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?
Partial sums of infinite series
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elementary-number-theory
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0What if k is large and the partial sum is arbitrary? How would I should that $b < q$ or $b > q$? – 2012-09-05
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0What if $k$ is represented by the p-series for an arbitrarily large p, would $b < q$? – 2012-09-05
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0No 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
2 Answers
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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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0You 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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0What if $k$ is represented by the p-series for an arbitrarily large p, would $b < q$? – 2012-09-05
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0What do you mean by "the $p$-series"? If you mean $\sum n^{-p}$ with $p$ an integer (so the partial sums are rational), then we don't expect $k$ to be rational. – 2012-09-05
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0Yes, but if I suppose $k$ to be rational, like in a proof by contradiction, for example, for $p = 5$ i.e. $\zeta (5)$. – 2012-09-05
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0Perhaps you are trying to prove $\zeta(5)$ irrational by assumming it rational and then saying that its denominator must exceed any denominator in the partial sums, but those are unbounded, contradiction. In the first place, if it were that easy, Euler (or Pythagoras!) would have done it. In the second place, it assumes something is true for $p$-series that isn't true for series in general, so it requires proving it for $p$-series - but then you're back to square one. – 2012-09-05
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0If $q$ 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