Alternating series test — Leibniz test, proof: Question bank

Notation reminder so nothing is used unearned: an alternating series is a sum whose terms flip sign every step. Call the $n$ -th term $a_{n}$ (so $a_{1}$ is the first term, $a_{2}$ the second, and so on — signs included). Its size is $b_{n} = ∣ a_{n} ∣ > 0$ , the absolute value with the sign stripped away. Then we write $\sum_{n = 1}^{\infty} (- 1)^{n - 1} b_{n}$ , where $(- 1)^{n - 1}$ is only the sign-flipper and $b_{n}$ carries all the real content. $s_{N}$ is the $N$ -th partial sum (the running total after $N$ terms), and $s$ is the true infinite sum when it exists.

By bracket-squeeze we mean the core proof move: grouping the terms into brackets like $(b_{1} - b_{2}) + (b_{3} - b_{4}) + \dots$ so each bracket is $\geq 0$ , which traps the partial sums between an increasing floor and a fixed ceiling $b_{1}$ — that squeeze is what forces convergence.

True or false — justify

b_{n} \to 0

, then

\sum (- 1)^{n - 1} b_{n}

converges.

False. Vanishing terms is only condition 2; without monotone decrease the bracket-squeeze in the proof can fail, so convergence is not guaranteed.

\sum (- 1)^{n - 1} b_{n}

converges, then

b_{n} \to 0

True. Any convergent series (alternating or not) must have terms tending to zero — this is the nth-term test read in reverse.

The Leibniz test can prove that a series diverges.

False. It is a one-way test: passing it proves convergence, but failing it (e.g.

b_{n}

not monotone) proves nothing on its own.

b_{n} \neq \to 0

, the alternating series diverges.

True, but by the nth-term test, not by Leibniz — terms that don't vanish can never sum to a finite limit.

The alternating harmonic series and the harmonic series have the same convergence behaviour.

False.

\sum (- 1)^{n - 1} / n

converges (to

ln 2

); the plain harmonic series

\sum 1/ n

diverges. The alternation rescues it.

Every convergent alternating series is only conditionally convergent.

False. Some converge absolutely, e.g.

\sum (- 1)^{n - 1} / n^{2}

, since

\sum 1/ n^{2}

converges. Alternation being present does not force conditional convergence.

The true sum equals the first term

b_{1}

False. The proof shows

s \leq b_{1}

; that is an upper bound, not the value. The true

s

lies strictly between

s_{1} = b_{1}

and

s_{2} = b_{1} - b_{2}

For a Leibniz series, the partial sums straddle the limit: consecutive

s_{N}

sit on opposite sides of

s

True. Odd partial sums overshoot and even ones undershoot (or vice versa), each jump shrinking, which is exactly why

s

is trapped between them.

The error bound

∣ s - s_{N} ∣ \leq b_{N + 1}

requires the terms to be decreasing, not just

\to 0

True. The straddling that gives the bound relies on monotonicity; without it the sum need not lie between consecutive partial sums.

Monotone decrease only needs to hold "eventually" (for large

n

), not from

n = 1

True. Finitely many early terms cannot affect convergence, so decrease from some

n = N

onward is enough — you just apply the test to the tail.

Spot the error

\sum (- 1)^{n - 1} \frac{n}{n + 1}

converges by Leibniz because it alternates and each term is less than 1."

The error: condition 2 is violated —

b_{n} = \frac{n}{n + 1} \to 1 \neq = 0

. Boundedness by 1 is irrelevant; the series diverges by the nth-term test.

b_{n} = 1, \frac{1}{3}, \frac{1}{2}, \frac{1}{5}, \frac{1}{4}, \dots \to 0

and alternates, so Leibniz gives convergence."

The error:

b_{n}

is not monotone decreasing (it wiggles up). The bracket argument

(b_{2 k - 1} - b_{2 k}) \geq 0

breaks, so Leibniz simply does not apply.

"Since Leibniz proved

s \leq b_{1}

, we conclude

s = b_{1}

when terms are tiny after the first."

The error: an inequality is not an equality.

s

is pinned between

s_{1}

and

s_{2} = b_{1} - b_{2}

, so

s < b_{1}

whenever

b_{2} > 0

"The series converges, so I can rearrange its terms freely without changing the sum."

The error: conditional convergence (like alternating harmonic) forbids free rearrangement — Riemann's theorem says you can make it sum to anything. Only absolutely convergent series rearrange safely (see Absolute vs Conditional Convergence).

"To check monotonicity I confirmed

b_{5} < b_{4}

, so

b_{n}

is decreasing."

The error: one comparison isn't a pattern. You need

b_{n + 1} \leq b_{n}

for all (large)

n

, usually shown via

f^{'} (x) \leq 0

for

f (x) = b_{x}

or an algebraic ratio/difference argument.

b_{n} \to 0

was enough to converge here, and

\sum 1/ n

has

1/ n \to 0

, so the harmonic series converges too."

The error: the alternation plus monotonicity did the work, not

b_{n} \to 0

alone. The plain harmonic series has no sign-flipping to trap partial sums, and it diverges.

Why questions

Why must we group as

(b_{1} - b_{2}) + (b_{3} - b_{4}) + \dots

rather than sum term by term?

Grouping exposes that each bracket is

\geq 0

(from decrease), which instantly shows the even partial sums are increasing — the first half of the Monotone Convergence setup.

Why do we regroup a second time as

b_{1} - (b_{2} - b_{3}) - \dots - b_{2 m}

To get an upper bound: every new bracket is subtracted and non-negative, forcing

s_{2 m} \leq b_{1}

. Increasing plus bounded above then gives convergence by the Monotone Convergence Theorem.

Why does showing the even partial sums converge not finish the proof by itself?

A subsequence converging doesn't force the whole sequence to converge; the odd sums could head elsewhere. We must show both go to the same

s

Why is

lim b_{n} = 0

exactly the ingredient that merges the odd and even limits?

Because

s_{2 m + 1} = s_{2 m} + b_{2 m + 1}

, and

b_{2 m + 1} \to 0

makes the odd limit equal the even limit

s + 0 = s

. Without it the two limits differ and

s_{N}

has no single limit.

Why does the error bound "first omitted term" fall out for free?

Since

s

lies between

s_{N}

and

s_{N + 1}

, and those differ by exactly

b_{N + 1}

, the distance from

s_{N}

s

cannot exceed that gap.

Why do we always check

\sum b_{n}

separately even after Leibniz succeeds?

Leibniz only certifies convergence, not absolute convergence. Testing

\sum b_{n}

tells you whether it's absolute or merely conditional — crucial for rearrangement and further manipulation.

Why is Leibniz so useful for power series endpoints?

At an endpoint the series often becomes alternating with decreasing terms, where absolute-convergence tests are inconclusive but Leibniz cleanly settles convergence.

Edge cases

b_{n}

is eventually constant (say

b_{n} = c > 0

for all large

n

), does Leibniz apply?

No. It's weakly decreasing but

b_{n} \to c \neq = 0

, so condition 2 fails and the series diverges by the nth-term test.

b_{n} = 0

for infinitely many

n

, is the series still "alternating"?

Borderline: the definition wants

b_{n} > 0

. Zeros break the strict sign-flip pattern; you'd first drop the zero terms and reindex before applying the test.

What if

b_{n + 1} = b_{n}

for some

n

(ties in the decrease)?

Fine — the test allows

b_{n + 1} \leq b_{n}

, so ties are permitted. Each bracket is still

\geq 0

(possibly

= 0

), and the argument goes through.

A series is alternating and

b_{n} \to 0

, but

b_{n}

increases on a finite stretch then decreases forever after. Converges?

Yes. Convergence depends only on the tail; apply Leibniz from where monotonicity begins, since the finite front changes the sum but not whether it converges.

Can an alternating series converge even though it fails BOTH Leibniz conditions?

Yes — Leibniz is sufficient, not necessary. A cleverly arranged non-monotone, still-vanishing series could converge; Leibniz just wouldn't be the tool that proves it.

If every

b_{n}

is the same as some convergent absolute series' terms, is conditional/absolute distinction still needed?

Yes — if

\sum b_{n}

converges the alternating version converges absolutely, so "conditional" wouldn't apply. The distinction always hinges on

\sum b_{n}

, never on the signs.

Recall One-line summary of every trap

Leibniz needs Down to Zero (decreasing and vanishing); it only proves convergence, gives an upper bound not the value, and never tells you absolute vs conditional — check $\sum b_{n}$ for that.

Connections

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