# Need counter example to series-related statement

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?

• an example is $a_n=\frac1{n^2}$ – J. W. Tanner Nov 26 '19 at 11:36
• do you mean $n \to \infty$ in the limit? – Multigrid Nov 26 '19 at 11:38
• @J.W.Tanner thanks! – user532874 Nov 26 '19 at 11:43

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 $$\sum\limits_{n=1}^\infty\dfrac1{n^2}=\dfrac{\pi^2}6$$.
• Off-topic. I have a T-shirt with the series of $\pi^2/6$! – manooooh Nov 26 '19 at 11:49