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Suppose some constant $k$ can be represented by a non-alternating convergent infinite series whose partial sums are rational, and we assume $k$ is rational, say $p/q$. If we consider a partial sum $s_k = a/b$ would $b < q$ always hold?

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    No need to comment on your own question - you can just edit stuff into the body of the question. But if the partial sum is arbitrary (but still rational), then it's arbitrary, and you won't be able to say anything about the relation between $b$ and $q$.2012-09-05

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$\sum_{k=1}^\infty\frac{1}{2^k}=\frac{1}{1}$ while $s_1=\frac{1}{2}.$

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    You could even point out that the denominator (in lowest terms) of $s_n$ is $2^n$, so not only are the denominators all bigger than that (in lowest terms) of $k$, but they aren’t even bounded!2012-09-05
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No. $7/16+0+0+0+0+0+0+0+0+1/16+0+0+0+0+\cdots=1/2$

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    If $q$ is a prime then the denominator of $\sum_1^qn^{-p}$ is divisible by $q^p$, hence is at least $q^p$. Since there are infinitely many primes, the denominators of the partial sums of the $p$-series are unbounded. I can't imagine any useful conditions guaranteeing $q\gt b$. If the series converges, and the terms are non-zero, then the denominators of the terms must go to inifnity, and I suspect the denominators of the partial sums must go to infinity as well - if that's true, you'll never get $b\lt q$ for all $k$.2012-09-06