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$\beta_n$ is a sequence of positive real numbers.It is known that $\beta_n = O(1/n)$ and $\beta_n\ne o(1/n)$. What can we say about the series $$\sum_{k=0}^n \beta_k$$

1 Answers 1

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The series may converge or diverge. Consider the examples

$$ \beta_n = \frac{1}{n} $$

and

$$ \beta_n = \begin{cases} 2^{-n} & \text{if } n \neq 2^m \text{ for any } m \in \mathbb Z, \\ 2^{-m} & \text{if } n = 2^m \text{ for some } m \in \mathbb Z. \end{cases} $$

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    I am not asking for convergence. I mean to know it has any particular asymptotics. Like $O(log(n))$. – 2017-01-29
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    It cannot converge – 2017-01-29
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    @RajeshDachiraju, the second example I gave in my answer converges. – 2017-01-29
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    Both co ditions ate simultaneous. It is O(1/n) and not o(1/n) at same time. – 2017-01-29
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    @RajeshDachiraju, yes, I know that. The second example satisfies both conditions *and* converges. – 2017-01-29
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    looks like there is some misconception about my notation or some misunderstanding. by $\beta_n \ne o(1/n)$ I intend $\beta_n$ does not decay faster than $1/n$. Let me know if anything wrong with my notation. – 2017-01-29
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    @RajeshDachiraju, that is not very clear. The formal definition of $\beta_n = o(1/n)$ is $\lim_{n \to \infty} |n\beta_n| = 0$, so $\beta_n \neq o(1/n)$ means $\lim_{n \to \infty} |n\beta_n|$ doesn't exist or is $> 0$. This is equivalent to $\limsup_{n \to \infty} |n\beta_n| > 0$. For the second example of my answer, $\limsup_{n\to\infty} |n\beta_n| = 1$, so $\beta_n \neq o(1/n)$ and $\beta_n = O(1/n)$. – 2017-01-29
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    Please see this : https://en.wikipedia.org/wiki/Big_O_notation – 2017-01-29
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    According to above link, your second example is not valid. – 2017-01-29
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    Thanks for the answer Antonio – 2017-01-30
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    @RajeshDachiraju Could you clarify what you mean by "your second example is not valid"? – 2017-01-30
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    I thought (1/2)^n decays rapidly than 1/n. I caught between exponential and hypebolic. – 2017-01-30
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    @RajeshDachiraju, but $(1/2)^n$ does decay more rapidly than $1/n$. The point is that $\beta_n$ is not equal to $(1/2)^n$ for all $n$, only for some. – 2017-01-30
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    @RajeshDachiraju, maybe writing it this way would make it clearer: $$\beta_n = \begin{cases} 1/n & \text{if } n = 2^{m},\, m \in \mathbb Z \\ (1/2)^n & \text{otherwise}. \end{cases}$$ – 2017-01-30
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    Hah i got it now. This makes more clear to understand. – 2017-01-30