Solution

Since the positive series converges,

a_{n} \to 0

. We construct the subsequence recursively. Choose

n (1)

arbitrarily. Once

n (1) < \dots < n (k - 1)

have been chosen, write the reduced partial sums as

s_{j} = \sum_{r = 1}^{j} a_{n (r)} = \frac{p _{j}}{q _{j}} (1 \leq j < k) .

Choose

n (k) > n (k - 1)

so far out that

0 < a_{n (k)} < 2^{- k} mi n_{1 \leq j < k} q_{j}^{- j} .

This is possible because

a_{n} \to 0

Let

α = \sum_{k = 1}^{\infty} a_{n (k)} .

For every fixed

j

, each later choice includes

q_{j}^{- j}

in its minimum, so

0 < α - s_{j} < q_{j}^{- j} \sum_{k = j + 1}^{\infty} 2^{- k} = 2^{- j} q_{j}^{- j} .

First,

α

is irrational. If

α = u / v

were reduced and rational, the strictly increasing rational sequence

s_{j} \to α

would have unbounded denominators

q_{j}

; only finitely many reduced fractions in a bounded interval have bounded denominator. Choose

j > 2

with

q_{j} > v

. Part (b) would give

∣ α - s_{j} ∣ > q_{j}^{- 2},

contrary to

∣ α - s_{j} ∣ < 2^{- j} q_{j}^{- j} < q_{j}^{- 2}

α

were algebraic of degree

d

, part (a) would give a constant

c > 0

with

∣ α - s_{j} ∣ > c q_{j}^{- d} .

For

j > d

our construction instead gives

∣ α - s_{j} ∣ < 2^{- j} q_{j}^{- j} \leq 2^{- j} q_{j}^{- d},

which contradicts the lower bound once

2^{- j} < c

. Hence

α

Solved by gpt-5.6-sol high.

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