Erdős Problem 1050
Statement and result.
The series \[
S=\sum_{j=1}^{\infty}\frac1{2^j-3}
\] is irrational. Its first two terms cancel, and the remaining terms form an absolutely convergent positive series. We reconstruct the integer-linear-form argument in Borwein's proof, with its Padé approximation estimates explicitly identified as external inputs.
The approximation input.
Put $L(x)=\sum_{j\geq1}x/(2^j-x)$, away from its poles. Borwein's Padé construction supplies polynomials $Q_N,P_N$, both of degree at most $N$, with the following properties. For $N\geq2$, put \[
D_N=\prod_{k=\lfloor N/2\rfloor}^{N}(2^k-1).
\] Then $Q_N\in\mathbb Z[x]$, $D_NP_N\in\mathbb Z[x]$, and there are absolute constants $C_1,C_2>0$, independent of $N$, such that \[
|Q_N(x)|\leq C_1\,2^{N^2}\quad(-1\leq x\leq1),
\tag{1}
\] \[
0<\left|L(x)-\frac{P_N(x)}{Q_N(x)}\right|
\leq C_2\,\frac{|x|^{2N}}{2^{N(N+1)}}
\quad(0<|x|\leq1).
\tag{2}
\] In particular the quotients in $2$ are defined. These are the denominator, size, and nonzero-error properties used in the proof of Theorem 4 on pages 257--258 of Borwein's paper; $2$ is its Theorem 3. We do not reprove the underlying Padé theory. There is a sign inconsistency in the scanned statement of Theorem 1(b): its printed plus sign in the denominator-clearing product is incompatible even with $N=3$. Here $D_N$ uses the minus factors in the proof of Theorem 4. Only their absolute values enter the estimates.
Moving to a small argument.
For $N\geq2$, set $x_N=3/2^N$ and \[
R_N=\sum_{j=1}^{N}\frac3{2^j-3}.
\] Changing index in the absolutely convergent tail gives the exact identity \[
L(3)-R_N=L(x_N).
\tag{3}
\] Applying $2$ at $x_N$ therefore gives \[
0<
\left|L(3)-R_N-\frac{P_N(x_N)}{Q_N(x_N)}\right|
\leq C_2\,9^N\,2^{-3N^2-N}.
\tag{4}
\]
Clearing all denominators.
Let \[
B_N=\prod_{j=1}^{N}(2^j-3),\qquad T_N=B_ND_N,
\] and define \[
V_N=2^{N^2}T_NQ_N(x_N),\qquad
U_N=2^{N^2}T_N\left(Q_N(x_N)R_N+P_N(x_N)\right).
\] Both are integers. Indeed, evaluating a degree-at-most-$N$ integral polynomial at $3/2^N$ requires at most the denominator $2^{N^2}$; $D_N$ clears the coefficients of $P_N$, while $B_N$ clears $R_N$. No factor in $B_ND_N$ is zero. The elementary product estimates give \[
|B_N|\leq2^{N(N+1)/2},\qquad
D_N\leq2^{\,3N^2/8+N+1},
\] and hence, with room to spare, \[
|T_N|\leq2^{\,7N^2/8+2N+1}.
\tag{5}
\] Multiplying $4$ by $2^{N^2}|T_NQ_N(x_N)|$ and using $1$, $5$, we obtain \[
0<|V_NL(3)-U_N|
\leq 2C_1C_2\,9^N\,2^{-N^2/8+N}
\longrightarrow0.
\tag{6}
\] The negative quadratic exponent dominates the linear terms.
Irrationality and assessment.
If $L(3)=a/b$ with integers $a,b$, $b\geq1$, the quantity $b(V_NL(3)-U_N)$ would be a nonzero integer. Its absolute value could not tend to zero, contradicting $6$. Thus $L(3)$, and therefore $S=L(3)/3$, is irrational. This is a complete proof relative to the explicitly stated Padé input, not a new independent derivation of that input. The denominator arithmetic, small-argument shift, and irrationality deduction are included. Primary source: P. B. Borwein,
On the irrationality of $\sum 1/(q^n+r)$, Journal of Number Theory 37 $1991$, 253--259,
https://www.cecm.sfu.ca/~pborwein/PAPERS/P51.pdf; statement:
https://www.erdosproblems.com/1050. Sources checked 24 September 2026. No novelty or local formal verification is claimed.
Completion estimate.
COMPLETION: 100%