How can I prove that $\lim_{k\to\infty}L(k+1,k)=\pi$?

Question

Define $L(k+1,k)=\frac{ \sum_{n=0}^{k+1} {\lfloor{\pi^{n}}\rfloor} }{ \sum_{n=0}^{k} {\lfloor{\pi^{n}}\rfloor} }$

${\lfloor{\pi^{0}}\rfloor}=1$,

${\lfloor{\pi^{1}}\rfloor}=3$,

${\lfloor{\pi^{2}}\rfloor}=9$,

${\lfloor{\pi^{3}}\rfloor}=31$,

${\lfloor{\pi^{4}}\rfloor}=97$,

${\lfloor{\pi^{5}}\rfloor}=306$,

${\lfloor{\pi^{6}}\rfloor}=961$,

${\lfloor{\pi^{7}}\rfloor}=3020$,

${\lfloor{\pi^{8}}\rfloor}=9488$,

${\lfloor{\pi^{9}}\rfloor}=29809$,

${\lfloor{\pi^{10}}\rfloor}=93648$

Then:

$\sum_{n=0}^{0} {\lfloor{\pi^{n}}\rfloor}=1$,

$\sum_{n=0}^{1} {\lfloor{\pi^{n}}\rfloor}=4$,

$\sum_{n=0}^{2} {\lfloor{\pi^{n}}\rfloor}=13$,

$\sum_{n=0}^{3} {\lfloor{\pi^{n}}\rfloor}=44$,

$\sum_{n=0}^{4} {\lfloor{\pi^{n}}\rfloor}=141$,

$\sum_{n=0}^{5} {\lfloor{\pi^{n}}\rfloor}=447$,

$\sum_{n=0}^{6} {\lfloor{\pi^{n}}\rfloor}=1408$,

$\sum_{n=0}^{7} {\lfloor{\pi^{n}}\rfloor}=4428$,

$\sum_{n=0}^{8} {\lfloor{\pi^{n}}\rfloor}=13916$,

$\sum_{n=0}^{9} {\lfloor{\pi^{n}}\rfloor}=43725$,

$\sum_{n=0}^{10} {\lfloor{\pi^{n}}\rfloor}=137373$

Then:

$L(k+1,k)=\frac{ \sum_{n=0}^{k+1} {\lfloor{\pi^{n}}\rfloor} }{ \sum_{n=0}^{k} {\lfloor{\pi^{n}}\rfloor} }$

$L(1,0)=\frac{ \sum_{n=0}^{1} {\lfloor{\pi^{n}}\rfloor} }{ \sum_{n=0}^{0} {\lfloor{\pi^{n}}\rfloor} }=\frac{4}{1}=4$

$L(2,1)=\frac{ \sum_{n=0}^{2} {\lfloor{\pi^{n}}\rfloor} }{ \sum_{n=0}^{1} {\lfloor{\pi^{n}}\rfloor} }=\frac{13}{4}=3.25$

$L(3,2)=\frac{ \sum_{n=0}^{3} {\lfloor{\pi^{n}}\rfloor} }{ \sum_{n=0}^{2} {\lfloor{\pi^{n}}\rfloor} }=\frac{44}{13}=3.384615...$

$L(4,3)=\frac{ \sum_{n=0}^{4} {\lfloor{\pi^{n}}\rfloor} }{ \sum_{n=0}^{3} {\lfloor{\pi^{n}}\rfloor} }=\frac{141}{44}=3.20454545...$

$L(5,4)=\frac{ \sum_{n=0}^{5} {\lfloor{\pi^{n}}\rfloor} }{ \sum_{n=0}^{4} {\lfloor{\pi^{n}}\rfloor} }=\frac{447}{141}=3.170212...$

$L(6,5)=\frac{ \sum_{n=0}^{6} {\lfloor{\pi^{n}}\rfloor} }{ \sum_{n=0}^{5} {\lfloor{\pi^{n}}\rfloor} }=\frac{1408}{447}=3.149888143...$

$L(7,6)=\frac{ \sum_{n=0}^{7} {\lfloor{\pi^{n}}\rfloor} }{ \sum_{n=0}^{6} {\lfloor{\pi^{n}}\rfloor} }=\frac{4428}{1408}=3.1448863636...$

$L(8,7)=\frac{ \sum_{n=0}^{8} {\lfloor{\pi^{n}}\rfloor} }{ \sum_{n=0}^{7} {\lfloor{\pi^{n}}\rfloor} }=\frac{13916}{4428}=3.14272809...$

$L(9,8)=\frac{ \sum_{n=0}^{9} {\lfloor{\pi^{n}}\rfloor} }{ \sum_{n=0}^{8} {\lfloor{\pi^{n}}\rfloor} }=\frac{43725}{13916}=3.142066685...$

$L(10,9)=\frac{ \sum_{n=0}^{10} {\lfloor{\pi^{n}}\rfloor} }{ \sum_{n=0}^{9} {\lfloor{\pi^{n}}\rfloor} }=\frac{137373}{43725}=3.14174957...$

So Is the following true and how can we prove it:

$\lim_{k\to\infty}L(k+1,k)=\pi$ ?

and is it good way to approxiamte $\pi$?

Answer 1

For all real numbers $a > 1$ is $$ \lim_{k \to \infty}\frac{ \sum_{n=0}^{k+1} {\lfloor{a^{n}}\rfloor} }{ \sum_{n=0}^{k} {\lfloor{a^{n}}\rfloor} } = \lim_{k \to \infty} \frac{\lfloor{a^{k+1}}\rfloor}{\lfloor{a^{k}}\rfloor} = a \, . $$ The first equality is an application of the Stolz-Cesàro theorem, and the second equality follows from $$ \frac{a^{k+1}-1}{a^{k}} \le \frac{\lfloor{a^{k+1}}\rfloor}{\lfloor{a^{k}}\rfloor} \le \frac{a^{k+1}}{a^{k}-1} $$ and the “squeeze theorem”.

Answer 2

$$L_{k+1, k}=1+\frac{\lfloor \pi^{k+1}\rfloor}{\sum_{n=0}^k\lfloor \pi^n\rfloor}$$

Since $\lfloor \pi^n\rfloor \sim \pi^n$, $$L_{k+1,k}\sim 1+\frac{\pi^{k+1}}{1+\pi+\cdots+\pi^k}$$ By geometric series, $$1+\pi+\cdots+\pi^k=\frac{\pi^{k+1}-1}{\pi-1}$$ So $$L_{k+1,k}\sim 1+\frac{\pi^{k+2}-\pi^{k+1}}{\pi^{k+1}-1}\sim1+\pi -1\sim \pi$$ For large $k$.

Answer 3

Note that $$-1 + \pi^n < \lfloor \pi^n \rfloor \leq \pi^n$$ so by summing from $n=0$ to $k$, we have $$-k-1 + \sum_{n=0}^k \pi^n < \sum_{n=0}^k\lfloor \pi^n\rfloor \leq \sum_{n=0}^k \pi^n \\ -k-1 + \frac{\pi^{k+1}-1}{\pi-1} < \sum_{n=0}^k\lfloor \pi^n \rfloor \leq \frac{\pi^{k+1}-1}{\pi-1} \\ 1 - \frac{(k+1)(\pi-1)+1}{\pi^{k+1}} < \frac{\sum_{n=0}^k\lfloor \pi^n\rfloor}{\pi^{k+1}/(\pi-1)} \leq 1 - \frac{1}{\pi^{k+1}}$$ so that $$\sum_{n=0}^k\lfloor\pi^n\rfloor \sim \frac{\pi^{k+1}}{\pi-1}.$$

Therefore $$\lim_{k\to\infty} L(k+1,k) = \lim_{k\to\infty}\frac{\pi^{k+2}/(\pi-1)}{\pi^{k+1}/(\pi-1)} = \pi$$