Show that $\lim\limits_{n\to\infty}\frac1{n}\sum\limits_{k=1}^{\infty}\left\lfloor\frac{n}{3^k}\right\rfloor=\frac{1}{2}$

Question

Show that $$\lim_{n\to\infty}\frac1n\sum_{k=1}^{\infty}\left\lfloor\dfrac{n}{3^k}\right\rfloor=\frac{1}{2}$$

I can do right hand. $$\sum_{k=1}^{\infty}\left\lfloor\dfrac{n}{3^k}\right\rfloor\le \sum_{k=1}^{\infty}\dfrac{n}{3^k}=\dfrac{n}{2}$$ But how to solve left hand?

Answer 1

$$0\leq \frac{n}{3^k}-\left\lfloor\frac{n}{3^k}\right\rfloor\leq 1$$ and the number of non-zero terms of the sum is bounded by $1+\log_3(n)$, hence:

$$\begin{eqnarray*} \sum_{k=1}^{+\infty}\left\lfloor\frac{n}{3^k}\right\rfloor=\sum_{k=1}^{\left\lceil\log_3(n)\right\rceil}\left\lfloor\frac{n}{3^k}\right\rfloor&\geq& -(1+\log_3(n))+\sum_{k=1}^{\left\lceil\log_3(n)\right\rceil}\frac{n}{3^k}\\&\geq&\frac{n}{2}-(1+\log_3(n))-\sum_{k>\left\lceil \log_3(n)\right\rceil}\frac{n}{3^k}\\&\geq&\frac{n}{2}-2\log(n)\end{eqnarray*}$$ for any $n$ big enough.