Worked examples — Series — partial sums, convergence definition

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The scenario matrix

Every series problem falls into one of these cells. Our job below is to hit each one at least once.

Cell	What makes it that case	Example we use
A Geometric, $|r|<1$, $r>0$	ratio positive and shrinking → converges	Ex 1
B Geometric, $|r|<1$, $r<0$	ratio negative and shrinking → still converges (alternating)	Ex 2
C Geometric, $|r|\ge 1$	ratio too big → diverges; formula forbidden	Ex 3
D Telescoping	terms cancel in a chain → converges to the leftover ends	Ex 4
E Terms $\to 0$ but diverges	necessary-not-sufficient trap (harmonic-style)	Ex 5
F Terms $\nrightarrow 0$	$n$-th term test kills it instantly	Ex 6
G Bounded but no limit	oscillating partial sums → diverges without going to $\infty$	Ex 7
H Real-world word problem	translate physical situation into a geometric series	Ex 8
I Exam twist: shifted index / mixed	recognise a geometric series hiding under a different start or a split	Ex 9

Read the matrix as a checklist. Each example below is stamped with its cell.

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Ex 1 — Cell A: geometric, positive shrinking ratio

Forecast: each term is half the one before. Guess: does the pile stop somewhere, and roughly where?

Look at the figure: each amber block is half the previous, stacked end-to-end. They visibly pile up toward a wall.

1. Identify $a$ and $r$. Matching $a{r}^{n-1}$: first term $a=3$, ratio $r=\frac{1}{2}$. Why this step? The geometric formula only makes sense once we name its two ingredients.
2. Check the passport $|r|<1$. Here $|\frac{1}{2}|=\frac{1}{2}<1$. ✅ Why this step? The closed form $\frac{a}{1-r}$ exists only because $r^N\to 0$, which needs $|r|<1$. Skip this and you might use a forbidden formula.
3. Apply the sum. $S=\frac{a}{1-r}=\frac{3}{1-\frac{1}{2}}=\frac{3}{\frac{1}{2}}=6.$ Why this step? This is the limit of $s_N=\frac{a(1-r^N)}{1-r}$ once $r^N\to 0$.

Verify: partial sums $s_1=3,s_2=4.5,s_3=5.25,s_4=5.625,\dots$ — each closer to $6$, never past it. The gap $6-s_N=6r^N$ shrinks to $0$. ✅

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Ex 2 — Cell B: geometric, negative shrinking ratio (alternating)

Forecast: the terms flip sign and shrink. Guess: does flipping ruin convergence, or does it still settle (and land between the first two partial sums)?

Look at the figure: the running total overshoots then undershoots the target, zig-zagging inward.

1. Identify $a$ and $r$. $a=8$, $r=-\frac{1}{3}$. Why this step? Same two ingredients; the sign lives inside $r$.
2. Check the passport. $|r|=\frac{1}{3}<1$. ✅ The absolute value is what matters, not the sign. Why this step? $r^N\to 0$ whenever $|r|<1$, even for negative $r$ — the sign just alternates while the size dies.
3. Apply the sum. $S=\frac{8}{1-(-\frac{1}{3})}=\frac{8}{\frac{4}{3}}=6.$ Why this step? Formula is identical; just be careful $1-r=1+\frac{1}{3}=\frac{4}{3}$.

Verify: $s_1=8$ (over), $s_2=5.333$ (under), $s_3=6.222$ (over), $s_4=5.926$ (under) — a shrinking zig-zag squeezing to $6$. ✅

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Ex 3 — Cell C: geometric with $|r|\ge 1$ (diverges)

Forecast: ratio $>1$ means each term is bigger. Guess: converge or diverge? And is $\frac{a}{1-r}$ allowed?

1. Identify $a$ and $r$. $a=2$, $r=\frac{3}{2}$.
2. Check the passport. $|\frac{3}{2}|=1.5\ge 1$. ❌ Formula forbidden. Why this step? The closed form was born from $r^N\to 0$. Here $r^N=(1.5{)}^N\to \infty$, so there is no finite limit to name.
3. Confirm via $s_N$. $s_N=\frac{2(1-1.5^N)}{1-1.5}=\frac{2(1-1.5^N)}{-0.5}=4(1.5^N-1)\to \infty .$ Why this step? Always fall back to the definition ($\lim s_N$) when the shortcut is illegal.

Verify: $s_1=2,s_2=5,s_3=9.5,s_4=16.25,\dots$ — accelerating upward, no ceiling. Diverges. ✅

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Ex 4 — Cell D: telescoping

Forecast: looks like the parent's $\frac{1}{n(n+1)}$ but shifted. Guess: does it still telescope, and does it sum to $1$ or to something smaller?

1. Partial fractions. $\frac{1}{(n+1)(n+2)}=\frac{1}{n+1}-\frac{1}{n+2}.$ Why this step? Writing each term as a difference makes consecutive terms cancel.
2. Write $s_N$ and cancel. $s_N=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots +\left(\frac{1}{N+1}-\frac{1}{N+2}\right)=\frac{1}{2}-\frac{1}{N+2}.$ Why this step? Every inner fraction appears once with $+$ and once with $-$ and dies; only the very first left piece and the very last right piece survive.
3. Take the limit. $s_N=\frac{1}{2}-\frac{1}{N+2}\to \frac{1}{2}.$ Why this step? Convergence is defined as $\lim s_N$; the tail $\frac{1}{N+2}\to 0$.

Verify: $s_1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\approx 0.1667$; $s_2=\frac{1}{2}-\frac{1}{4}=0.25$; $s_3=0.3$; climbing to $0.5$. ✅ (Smaller than the parent's $1$ because we dropped the first big term.)

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Ex 5 — Cell E: terms $\to 0$ but the series diverges

Forecast: terms shrink to $0$. Guess: converge (crumbs get tiny) or diverge (harmonic in disguise)?

1. Check terms. $a_n=\frac{1}{2n}\to 0$. Why this step? The $n$-th term test can only rule out convergence; here it passes, so it tells us nothing — we must dig deeper.
2. Factor out the constant. $\sum \frac{1}{2n}=\frac{1}{2}\sum \frac{1}{n}.$ Why this step? Multiplying a divergent series by a nonzero constant can't rescue it — $\frac{1}{2}\cdot \infty$ is still $\infty$.
3. Invoke the known result. $\sum \frac{1}{n}$ is the Harmonic series and diverges. So half of infinity is still infinity → diverges. Why this step? Reusing an established divergent benchmark saves re-deriving the grouping argument.

Verify: grouping like the parent, $\frac{1}{2}(1+\frac{1}{2}+(\frac{1}{3}+\frac{1}{4})+\cdots )$ — each bracket exceeds $\frac{1}{2}$, so $s_N$ grows past every bound. Diverges. ✅ This is the necessary-not-sufficient trap made concrete.

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Ex 6 — Cell F: terms do NOT go to zero

Forecast: the fraction settles near $3$ for large $n$. Guess: what does adding "roughly $3$ forever" do?

1. Compute $\lim a_n$. $\underset{n\to \infty }{\lim }\frac{3n+1}{n+2}=\lim \frac{3+\frac{1}{n}}{1+\frac{2}{n}}=\frac{3}{1}=3.$ Why this step? The n-th Term Test for Divergence says: if $a_n\nrightarrow 0$, the series diverges — so the first thing to check is whether the terms vanish.
2. Compare to $0$. $3\ne 0$, so $a_n\nrightarrow 0$. ❌ Why this step? Adding numbers that stay near $3$ forever obviously blows up — the partial sums grow like $\approx 3N$.
3. Conclude. By the $n$-th term test, the series diverges immediately. No further work needed.

Verify: $a_{100}=\frac{301}{102}\approx 2.951$, $a_{1000}\approx 2.995$ — hovering near $3$, never near $0$. Diverges. ✅

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Ex 7 — Cell G: bounded, oscillating, no limit

Forecast: it never runs off to infinity. Guess: does "staying bounded" mean it converges?

1. List the partial sums. $s_1=1,s_2=0,s_3=1,s_4=0,\dots$ Why this step? Convergence is a statement about $(s_N)$, so look directly at that sequence.
2. Does $(s_N)$ have a limit? It bounces $1,0,1,0,\dots$ forever — no single number it approaches. Why this step? A limit must be one value that the tail stays arbitrarily close to; $1$ and $0$ are $\epsilon =0.4$ apart, so no band of half-width $0.4$ can catch both.
3. Conclude. Bounded but no limit ⇒ diverges. (Also: $a_n=(-1{)}^{n-1}\nrightarrow 0$, so the term test agrees.)

Verify: $s_N$ takes only the values $\{0,1\}$; the set of limit-candidates is not a single point, so $\lim s_N$ does not exist. Diverges. ✅ Divergence does not require $\to \infty$.

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Ex 8 — Cell H: real-world word problem (bouncing ball)

Forecast: it falls, bounces up, falls, bounces up... Guess: infinite bounces — infinite distance, or a finite total?

Look at the figure: the down-drop is counted once; every bounce height is counted twice (up then down).

1. Split the distance. Initial fall $=10$. After that, each bounce to height $h$ contributes $2h$ (up + down). Why this step? The up and down of one bounce are equal, so grouping them as $2h$ keeps the geometric pattern clean.
2. Build the bounce series. Peak heights: $10(0.6),10(0.6{)}^2,\dots$. Total bounce distance $D_{\text{bounce}}=2{\sum }_{n=1}^{\infty }10(0.6{)}^n=20{\sum }_{n=1}^{\infty }(0.6{)}^n.$ Why this step? The peaks form a geometric sequence with ratio $r=0.6$; factoring the $2$ and $10$ out exposes it.
3. Sum the geometric part. $\sum _{n=1}^{\infty }(0.6{)}^n=\frac{0.6}{1-0.6}=\frac{0.6}{0.4}=1.5$ (passport: $|0.6|<1$ ✅). So $D_{\text{bounce}}=20\times 1.5=30$. Why this step? This is $\sum a{r}^{n-1}$ starting at $n=1$ with $a=0.6,r=0.6$ — or just the standard "first term over $1-r$", first term $=0.6$.
4. Add the initial drop. Total $=10+30=40$ m. Why this step? The very first fall was not a bounce, so it's added separately, counted once.

Verify: partial totals $10,10+2(6)=22,22+2(3.6)=29.2,29.2+2(2.16)=33.52,\cdots \to 40$ m. Units are metres throughout. Finite total despite infinitely many bounces. ✅

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Ex 9 — Cell I: exam twist (shifted start + a split)

Forecast: it is geometric, but the index games make it easy to mis-plug. Guess: is the answer the same as if it started at $n=1$?

1. Write out the first few terms to find the real first term. $5(\frac{1}{4}{)}^3+5(\frac{1}{4}{)}^4+5(\frac{1}{4}{)}^5+\cdots$ The actual first term is $a=5(\frac{1}{4}{)}^3=\frac{5}{64}$; ratio $r=\frac{1}{4}$. Why this step? The "$a$" in $\frac{a}{1-r}$ means whatever the series' true first term is, regardless of what $n$ that corresponds to.
2. Check the passport. $|\frac{1}{4}|<1$ ✅. Why this step? Shifting the start index never changes the ratio, so convergence is unaffected — only the total changes.
3. Sum from the true first term. $S=\frac{a}{1-r}=\frac{5/64}{1-\frac{1}{4}}=\frac{5/64}{3/4}=\frac{5}{64}\cdot \frac{4}{3}=\frac{5}{48}.$ Why this step? Once $a$ and $r$ are correct, the standard formula applies unchanged.

Verify (alternative — subtract the missing terms): The full series from $n=0$ would be ${\sum }_{n=0}^{\infty }5(\frac{1}{4}{)}^n=\frac{5}{1-\frac{1}{4}}=\frac{20}{3}$. Subtract $n=0,1,2$: $5+\frac{5}{4}+\frac{5}{16}=\frac{80+20+5}{16}=\frac{105}{16}$. Then $\frac{20}{3}-\frac{105}{16}=\frac{320-315}{48}=\frac{5}{48}$. ✅ Two routes, same answer.

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Active recall

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Connections

• Parent: partial sums & convergence
• Geometric series — Cells A, B, C, H, I all live here
• Telescoping series — Cell D
• Harmonic series — Cell E's benchmark
• n-th Term Test for Divergence — Cells F, G
• Sequences — limits and convergence — every verdict is a statement about the sequence $(s_N)$
• p-series and the Integral Test, Comparison Test, Ratio & Root Tests — the next tools when no closed form exists
• Power series and radius of convergence — geometric series with a variable $r=x$