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Let $\displaystyle a_n=(2^n+3^n)^\frac{1}{n}$. Compute the limit $$\displaystyle \lim_{n\to\infty}\dfrac{n}{\displaystyle \sum_{i=1}^{n}\dfrac{1}{a_i}}$$

By squeeze theorem, limit of $a_n$ is $3$. Further the sequence $\displaystyle \sum_{i=1}^{n}\dfrac{1}{a_i}$ is strictly monotonic and diverges to infinity since the limit of $n^{th}$ term of the series $\displaystyle \sum_i\dfrac{1}{a_i}$ is not $0$ infact $1/3$. Hence by Cesaro Stolz theorem for $\cdot/\infty$ case, $$\displaystyle \lim_{n\to\infty}\dfrac{n}{\displaystyle \sum_{i=1}^{n}\dfrac{1}{a_i}}=3.$$

Is my reasoning correct? Are there more easier ways to solve this?

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    $\begingroup$ What you have done is correct and it seems to be the best way to answer the question. $\endgroup$ – Kavi Rama Murthy Dec 13 '19 at 23:28
  • $\begingroup$ @KaboMurphy Thank you Sir. $\endgroup$ – Yadati Kiran Dec 13 '19 at 23:29
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    $\begingroup$ This works for any mean and any sequence that approaches a limit. $\endgroup$ – marty cohen Dec 13 '19 at 23:43
  • $\begingroup$ @martycohen Does that mean I can employ Cauchy's first theorem on limits here? $\endgroup$ – Yadati Kiran Dec 14 '19 at 0:03
  • $\begingroup$ @YadatiKiran:How you used squeeze theorem to show $a_n \rightarrow 0$? $\endgroup$ – Styles Sep 29 '20 at 16:43
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$a_n = 3\Big(1+(\frac23)^n\Big)^{1/n}=3\exp\Big(\frac1n\ln(1+(2/3)^n)\Big)\xrightarrow{n\rightarrow\infty}3$. The rest follows from Cesaro sums: $\lim_{n\rightarrow\infty}\frac1n\sum^n_{k=1}\frac{1}{a_k}=\frac{1}{3}$

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