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?
Asked
Active
Viewed 508 times
2 Answers
7
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 $1$.)
Jonas Meyer
- 53,602
6
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.)
Qiaochu Yuan
- 419,620
-
1Here is a reference: http://www.jstor.org/stable/2687153 – Jonas Meyer Nov 08 '10 at 01:01
-
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 $g(n) = (2/3)^j f(n)$ – Zarrax Nov 08 '10 at 02:09
-
@Zaricuse: I guess you meant $k_1 < k_2 < \cdots$. – Hans Lundmark Nov 08 '10 at 06:33
-
yeah $k_1 < k_2 < ...$ and $k_j \leq n < k_{j+1}$ – Zarrax Nov 08 '10 at 06:42