Alternating series test — Leibniz test, proof: Visual walkthrough

Step 0 — What we are even looking at

So $s_{N} = b_{1} - b_{2} + b_{3} - b_{4} + \dots$ , stopping after $N$ hops. Each $\pm$ is the direction of that hop. Look at the picture: we live on a horizontal number line, and each hop is an arrow whose length is $b_{n}$ and whose direction alternates.

Figure — Alternating series test — Leibniz test, proof

We assume, exactly as the test demands:

Decreasing: $b_{n + 1} \leq b_{n}$ — each hop is no longer than the one before.
Vanishing: $b_{n} \to 0$ — hops shrink all the way to nothing.

Step 1 — Plot the partial sums as landing spots

WHAT: Instead of arrows, mark just the landing spot after each hop: $s_{1}, s_{2}, s_{3}, \dots$

WHY: The question "does the series converge?" is exactly the question "do the landing spots $s_{N}$ settle onto one point?" Converting hops to spots turns a walking problem into a sequence-of-points problem, which we know how to test.

PICTURE: Watch how the spots bounce right, left, right, left — but the bounces get shorter. $s_{1}$ is far right, $s_{2}$ jumps left (by $b_{2}$ ), $s_{3}$ nudges right (by $b_{3} < b_{2}$ ), and so on.

Step 2 — Split the spots into two teams: even and odd

WHAT: Colour the even spots one way, odd spots another.

WHY: Each team turns out to be monotone (moves steadily one direction), and monotone sequences are the ones we can pin down with the Monotone Convergence Theorem. The mixed zig-zag is not monotone, so we cannot use that theorem on it directly — hence the split.

PICTURE: The even team creeps rightward (up the number line); the odd team creeps leftward. They are squeezing toward each other.

Step 3 — The even team only moves right (it is increasing)

WHAT: Group the terms two at a time. Each bracket is (a bigger hop) minus (a smaller-or-equal hop).

WHY: Because $b_{n}$ is decreasing, $b_{2 k - 1} \geq b_{2 k}$ , so every bracket is $\geq 0$ . Going from $s_{2 m}$ to $s_{2 m + 2}$ just tacks on one more non-negative bracket. Adding something $\geq 0$ can only push you right. Therefore the even spots never move left: $s_{2} \leq s_{4} \leq s_{6} \leq \dots$

PICTURE: Each new pair of hops is a green block of non-negative width; stacking blocks marches the even spot steadily rightward.

Step 4 — But the even team can never pass $b_{1}$ (it is bounded)

WHAT: Re-bracket the same sum starting from $b_{1}$ , pairing $b_{2}$ with $b_{3}$ , etc.

Worked example The regrouping, done term-by-term (nothing added or removed)

Start from the original signed sum and shift the bracket boundaries one step to the right. Watch that every $b_{k}$ keeps its own sign — we only choose where the brackets open and close: $s_{2 m} = b_{1} - b_{2} + b_{3} - b_{4} + b_{5} - \dots + b_{2 m - 1} - b_{2 m}$ Peel off $b_{1}$ at the front and $- b_{2} m$ at the back, and pair the interior terms as $(- b_{2} + b_{3}), (- b_{4} + b_{5}), \dots$ : $s_{2 m} = b_{1} + (- b_{2} + b_{3}) + (- b_{4} + b_{5}) + \dots + (- b_{2 m - 2} + b_{2 m - 1}) - b_{2 m}$ Now factor a minus out of each interior bracket, turning $(- b_{2 k} + b_{2 k + 1})$ into $- (b_{2 k} - b_{2 k + 1})$ : $s_{2 m} = b_{1} - (b_{2} - b_{3}) - (b_{4} - b_{5}) - \dots - (b_{2 m - 2} - b_{2 m - 1}) - b_{2 m}$ Count the terms: $b_{1}$ appears once (+), $b_{2 m}$ once (−), and each middle $b_{k}$ once with its original sign — identical to the start, just re-bracketed.

WHY: Each bracket $(b_{2 k} - b_{2 k + 1})$ is $\geq 0$ (decreasing $b_{n}$ ), the leftover $b_{2 m} \geq 0$ , and every one of them is subtracted from $b_{1}$ . Subtracting non-negative things from $b_{1}$ can only leave you at or below $b_{1}$ : $s_{2 m} \leq b_{1} for every m .$

So the even spots are fenced in: they can march right (Step 3) but a wall at $b_{1}$ stops them.

PICTURE: The amber wall at $b_{1}$ ; the even spots approach it from the left but never cross.

Step 5 — An increasing, fenced-in team must settle (Monotone Convergence)

WHAT: Combine Step 3 (increasing) with Step 4 (bounded above by $b_{1}$ ).

WHY: Increasing + bounded above ⟹ a limit exists. Name it $s$ : $lim_{m \to \infty} s_{2 m} = s, s \leq b_{1} .$ Here $s$ is the settling-point of the even team, and it sits at or below the wall.

PICTURE: The even dots pile up against an invisible ceiling $s$ (just under $b_{1}$ ), getting arbitrarily close.

Step 6 — The odd team lands on the same spot (this is where $b_{n} \to 0$ enters)

WHAT: Take the even limit and add the very next hop.

WHY — the only place we use vanishing: If the hops did not shrink to zero, the odd team would settle a fixed gap $b_{2 m + 1}$ to the right of $s$ — two different limits, no convergence. But $b_{n} \to 0$ closes that gap: $lim_{m \to \infty} s_{2 m + 1} = s + 0 = s .$ Even team → $s$ . Odd team → $s$ . Same point.

PICTURE: The shrinking amber bridge $b_{2 m + 1}$ between the left (even) and right (odd) dots collapses to zero length.

Step 7 — Free bonus: the error bound, seen at a glance

WHAT: Notice that $s$ always lies between two consecutive spots $s_{N}$ and $s_{N + 1}$ .

WHY: In each parity the true sum $s$ is caught between the two consecutive spots $s_{N}$ and $s_{N + 1}$ , and the distance between those neighbours is exactly the length of the next hop, $∣ s_{N + 1} - s_{N} ∣ = b_{N + 1}$ . A point trapped inside an interval cannot be farther from either end than the interval's width — so your distance to $s$ can't beat $b_{N + 1}$ .

PICTURE: The true sum $s$ pinned inside the shrinking interval $[s_{2}, s_{1}], [s_{2}, s_{3}], \dots$ ; each interval's width is the next $b$ .

Step 8 — The degenerate cases (what breaks if a hypothesis dies)

PICTURE: Left panel — hops of near-constant length ( $b_{n} \to 1$ ), spots keep bouncing forever, never settle. Right panel — non-monotone hops make a spot jump the "wrong" way, breaking the fence.

The one-picture summary

Everything above in a single frame: two monotone teams (even ↑, odd ↓) squeezing toward a shared limit $s \leq b_{1}$ , the collapsing bridge $b_{2 m + 1}$ , and the error bound $b_{N + 1}$ as the width of the final trap.

Recall Feynman: the whole walkthrough in plain words

You hop along a line: a big step forward, a smaller step back, a smaller step forward, back again — each step shorter than the last. Mark where you land each time. The "back" landings (even) creep steadily to the right but can never pass your very first landing $b_{1}$ — that's a wall. Anything that only goes right yet hits a wall must stop somewhere; call it $s$ . The "forward" landings (odd) are always just one tiny step to the right of a back landing, and since the steps shrink to nothing, they end up at the same $s$ . Both kinds of landing pile onto one point — that point is the sum. And wherever you stop, the true answer is no farther than your next step size $b_{N + 1}$ . That's the entire test: hops down to zero ⟹ you land somewhere, within one hop of the truth. For the alternating harmonic series $1 - \frac{1}{2} + \frac{1}{3} - \dots$ that landing spot happens to be $ln 2$ .

Connections

Alternating series test — Leibniz test, proof (parent)
Convergence of Series — overview
Absolute vs Conditional Convergence
Monotone Convergence Theorem
nth-Term Test for Divergence
Harmonic Series
Power Series — radius of convergence