# What is $\lim_{n\to\infty} \frac{n}{\sum_{i=1}^n \frac1{a_i}}$ if $a_n=(2^n+3^n)^\frac{1}{n}$?

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

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, $$\displaystyle \lim_{n\to\infty}\dfrac{n}\sum_{i=1}^{n}\dfrac{1}{a_i}}=3.$$

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

• What you have done is correct and it seems to be the best way to answer the question. Dec 13, 2019 at 23:28
• @KaboMurphy Thank you Sir. Dec 13, 2019 at 23:29
• This works for any mean and any sequence that approaches a limit. Dec 13, 2019 at 23:43
• @martycohen Does that mean I can employ Cauchy's first theorem on limits here? Dec 14, 2019 at 0:03
• @YadatiKiran:How you used squeeze theorem to show $a_n \rightarrow 0$? Sep 29, 2020 at 16:43

$$a_n = 3\Big(1+(\frac23)^n\Big)^{1/n}=3\exp\Big(\frac1n\ln(1+(2/3)^n)\Big)\xrightarrow{n\rightarrow\infty}3$$. The rest follows from Cesaro sums: $$\lim_{n\rightarrow\infty}\frac1n\sum^n_{k=1}\frac{1}{a_k}=\frac{1}{3}$$