{2 ^{k - 1}}) - \frac{1}{2 ^{0}} - \frac{2}{2 ^{1}} = 4 - 1 - 1 = 2.$ Why this step? Rather than re-deriving, lop off the known missing head from the known full sum — fewer chances to mislabel $a$ .
Formula route — shift the index. Put $j = k - 2$ , so as $k = 3, 4, 5, \dots$ the new counter $j = 1, 2, 3, \dots$ starts cleanly at $1$ . Substituting $k = j + 2$ : $S = \sum_{j = 1}^{\infty} \frac{( j + 2 )}{2 ^{(j + 2) - 1}} = \sum_{j = 1}^{\infty} \frac{j + 2}{2 ^{j + 1}} .$ Why this step? The infinite AGP formula demands a series that starts at $j = 1$ ; re-indexing gives that, and it exposes the new first coefficient $j + 2$ (which is $3$ at $j = 1$ , NOT $1$ ).
Isolate the constant factor. Write the denominator as $2^{j + 1} = 2^{2} \cdot 2^{j - 1} = 4 \cdot 2^{j - 1}$ , so $S = \sum_{j = 1}^{\infty} \frac{j + 2}{4 \cdot 2 ^{j - 1}} = \frac{1}{4} \sum_{j = 1}^{\infty} (j + 2) (\frac{1}{2})^{j - 1} .$ Now it is a standard AGP in $j$ with $a = 3$ (value of $j + 2$ at $j = 1$ ), $d = 1$ , $r = \frac{1}{2}$ , scaled by $\frac{1}{4}$ . Why this step? Pulling out the single constant $\frac{1}{4}$ leaves a textbook AGP starting at $j = 1$ , ready for the boxed infinite formula — and pins down $a = 3, d = 1, r = \frac{1}{2}$ unambiguously.
Apply the infinite formula. $j \geq 1 \sum (j + 2) (\frac{1}{2})^{j - 1} = \frac{a}{1 - r} + \frac{d r}{( 1 - r ) ^{2}} = \frac{3}{1/2} + \frac{1 \cdot \frac{1}{2}}{( 1/2 ) ^{2}} = 6 + 2 = 8.$ Then $S = \frac{1}{4} \cdot 8 = 2.$ Why this step? We finally feed the clean $a, d, r$ into the master formula, then undo the $\frac{1}{4}$ scaling to recover $S$ .

Verify: both routes give $2$ ; partial sums $0.75, 1.25, 1.5625, \dots \to 2$ . ✓

C7 · Real-world word problem

Worked example Example 7 — a fading, climbing staircase of savings

On day $1$ you deposit ₹ $1$ . On day $k$ your intended deposit is ₹ $k$ (it grows), but inflation shrinks each rupee's value by a factor $\frac{4}{5}$ per day, so the real value of day $k$ 's deposit is $k \cdot (\frac{4}{5})^{k - 1}$ . Over infinitely many days, what is the total real value?

Forecast: intended deposits $1, 2, 3, \dots$ grow, but the $\frac{4}{5}$ decay wins eventually. A finite total exists. Guess it's a decent multiple of the first deposit — maybe around ₹ $20$ ?

Build the model. Real value on day $k$ is $k r^{k - 1}$ with $a = 1, d = 1, r = \frac{4}{5}$ . Why this step? "Grows linearly, decays geometrically" is precisely an AGP; naming $a, d, r$ turns the story into maths.
Convergence. $∣ r ∣ = \frac{4}{5} < 1$ : the total real value is finite (linear climb loses to geometric decay — see Limits: exponential vs polynomial growth). Why this step? A real bank total must be a number; the $∣ r ∣ < 1$ check guarantees the model gives one.
Sum. With $a = d = 1$ the two AGP fractions merge: $\sum_{k \geq 1} k r^{k - 1} = \frac{1}{( 1 - r ) ^{2}}$ (the parent's identity). Here $1 - r = \frac{1}{5}$ , so $S = \frac{1}{( 1/5 ) ^{2}} = 25.$ Why this step? Recognising the $a = d = 1$ special case lets us use the one-fraction shortcut $\frac{1}{( 1 - r ) ^{2}}$ instead of adding two terms — faster and less error-prone.

Verify: partial real totals $1, 2.6, 4.52, 6.56, \dots$ climb steadily; the geometric formula caps them at ₹ $25$ . Units: rupees throughout, dimensionally consistent. ✓

Reading the figure below. The horizontal axis is the day number $k$ ; the vertical axis is the running real total in rupees, i.e. the cumulative sum of the first $k$ deposits' real values. The magenta curve is that running total, and the violet dashed line marks the limit $25$ . The shaded orange gap between the curve and the dashed line is the "money still to come" — watch it shrink every day: this shrinking gap is the visual meaning of convergence, showing the infinite pile really does stop at ₹ $25$ .

C8 · Exam twist — prove the identity by calculus

Worked example Example 8 — where does

$\frac{1}{( 1 - x ) ^{2}}$ come from? Show that $n = 1 \sum \infty n x^{n - 1} = \frac{1}{( 1 - x ) ^{2}}$ for $∣ x ∣ < 1$ , using the GP sum and a derivative.

Forecast: the AGP $\sum n x^{n - 1}$ looks like the derivative of something. Which known series, when differentiated term by term, produces the coefficients $1, 2, 3, \dots$ ?

Start from the GP. For $∣ x ∣ < 1$ , $n = 0 \sum \infty x^{n} = 1 + x + x^{2} + x^{3} + \dots = \frac{1}{1 - x}$ (see Sum to infinity of GP). Why this step? We tame the hard AGP by relating it to the easy GP whose sum we already trust.
WHY a derivative, not another tool? Differentiating $x^{n}$ gives $n x^{n - 1}$ — exactly the arithmetic coefficient $n$ we want to manufacture. No other single operation drops the exponent into the coefficient like this. Why this step? The derivative is the unique tool that converts a geometric term into an arithmetic×geometric term — it literally builds the AGP.
Differentiate both sides. Left side, term by term: $\frac{d}{d x} \sum_{n \geq 0} x^{n} = \sum_{n \geq 0} \frac{d}{d x} x^{n} = \sum_{n \geq 1} n x^{n - 1}$ (the $n = 0$ term is the constant $1$ , whose derivative is $0$ , so the sum now starts at $n = 1$ ). Right side: $\frac{d}{d x} \frac{1}{1 - x} = \frac{d}{d x} (1 - x)^{- 1} = (- 1) (1 - x)^{- 2} \cdot (- 1) = \frac{1}{( 1 - x ) ^{2}} .$ Why this step? Term-by-term differentiation is legal inside the radius $∣ x ∣ < 1$ (see Power series and generating functions); the chain rule on the right supplies the two minus signs that cancel to give a positive result.
Read off the identity. Equating the two differentiated sides: $\sum_{n = 1}^{\infty} n x^{n - 1} = \frac{1}{( 1 - x ) ^{2}} .$ Why this step? Differentiating an equality preserves it, so the left and right derivatives must be equal — that IS the AGP identity, now proved rather than memorised.

Verify: set $x = \frac{1}{2}$ : left $= \sum n / 2^{n - 1} = 4$ (parent's Example 2); right $= \frac{1}{( 1/2 ) ^{2}} = 4$ . ✓ And $x = \frac{1}{3}$ : right $= \frac{1}{( 2/3 ) ^{2}} = \frac{9}{4} = 2.25$ ; partial sums of $\sum n / 3^{n - 1}$ approach $2.25$ . ✓

Recall

Connections

Arithmetic-geometric progression — finding sum — the parent machine every example here drives.
Geometric Progression — sum — the leftover engine used in C1–C4.
Arithmetic Progression — the degenerate cell C5 collapses to this.
Sum to infinity of GP — the $∣ r ∣ < 1$ licence and Example 8's starting point.
Power series and generating functions — justifies term-by-term differentiation in C8.
Limits: exponential vs polynomial growth — why the tail vanishes in C1, C2, C7.