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I'm trying to find out whether the series $$\sum\limits_{n=1}^{\infty}(-1)^n\ln\left[\frac{8n+2}{7n+1}\right]$$ converges or not, but the alternating series test seems not to apply. What other tests can I use? Does this series converges or diverges?

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    $\begingroup$ The terms don't even tend to zero, so..... $\endgroup$ – user296602 May 10 '16 at 21:31
  • $\begingroup$ @T.Bongers The term inside the floor function goes to $8/7$ which means that the limit will be $\ln 1 = 0$ $\endgroup$ – Ant May 10 '16 at 21:32
  • $\begingroup$ @Ant How can that be possible? The limit is $\;\log\frac87\neq\log1=0\;$, isn't it? $\endgroup$ – DonAntonio May 10 '16 at 21:33
  • $\begingroup$ @Joanpemo The square brackets used this way usually indicate the floor function.. If they are just parenthesis indeed you're correct $\endgroup$ – Ant May 10 '16 at 21:33
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    $\begingroup$ @Ant Those look like brackets to me, not $\lfloor$ and $\rfloor$. $\endgroup$ – user296602 May 10 '16 at 21:33
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$$\log\frac{8n+2}{7n+1}\xrightarrow[n\to\infty]{}\log\frac87\neq0$$

and thus

$$(-1)^n\log\frac{8n+2}{7n+1}\rlap{\;\;\;\;/}\xrightarrow[n\to\infty]{}0\implies\text{ the series cannot converge}$$

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  • $\begingroup$ you forgot the floor function $\endgroup$ – Ant May 10 '16 at 21:32
  • $\begingroup$ @Ant Thank you, but excuse me: what floor function? For me those are square parentheses...Perhaps the asker can clear this out. $\endgroup$ – DonAntonio May 10 '16 at 21:34
  • $\begingroup$ @Ant Thank you. Yes, but it really can cause confusion. $\endgroup$ – DonAntonio May 10 '16 at 21:36

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