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Prove the following statement is false by providing a counter-example

If $\lim_{n \to \infty}|\frac {a_{n+1}}{a_n}| =1$ then $\sum_{n=1}^\infty a_n $ diverges.

Can anyone think of the simplest series possible where $\lim_{x \to \infty}|\frac {a_{n+1}}{a_n}| =1$ and $\sum_{n=1}^\infty a_n $ converges?

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    $\begingroup$ an example is $a_n=\frac1{n^2}$ $\endgroup$ – J. W. Tanner Nov 26 '19 at 11:36
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    $\begingroup$ do you mean $n \to \infty$ in the limit? $\endgroup$ – Multigrid Nov 26 '19 at 11:38
  • $\begingroup$ @J.W.Tanner thanks! $\endgroup$ – user532874 Nov 26 '19 at 11:43
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If $a_n=\dfrac 1{n^2}$, then $\lim\limits_{n\to\infty}\left|\dfrac {a_{n+1}}{a_n}\right|=\lim\limits_{n\to\infty}\dfrac{n^2}{(n+1)^2}=1$, but famously $\sum\limits_{n=1}^\infty\dfrac1{n^2}=\dfrac{\pi^2}6$.

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    $\begingroup$ Off-topic. I have a T-shirt with the series of $\pi^2/6$! $\endgroup$ – manooooh Nov 26 '19 at 11:49

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