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Is there some increasing function $f(n)$ that grows slower than $n^{c}$ for some $c > 1$ such that $\sum_{n=1}^{\infty} \frac{1}{f(n)}$ converges?

2 Answers 2

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Yes, $f(n)=n(\log(n))^2$. (Of course you can ignore the first term where the reciprocal is undefined, and the finitely many terms where it may be bigger than $n^c$ for $c$ close to $1$.)

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The answer to your title question, however, is no. That is, given an increasing function $f(n) : \mathbb{N} \to \mathbb{R}$ such that $\sum \frac{1}{f(n)}$ converges, there exists an increasing function $g(n)$ such that

  • $\sum \frac{1}{g(n)}$ converges, and
  • $\lim_{n \to \infty} \frac{g(n)}{f(n)} = 0$.

(Unfortunately the proof escapes me at the moment.)

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    Here is a reference: http://www.jstor.org/stable/26871532010-11-08
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    It goes like this.. let $k_1 > k_2 > ...$ such that $\sum_{n \geq k_j} {1 \over f(n)} < 3^{-j}$. Let $A_j$ be the set of $n$ with $k_j \geq n > k_{j+1}$. Then for $n \in A_j$ let $g(n) = (2/3)^j f(n)$2010-11-08
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    @Zaricuse: I guess you meant $k_1 < k_2 < \cdots$.2010-11-08
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    yeah $k_1 < k_2 < ...$ and $k_j \leq n < k_{j+1}$2010-11-08