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Let $\alpha_n$ and $\beta_n$ be two sequences of real numbers. The sequence $\beta_n$ is absolutely summable, i.e $$\lim_{n\to\infty}\sum_{k=0}^n \left|\beta_k\right| = l$$ while that is not the case with the former sequence, and $$\sum_{k=0}^n \left|\alpha_k\right| \sim g(n)$$ where $g(n)$ is a strictly increasing function of $n$. Also given $\lim_{n \to \infty}g(n)\beta_n = 0$

Let $\gamma_n$ be another sequence, where $\gamma_n = \left|\alpha_n + \beta_n\right|$ then I'd like to prove

$$\sum_{k=0}^n \left|\gamma_k\right| \sim g(n)$$

I am hoping it would be a straightforward application of some theorem that I am not aware of, either to prove or disprove.

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    What does $\sim$ mean? – 2017-02-01
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    @GuachoPerez : $x_n \sim y_n$ implies $$\lim_{n \to \infty}\frac{x_n}{y_n} = 1$$ – 2017-02-01
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    I assume you mean $g(n)$ increases without bound. – 2017-02-01
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    @Fimpellizieri : yes – 2017-02-01

1 Answers 1

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On the one hand, as $n \to \infty$:

$$\frac{\sum_{k=1}^n|\alpha_k+\beta_k|}{g(n)}\leq\frac{\sum_{k=1}^n|\alpha_k|}{g(n)}+\frac{\sum_{k=1}^n|\beta_k|}{g(n)}\longrightarrow 1$$

On the other:

$$\frac{\sum_{k=1}^n|\alpha_k+\beta_k|}{g(n)}\geq \frac{\sum_{k=1}^n\Big||\alpha_k|-|\beta_k|\Big|}{g(n)}\geq \frac{\sum_{k=1}^n|\alpha_k|-|\beta_k|}{g(n)}=\frac{\sum_{k=1}^n|\alpha_k|}{g(n)}-\frac{\sum_{k=1}^n|\beta_k|}{g(n)}\longrightarrow 1$$

So by the squeeze theorem, what you'd like to prove is true.

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    Nice Answer Fimpellizieri. Thanks. Hope this can be a handy tool while dealing with asymptotic estimates. One can just quote this post. – 2017-02-01
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    You are welcome! Glad to help. – 2017-02-01