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$\prod_{j \geq 0} (1 + (- 1)^{j} / \sqrt{j + 2})$ has partial products tending to $0$ although $\sum_{j \geq 0} (- 1)^{j} / \sqrt{j + 2}$ converges

Statement refuted

Refuted claim: if $\sum p_{k}$ converges then $\prod (1 + p_{k})$ converges (Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors, Series, partial sums, convergence and the sum, divergence, and the tail series).

For nonnegative $p_{k}$ this is true, and is For $p_{k} \geq 0$ the product $\prod (1 + p_{k})$ converges iff $\sum p_{k}$ converges, with $1 + \sum_{k < n} p_{k} \leq \prod_{k < n} (1 + p_{k}) \leq 1 / (1 - \sum_{k < n} p_{k})$ when $\sum_{k < n} p_{k} < 1$ ; for $0 \leq p_{k} < 1$ the product $\prod (1 - p_{k})$ converges iff $\sum p_{k}$ converges and its partial products tend to $0$ otherwise; and $\sum ∣ p_{k} ∣$ convergent implies $\prod (1 + p_{k})$ convergent. For signed $p_{k}$ it is false, and the witness is

$p_{j} : = \frac{(- 1)^{j}}{\sqrt{ι (j + 2)}} (j \in N),$

with $\sqrt{}$ the nonnegative square root (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ). The series $\sum_{j} p_{j}$ converges by the alternating series test. The factors $1 + p_{j}$ are all positive, since $∣ p_{j} ∣ \leq 1 / \sqrt{2} < 1$ ; nevertheless the partial products

$Π_{m} = \prod_{j < m} (1 + \frac{(- 1)^{j}}{\sqrt{j + 2}})$

tend to $0$ , so no tail of the product has partial products with a nonzero limit and the product diverges.

The mechanism, and why no logarithm is needed. Consecutive factors are paired. With $a = ι (2 i + 2)$ and $b = ι (2 i + 3)$ , so that $b - a = 1$ ,

$(1 + \frac{1}{\sqrt{a}}) (1 - \frac{1}{\sqrt{b}}) = 1 - \frac{1}{\sqrt{a b}} (1 - \frac{1}{\sqrt{a} + \sqrt{b}}) \leq 1 - \frac{1}{ι (4 i + 6)},$

and $\sum_{i} 1 / ι (4 i + 6)$ diverges. So the even partial products are dominated by $\prod_{i < n} (1 - 1 / ι (4 i + 6))$ , which tends to $0$ by For $p_{k} \geq 0$ the product $\prod (1 + p_{k})$ converges iff $\sum p_{k}$ converges, with $1 + \sum_{k < n} p_{k} \leq \prod_{k < n} (1 + p_{k}) \leq 1 / (1 - \sum_{k < n} p_{k})$ when $\sum_{k < n} p_{k} < 1$ ; for $0 \leq p_{k} < 1$ the product $\prod (1 - p_{k})$ converges iff $\sum p_{k}$ converges and its partial products tend to $0$ otherwise; and $\sum ∣ p_{k} ∣$ convergent implies $\prod (1 + p_{k})$ convergent; the odd ones differ from them by one bounded factor.

Facts & Assumptions

Given: The alternating sequence $(ε_{j})$ with index maps $e_{i} = 2 i$ and $o_{i} = 2 i + 1$ ; the sequence $p_{j} = ε_{j} / \sqrt{ι (j + 2)}$ ; the factors $f_{j} : = 1 + p_{j}$ ; and the partial products $Π_{m} = \prod_{j < m} f_{j}$ .

[L1]

The alternating sequence: $ε_{e_{i}} = 1$ , $ε_{o_{i}} = - 1$ , $∣ ε_{j} ∣ = 1$ , $e_{0} = 0$ , $e_{i + 1} = e_{i} + 2$ , $o_{i} = e_{i} + 1$ , and $N$ is the disjoint union of the two ranges (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ).

[L2]

Square roots: every $t \geq 0$ has a unique $\sqrt{t} \geq 0$ with $(\sqrt{t})^{2} = t$ ; $\sqrt{u v} = \sqrt{u} \sqrt{v}$ and $\sqrt{}$ is strictly increasing on the nonnegative reals; and $\sqrt{t} = t^{1 / 2}$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Rational powers $a^{r}$ of a positive base).

[L3]

The canonical naturals are positive for $n \geq 1$ , strictly increasing, with $ι (m + n) = ι (m) + ι (n)$ and $ι (m n) = ι (m) ι (n)$ for $m, n \geq 1$ ; reciprocation reverses the order on the positives; and for every real $ε > 0$ there is $n \geq 1$ with $1 / ι (n) < ε$ (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L4]

The alternating series test (The alternating series test: if $(b_{k})$ is nonincreasing with $b_{k} \to 0$ then $\sum_{k} (- 1)^{k} b_{k}$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_{n}$ , Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, Limits and Cauchy sequences of reals).

[L5]

AM-GM for two nonnegative reals: $u v \leq ((u + v) / 2)^{2}$ (The arithmetic mean, geometric mean inequality).

[L6]

Finite products: $\prod_{j < 0} x_{j} = 1$ , $\prod_{j < n + 1} x_{j} = (\prod_{j < n} x_{j}) x_{n}$ , splitting at an intermediate index, and a finite product of positive factors is positive (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L7]

The principle of induction on $N$ (The principle of mathematical induction).

[L8]

For $0 \leq q_{i} < 1$ with $\sum q_{i}$ divergent, the partial products of $\prod (1 - q_{i})$ tend to $0$ (For $p_{k} \geq 0$ the product $\prod (1 + p_{k})$ converges iff $\sum p_{k}$ converges, with $1 + \sum_{k < n} p_{k} \leq \prod_{k < n} (1 + p_{k}) \leq 1 / (1 - \sum_{k < n} p_{k})$ when $\sum_{k < n} p_{k} < 1$ ; for $0 \leq p_{k} < 1$ the product $\prod (1 - p_{k})$ converges iff $\sum p_{k}$ converges and its partial products tend to $0$ otherwise; and $\sum ∣ p_{k} ∣$ convergent implies $\prod (1 + p_{k})$ convergent).

[L9]

$\sum_{k \geq 1} 1 / k^{p}$ diverges at $p = 1$ ; direct comparison in its divergence form; and $\sum c x_{k}$ diverges when $c \neq 0$ and $\sum x_{k}$ diverges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ , If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ , Convergent series add and scale termwise, Integer powers $a^{m}$ , Series, partial sums, convergence and the sum, divergence, and the tail series).

[L10]

The squeeze theorem (The squeeze theorem).

[L11]

Convergence of an infinite product (Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors).

Counterexample

technique · direct

1.1

For every $j$ , $ι (j + 2) \geq ι (2) = 2$ , so $\sqrt{ι (j + 2)} \geq \sqrt{2} > 1$ and $∣ p_{j} ∣ = 1 / \sqrt{ι (j + 2)} \leq 1 / \sqrt{2} < 1$ ; hence every factor satisfies $0 < f_{j} \leq 1 + 1 / \sqrt{2} < 2$ , and every $Π_{m}$ is positive.

givenL1L2L3L6

1.2

Fix $i$ and put $a : = ι (2 i + 2)$ , $b : = ι (2 i + 3)$ , so $b - a = 1$ and $a + b = ι (4 i + 5)$ , both positive. By [L1], $f_{2 i} = 1 + 1 / \sqrt{a}$ and $f_{2 i + 1} = 1 - 1 / \sqrt{b}$ .

L1L3

1.3

An induction gives that finite products are monotone in nonnegative factors: if $0 \leq x_{i} \leq y_{i}$ for all $i < n$ then $\prod_{i < n} x_{i} \leq \prod_{i < n} y_{i}$ , since both products are nonnegative and $\prod_{i < n + 1} x_{i} = (\prod_{i < n} x_{i}) x_{n} \leq (\prod_{i < n} y_{i}) x_{n} \leq (\prod_{i < n} y_{i}) y_{n}$ .

L6L7

2.1

The sequence $β_{j} : = 1 / \sqrt{ι (j + 2)}$ is positive, nonincreasing and converges to $0$ : monotonicity from $0 < ι (j + 2) < ι (j + 3)$ and strict increase of the square root, and convergence because, given a rational $ε > 0$ , an $n \geq 1$ with $1 / ι (n) < ε^{2}$ gives $ι (j + 2) > (1 / ε)^{2}$ and so $β_{j} < ε$ for every $j \geq n$ .

step 1.1L2L3

2.2

Since $\sqrt{b} - \sqrt{a} = (b - a) / (\sqrt{a} + \sqrt{b}) = 1 / (\sqrt{a} + \sqrt{b})$ , one has $\frac{1}{\sqrt{a}} - \frac{1}{\sqrt{b}} = \frac{\sqrt{b} - \sqrt{a}}{\sqrt{a} \sqrt{b}} = \frac{1}{\sqrt{a b} (\sqrt{a} + \sqrt{b})}$ , so $P_{i} : = f_{2 i} f_{2 i + 1} = 1 + \frac{1}{\sqrt{a}} - \frac{1}{\sqrt{b}} - \frac{1}{\sqrt{a b}} = 1 - \frac{1}{\sqrt{a b}} (1 - \frac{1}{\sqrt{a} + \sqrt{b}})$ .

step 1.2L2algebra

2.3

Here $\sqrt{a} \geq \sqrt{2} > 1$ and $\sqrt{b} \geq \sqrt{3} > 1$ , so $\sqrt{a} + \sqrt{b} > 2$ and $1 - 1 / (\sqrt{a} + \sqrt{b}) > 1 / 2$ ; and by [L5], $\sqrt{a b} \leq (a + b) / 2 = ι (4 i + 5) / 2$ , so $1 / \sqrt{a b} \geq 2 / ι (4 i + 5)$ .

step 1.2L2L3L5

2.4

An induction gives $Π_{2 n} = \prod_{i < n} P_{i}$ for every $n$ : at $n = 0$ both are the empty product $1$ , and $Π_{2 (n + 1)} = Π_{2 n} f_{2 n} f_{2 n + 1} = Π_{2 n} P_{n}$ .

step 1.2L6L7

3.1

By the alternating series test $\sum_{j} p_{j} = \sum_{j} ε_{j} β_{j}$ converges.

step 2.1L4

3.2

Combining, $P_{i} \leq 1 - (2 / ι (4 i + 5)) \cdot \frac{1}{2} = 1 - 1 / ι (4 i + 5) \leq 1 - q_{i}$ , where $q_{i} : = 1 / ι (4 i + 6)$ , using $ι (4 i + 5) < ι (4 i + 6)$ ; and $0 < P_{i}$ by step 1.1, while $0 < q_{i} < 1$ .

step 1.1step 2.2step 2.3L3

4.1

Hence $0 < Π_{2 n} = \prod_{i < n} P_{i} \leq \prod_{i < n} (1 - q_{i})$ for every $n$ .

step 1.1step 3.2step 2.4step 1.3

4.2

The series $\sum_{i} q_{i}$ diverges: $6 (i + 1) = 6 i + 6 \geq 4 i + 6$ , so $ι (4 i + 6) \leq 6 ι (i + 1)$ and $q_{i} \geq \frac{1}{6} \cdot \frac{1}{ι (i + 1)}$ ; the series $\sum_{i} \frac{1}{6} \cdot 1 / ι (i + 1)$ diverges, being a nonzero multiple of the harmonic series, so $\sum_{i} q_{i}$ diverges by comparison.

step 3.2L3L9

5.1

By [L8] applied to $(q_{i})$ , the partial products $\prod_{i < n} (1 - q_{i})$ tend to $0$ ; with step 4.1 and the squeeze, $Π_{2 n} \to 0$ .

step 4.1step 4.2L8L10

6.1

Also $Π_{2 n + 1} = Π_{2 n} f_{2 n}$ with $0 < f_{2 n} < 2$ , so $0 < Π_{2 n + 1} < 2 Π_{2 n}$ and $Π_{2 n + 1} \to 0$ as well.

step 1.1step 5.1L6L10

7.1

Therefore $Π_{m} \to 0$ : given a rational $ε > 0$ , choose $N$ with $Π_{2 n} < ε / 2$ for all $n \geq N$ ; then for $m \geq 2 N + 1$ , writing $m$ as $2 n$ or $2 n + 1$ according to the partition of $N$ by the two index maps, in either case $n \geq N$ and $Π_{m} \leq 2 Π_{2 n} < ε$ .

step 1.1step 5.1step 6.1L1

8.1

For every $N^{'}$ the $N^{'}$ -th tail products satisfy $\prod_{j = N^{'}}^{N^{'} + n - 1} f_{j} = Π_{N^{'} + n} / Π_{N^{'}}$ with $Π_{N^{'}} > 0$ fixed, so they tend to $0$ too; no tail has partial products with a nonzero limit, and $\prod_{j} (1 + p_{j})$ diverges.

step 1.1step 7.1L6L11

9.1

So $\sum_{j} p_{j}$ converges while $\prod_{j} (1 + p_{j})$ diverges, and the refuted claim fails; the hypothesis it is missing is a sign condition, or absolute convergence of $\sum p_{j}$ , as in For $p_{k} \geq 0$ the product $\prod (1 + p_{k})$ converges iff $\sum p_{k}$ converges, with $1 + \sum_{k < n} p_{k} \leq \prod_{k < n} (1 + p_{k}) \leq 1 / (1 - \sum_{k < n} p_{k})$ when $\sum_{k < n} p_{k} < 1$ ; for $0 \leq p_{k} < 1$ the product $\prod (1 - p_{k})$ converges iff $\sum p_{k}$ converges and its partial products tend to $0$ otherwise; and $\sum ∣ p_{k} ∣$ convergent implies $\prod (1 + p_{k})$ convergent.

step 3.1step 8.1L8L11∎

Remarks

The pairing is what replaces the logarithm. The classical argument writes $\log (1 + p_{j}) = p_{j} - p_{j}^{2} / 2 + \dots$ and observes that $\sum p_{j}$ converges while $\sum p_{j}^{2}$ diverges, so the logarithms sum to $- \infty$ . That expansion is not available at this point in the reading order. Pairing consecutive factors reproduces the same effect with one algebraic identity: the first-order terms cancel to size $1 / (\sqrt{a b} (\sqrt{a} + \sqrt{b}))$ , of order $i^{- 3 / 2}$ , while the cross term $- 1 / \sqrt{a b}$ , of order $i^{- 1}$ , survives, and its sum diverges.
Absolute convergence would settle it the other way. Here $\sum_{j} ∣ p_{j} ∣ = \sum_{j} 1 / \sqrt{ι (j + 2)}$ diverges, so claim 4 of For $p_{k} \geq 0$ the product $\prod (1 + p_{k})$ converges iff $\sum p_{k}$ converges, with $1 + \sum_{k < n} p_{k} \leq \prod_{k < n} (1 + p_{k}) \leq 1 / (1 - \sum_{k < n} p_{k})$ when $\sum_{k < n} p_{k} < 1$ ; for $0 \leq p_{k} < 1$ the product $\prod (1 - p_{k})$ converges iff $\sum p_{k}$ converges and its partial products tend to $0$ otherwise; and $\sum ∣ p_{k} ∣$ convergent implies $\prod (1 + p_{k})$ convergent does not apply. That claim is exactly the hypothesis under which a signed product is safe.
The refinement that decides every case is deferred. For signed $(p_{j})$ with $\sum p_{j}$ convergent, the classical criterion is convergence of $\sum p_{j}^{2}$ ; for this witness $p_{j}^{2} = 1 / ι (j + 2)$ and that series diverges, which is consistent with what is proved above. The criterion itself needs the logarithm and is recorded in Selected sums and products on this page that are proved to exist without being evaluated, and what their evaluation waits for.

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Sources

Infinite product (Wikipedia) (standard reference, not scraped)
Alternating series test (Wikipedia) (standard reference, not scraped)
Thomson, Bruckner, and Bruckner, Elementary Real Analysis (standard reference, not scraped)
D. Dikranjan, Analysis 478, Chapter 6 (standard reference, not scraped)

∏j≥0(1+(−1)j/j+2)∏j≥0​(1+(−1)j/j+2​) has partial products tending to 00 although ∑j≥0(−1)j/j+2∑j≥0​(−1)j/j+2​ converges