Alternating series test — Leibniz test, proof

WHAT is an alternating series?

HOW to derive / prove it (from scratch)

We build the proof by squeezing the partial sums. Let $s_N = \sum_{n=1}^N (-1)^{n-1}b_n$ .

Step 1 — Look at even partial sums $s_{2m}$ . $s_{2m} = (b_1 - b_2) + (b_3 - b_4) + \cdots + (b_{2m-1} - b_{2m}).$ Why this grouping? Because $b_n$ is decreasing, each bracket $\ge 0$ . Adding more non-negative brackets means $s_{2m}$ increases as $m$ grows. So $\{s_{2m}\}$ is increasing.

Step 2 — Show $s_{2m}$ is bounded above. Regroup differently: $s_{2m} = b_1 - (b_2 - b_3) - (b_4 - b_5) - \cdots - b_{2m}.$ Why regroup? Every bracket here is $\ge 0$ , and we subtract them, plus subtract $b_{2m}\ge0$ . So $s_{2m} \le b_1$ .

Step 3 — Conclude even sums converge. An increasing sequence bounded above converges (Monotone Convergence Theorem). Call the limit $s$ : $\lim_{m\to\infty} s_{2m} = s, \qquad s \le b_1.$

Step 4 — Tie in the odd sums. Note $s_{2m+1} = s_{2m} + b_{2m+1}$ . Take limits: $\lim_{m\to\infty} s_{2m+1} = s + \lim_{m\to\infty} b_{2m+1} = s + 0 = s.$ Why does this finish it? Even and odd partial sums share the same limit $s$ . If both subsequences of the partial sums go to $s$ , the whole sequence $s_N \to s$ . Hence the series converges. $\blacksquare$

Figure — Alternating series test — Leibniz test, proof

Bonus: the error bound (free gift from the proof)

Because partial sums oscillate around $s$ , jumping by less each time:

Worked examples

Common mistake Steel-manning the classic errors

Mistake A: "Terms go to 0, so it converges." Why it feels right: For alternating series we just used $b_n\to0$ — feels sufficient. The fix: $b_n\to0$ alone is never enough in general (harmonic series $\sum 1/n$ diverges!). Leibniz also demands decreasing + alternation.

Mistake B: "Leibniz tells me the sum equals $b_1$ ." Why it feels right: We proved $s\le b_1$ . The fix: $b_1$ is only an upper bound, not the value. The true $s$ sits between $s_1$ and $s_2$ .

Mistake C: "Converges ⟹ converges absolutely." Why it feels right: Convergence usually means "well-behaved." The fix: Leibniz often gives only conditional convergence (alternating harmonic). Always check $\sum b_n$ separately for absolute convergence.

Recall Feynman: explain it to a 12-year-old

Imagine hopping toward a flag. First a big hop forward, then a smaller hop back, then a smaller hop forward, then smaller back... Each hop is tinier than the last, and they keep getting super tiny. You can't keep bouncing forever in a smaller and smaller space — you have to settle on one spot. That spot is the sum! And wherever you are, the flag (true answer) is no farther away than your next hop size.

Active-recall flashcards

State the two conditions of the Leibniz (alternating series) test.

(1)

b_{n+1}\le b_n

(decreasing) and (2)

b_n\to 0

, where

b_n=|a_n|>0

Why must

b_n

be decreasing in the proof?

It makes the grouped brackets

(b_{2k-1}-b_{2k})\ge0

, giving monotone partial sums bounded above.

Where is the condition

b_n\to0

used in the proof?

To show even and odd partial sums share the same limit:

s_{2m+1}=s_{2m}+b_{2m+1}\to s+0

What is the alternating series error bound?

|s-s_N|\le b_{N+1}

— error is at most the first omitted term.

Does

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

converge? Absolutely?

Converges (to

\ln2

) but only conditionally, since

\sum 1/n

diverges.

b_n\to0

but is not monotone, can you conclude convergence by Leibniz?

No — monotonicity is required; the test does not apply.

b_n\not\to0

, what can you say?

Series diverges by the

n

th-term test; Leibniz is irrelevant.

Why is the true sum between two consecutive partial sums?

Partial sums alternate overshoot/undershoot the limit, with shrinking jumps

b_{N+1}

Connections

Convergence of Series — overview
Absolute vs Conditional Convergence
Monotone Convergence Theorem
nth-Term Test for Divergence
Harmonic Series
Power Series — radius of convergence (endpoint behaviour often uses Leibniz)

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, alternating series woh hoti hai jisme terms baari-baari plus aur minus hote hain: $b_1 - b_2 + b_3 - b_4 + \dots$ . Leibniz test bolta hai ki agar do cheezein sach hain — (1) terms ka size $b_n$ lagaataar ghat raha hai (decreasing), aur (2) $b_n$ aakhir mein zero ki taraf ja raha hai — toh yeh series converge kar jaayegi. Intuition simple hai: socho tum ek flag ki taraf chhalaang maar rahe ho, pehle bada jump aage, phir chhota jump peeche, phir aur chhota aage... jumps chhote hote ja rahe hain, toh tum ek hi point pe settle ho jaaoge. Wahi point series ka sum hai.

Proof ka core idea partial sums ko squeeze karna hai. Even partial sums $s_{2m}$ ko brackets mein todo: kyunki $b_n$ decreasing hai, har bracket positive hai, toh $s_{2m}$ increasing hai. Doosre tarike se brackets banao toh dikh jaata hai $s_{2m} \le b_1$ — yaani bounded above. Increasing + bounded = converge (Monotone Convergence Theorem). Phir odd sums $s_{2m+1} = s_{2m} + b_{2m+1}$ ka limit bhi same $s$ aata hai kyunki $b_{2m+1}\to 0$ . Dono mil gaye, toh poori series $s$ pe converge ho gayi.

Ek bahut useful bonus: error bound $|s - s_N| \le b_{N+1}$ . Matlab agar tum $N$ terms tak rok do, toh galti agle term se zyada nahi hogi. Exam mein "kitne terms chahiye accuracy ke liye" type questions isi se solve hote hain.

Sabse common galti: "terms zero ki taraf ja rahe hain isliye converge ho gaya." Yeh galat hai! $\sum 1/n$ (harmonic) bhi zero ki taraf jaata hai par diverge karta hai. Leibniz mein decreasing condition aur alternation dono zaroori hain. Aur yaad rakho — Leibniz se aksar sirf conditional convergence milti hai; absolute convergence ke liye $\sum b_n$ alag se check karo.

Alternating series test — Leibniz test, proof

WHAT is an alternating series?

HOW to derive / prove it (from scratch)

Bonus: the error bound (free gift from the proof)

Worked examples

Active-recall flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus III — Sequences & Series

Connections