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Let's say I have a sequence $a_n \ge 0$ such that I know:

$$b \log n - C \le \sum_{i=1}^n a_i \le b \log n + C$$

for some constants $b$ and $C$ larger than 0.

How can I prove that:

$$a_n = \frac{b}{n} + o(1)\ ?$$

This intuitively seems correct because we know that for the harmonic series we get $\sum_{i=1}^n \frac{1}{i} = \log n + o(1)$, but I am not completely sure how to show the reverse.

  • 4
    Perhaps $b/n + o(1)$ is not what you intend? Since $b/n$ is, itself, $o(1)$.2012-02-23
  • 0
    @GEdgar: Exactly what I was about to comment!2012-02-23
  • 1
    I wonder if something along the lines of $a_n=\frac{b}{n}+o(\frac{1}{n})$ can be saved if $a_n$ is required to be nonincreasing?2012-02-24

2 Answers 2