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Determine the convergence of following series. $$ \sum_{n=1}^\infty a_n$$ where $$ a_n = \begin{cases} \bigg( \dfrac{2+3n}{5+6n} \bigg)^n, & \text{if $n$ is even} \\ 5 & \text{if $n$ is odd} \end{cases} $$

As I think this series obviously divergence series because the when $n$ is odd $a_n=5$, so all the partial sum of odd terms divergence

But how can I prove it using Root test or Ratio test?

I used root test then I got

$L=\lim \sup |5^\frac{1}{n}|=1$ so test is inconclusive

Can anyone give some idea about this problem?

Thank you!

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    $\begingroup$ $a_n$ does not tend to $0$ so the series does not converge. $\endgroup$ Jun 27 at 23:15
  • $\begingroup$ @KaviRamaMurthy Is there a way to prove it by using root test or ratio test? $\endgroup$
    – ALMEra
    Jun 27 at 23:17
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    $\begingroup$ No test is required except what I mentioned in above comment. Always look at $\lim a_n$ before applying any test for convergence. $\endgroup$ Jun 27 at 23:18
  • $\begingroup$ @KaviRamaMurthy Thank you $\endgroup$
    – ALMEra
    Jun 27 at 23:19
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If $n$ is even,$$a_n<\left(\frac{2+3n}{4+6n}\right)^n=\frac1{2^n},$$and therefore$$\frac{a_{n+1}}{a_n}>5\times2^n.$$So, by the ratio test, the series diverges.

But this is a complete waste of time. Since you don't have $\lim_{n\to\infty}a_n=0$, the series diverges. That's all.

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