Consider the sequence $\{a_n\}_{n=1}^{\infty}$ and $\{b_n\}_{n=1}^{\infty}$ defined by $a_n=(2^n+3^n)^{\frac{1}{n}}$ and $b_n=\dfrac{n}{\displaystyle\sum_{i=1}^{n}\frac{1}{a_i}}$,then what is the limit of $\{b_n\}_{n=1}^{\infty}$?
Solution: We have $$b_n=\dfrac{n}{\displaystyle\sum_{i=1}^{n}\frac{1}{a_i}}$$ $$\implies b_n=\dfrac{n}{\displaystyle\sum_{i=1}^{n}\frac{1}{(2^i+3^i)^{\frac{1}{i}}}}$$
$$\implies b_n=\dfrac{n}{\displaystyle\sum_{i=1}^{n}\frac{1}{3\big[\big(\frac{2}{3}\big)^i+1\big]^{\frac{1}{i}}}}$$
$$\implies b_n=\dfrac{3n}{\displaystyle\sum_{i=1}^{n}\frac{1}{\big[\big(\frac{2}{3}\big)^i+1\big]^{\frac{1}{i}}}}$$
On taking limit as $n\rightarrow \infty$,we get
$\displaystyle\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\dfrac{3n}{\displaystyle\sum_{i=1}^{n}\frac{1}{\big[\big(\frac{2}{3}\big)^i+1\big]^{\frac{1}{i}}}}$.
Now, I got stuck here, I'm not getting how to handle the denominator...Please give suggestions(NOT ANSWER)
Thank you!!
Hint: Prove that $a_n \to 3$. Now use the fact that if a sequence $(c_n)$ converges so does the sequence of averages $\frac {c_1+c+_2+...+c_n} n$.