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Hi is it possible to use Abel's Theorem for series to show the following:

If $\{x_{n}\}_{1}^{\infty}$ is a positive sequence such that $\lim\limits_{n \rightarrow \infty}x_{n} = 0$ and $\sum_{n=1}^{\infty}x_{n}$ diverges. Prove that the series $$\sum_{n=1}^{\infty}\frac{x_{n}}{x_{n}+1}$$ diverges. I have already proved this result by other means but am interested if you can use Abel's Theorem or the Integral Test to show this. Thanks.

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Since $\displaystyle \lim \limits_{n \to +\infty} x_{n} = 0$ and $(x_{n})_{n \in \mathbb{N}^{\ast}}$ is a sequence of positive real numbers, note that :

$$ \frac{x_n}{1+x_n} \, \mathop{\sim} \limits_{n \to +\infty} \, x_{n} $$

Therefore, the series $\displaystyle \sum_{n \in \mathbb{N}^{\ast}} \frac{x_n}{1+x_{n}}$ and $\displaystyle \sum_{n \in \mathbb{N}^{\ast}} x_{n}$ have the same behavior. Since $\displaystyle \sum_{n \in \mathbb{N}^{\ast}} x_{n}$ is divergent,$\displaystyle \sum_{n \in \mathbb{N}^{\ast}} \frac{x_n}{1+x_{n}}$ diverges too.

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  • $\begingroup$ Thanks for your response. I am though specifically interested in knowing if we can use those two tests to show the divergence of the series. $\endgroup$ – user116403 Oct 6 '14 at 10:13
  • $\begingroup$ I do not see how to apply the integral test here since it only concerns series of the form $\displaystyle \sum_{n \in \mathbb{N}} f(n)$ where $f$ is a decreasing, non-negative function. Here, we do not know $x_n$ explicitly. Therefore, $\displaystyle \sum_{n \in \mathbb{N}^{\ast}} \frac{x_n}{1+x_{n}}$ is not of the form $\displaystyle \sum_{n \in \mathbb{N}} f(n)$. $\endgroup$ – jibounet Oct 6 '14 at 10:18
  • $\begingroup$ Yeah I think so too. Specifically since we can't say whether it is decreasing? On a side point does it follow necessarily that if $\{x_{n}\}$ is a positive sequence such that $\lim_{n \rightarrow \infty}x_{n} = 0$ then $\{x_{n}\}$ is a decreasing sequence? $\endgroup$ – user116403 Oct 6 '14 at 10:20
  • $\begingroup$ Not necessarily. Consider $\displaystyle x_{n} = \frac{1+\sin(n)}{n}$. $x_{n}$ is positive and $\displaystyle \lim \limits_{n \to +\infty} x_{n} = 0$ but $(x_{n})$ is not decreasing. $\endgroup$ – jibounet Oct 6 '14 at 10:25
  • $\begingroup$ @jibounet I don't think it has to be given explicitly. You have $f(n) := \frac{x_{n}}{1 + x_{n}}$. I think it's more important if it satisifes the conditions of as you stated non-negative and decreasing. $\endgroup$ – user100431 Oct 6 '14 at 10:26

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