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  "body": "\n## Statement\n\nLet $(n_k)_{k \\ge 1}$ be a strictly increasing sequence of positive integers, and assume there is a real $q \u003e 1$ with $n_{k+1} / n_k \\ge q$ for all $k$.\nLet $(a_k)_{k \\ge 1}$ be complex numbers such that the power series\n$$f(x) = \\sum_{k \\ge 1} a_k x^{n_k}$$\nconverges for $0 \\le x \u003c 1$, and assume $f(x) \\to s$ as $x \\to 1^-$.\nThen the series $\\sum_{k \\ge 1} a_k$ [[converges|0CVVKY7GY2]] and its sum is $s$.\n\n## Remarks\n\nIn the language of [[Abel summation|EZXPY4A19C]], the theorem says that a series with Hadamard gaps that is Abel summable is convergent.\nSo the gap condition acts as a Tauberian condition, and no condition on the size of the coefficients $a_k$ is needed.\nThe proof is considerably harder than that of [[Tauber's theorem|8SK3GQ4S0M]].\nThe gap hypothesis cannot be dropped: $\\sum_{n \\ge 0} (-1)^n x^n = 1/(1 + x) \\to 1/2$ as $x \\to 1^-$, but $\\sum (-1)^n$ diverges.\nAs a consequence, $f(x) = \\sum_{k \\ge 0} (-1)^k x^{2^k}$ has no limit as $x \\to 1^-$, because $\\sum_k (-1)^k$ diverges.\n\n## Explanations\n\n### L1 human\n\nA power series with very sparse exponents, such as $x^2, x^4, x^8, \\dots$, cannot hide divergence behind smoothing.\nIf its value approaches a limit as $x$ goes up to $1$, then the series of coefficients already converges to that limit.\nFor ordinary series this is false, so the theorem shows how strong a constraint sparse exponents are.\n\n### L2 human\n\n[[Abel's theorem|0S4EMDQS2F]] goes from convergence to Abel summability, and Tauberian theorems go back under extra hypotheses.\nUsual Tauberian conditions bound the terms, as in $n a_n = O(1)$.\nHere the extra hypothesis is only on where the nonzero terms sit.\nThe same gap condition appears in the [[Hadamard gap theorem|DA8M8VZZHC]], where it forces the unit circle to be a natural boundary.\n",
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